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zwhe99

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simplerl-OlympiadBench

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[ "We deduce that $4<x<5$.\n\nOtherwise, if $x \\leq 4, x\\lfloor x\\rfloor \\leq 16$, and if $x \\geq 5, x\\lfloor x\\rfloor \\geq 25$.\n\nTherefore $\\lfloor x\\rfloor=4$\n\nSince $x\\lfloor x\\rfloor=17$\n\n$$\n\\begin{aligned}\n4 x & =17 \\\\\nx & =4.25\n\\end{aligned}\n$$" ]

[ "4.25" ]

[ "If we remove the cubes which have red paint, we are left with a smaller cube with measurements, $(n-1) \\times(n-1) \\times(n-1)$\n\nThus, $(n-1)^{3}=125$\n\n$$\nn=6 \\text {. }\n$$" ]

[ "6" ]

[ "Suppose that Thurka bought $x$ goats and $y$ helicopters.\n\nThen $19 x+17 y=201$.\n\nSince $x$ and $y$ are non-negative integers, then $19 x \\leq 201$ so $x \\leq 10$.\n\nIf $x=10$, then $17 y=201-19 x=11$, which does not have an integer solution because 11 is not divisible by 17 .\n\nIf $x=9$, then $17 y=201-19 x=30$, which does not have an integer solution.\n\nIf $x=8$, then $17 y=201-19 x=49$, which does not have an integer solution.\n\nIf $x=7$, then $17 y=201-19 x=68$, so $y=4$.\n\nTherefore, $19(7)+17(4)=201$, and so Thurka buys 7 goats and 4 helicopters.\n\n(We can check that $x=0,1,2,3,4,5,6$ do not give values of $y$ that work.)" ]

[ "7,4" ]

[ "Manipulating algebraically,\n\n$$\n\\begin{aligned}\n(x+8)^{4} & =(2 x+16)^{2} \\\\\n(x+8)^{4}-2^{2}(x+8)^{2} & =0 \\\\\n(x+8)^{2}\\left((x+8)^{2}-2^{2}\\right) & =0 \\\\\n(x+8)^{2}((x+8)+2)((x+8)-2) & =0 \\\\\n(x+8)^{2}(x+10)(x+6) & =0\n\\end{aligned}\n$$\n\nTherefore, $x=-8$ or $x=-10$ or $x=-6$.", "Manipulating algebraically,\n\n$$\n\\begin{aligned}\n(x+8)^{4} & =(2 x+16)^{2} \\\\\n(x+8)^{4}-2^{2}(x+8)^{2} & =0 \\\\\n(x+8)^{2}\\left((x+8)^{2}-2^{2}\\right) & =0 \\\\\n(x+8)^{2}\\left(x^{2}+16 x+64-4\\right) & =0 \\\\\n(x+8)^{2}\\left(x^{2}+16 x+60\\right) & =0 \\\\\n(x+8)^{2}(x+10)(x+6) & =0\n\\end{aligned}\n$$\n\nTherefore, $x=-8$ or $x=-10$ or $x=-6$.", "Since $(x+8)^{4}=(2 x+16)^{2}$, then $(x+8)^{2}=2 x+16$ or $(x+8)^{2}=-(2 x+16)$.\n\nFrom the first equation, $x^{2}+16 x+64=2 x+16$ or $x^{2}+14 x+48=0$ or $(x+6)(x+8)=0$. From the second equation, $x^{2}+16 x+64=-2 x-16$ or $x^{2}+18 x+80=0$ or $(x+10)(x+8)=0$.\n\nTherefore, $x=-8$ or $x=-10$ or $x=-6$." ]

[ "-6,-8,-10" ]

[ "We use the fact that $g(x)=g\\left(f\\left(f^{-1}(x)\\right)\\right)$.\n\nSince $f(x)=2 x+1$, then to determine $f^{-1}(x)$ we solve $x=2 y+1$ for $y$ to get $2 y=x-1$ or $y=\\frac{1}{2}(x-1)$. Thus, $f^{-1}(x)=\\frac{1}{2}(x-1)$.\n\nSince $g(f(x))=4 x^{2}+1$, then\n\n$$\n\\begin{aligned}\ng(x) & =g\\left(f\\left(f^{-1}(x)\\right)\\right) \\\\\n& =g\\left(f\\left(\\frac{1}{2}(x-1)\\right)\\right) \\\\\n& =4\\left(\\frac{1}{2}(x-1)\\right)^{2}+1 \\\\\n& =4 \\cdot \\frac{1}{4}(x-1)^{2}+1 \\\\\n& =(x-1)^{2}+1 \\\\\n& =x^{2}-2 x+2\n\\end{aligned}\n$$", "We use the expressions for $f(x)$ and $g(f(x))$ to construct $g(x)$.\n\nSince $f(x)$ is linear and $g(f(x))$ is quadratic, then it is likely that $g(x)$ is also quadratic.\n\nSince $f(x)=2 x+1$, then $(f(x))^{2}=4 x^{2}+4 x+1$.\n\nSince $g(f(x))$ has no term involving $x$, then we subtract $2 f(x)$ (to remove the $4 x$ term) to get\n\n$$\n(f(x))^{2}-2 f(x)=\\left(4 x^{2}+4 x+1\\right)-2(2 x+1)=4 x^{2}-1\n$$\n\nTo get $g(f(x))$ from this, we add 2 to get $4 x^{2}+1$.\n\nTherefore, $g(f(x))=(f(x))^{2}-2 f(x)+2$, and so an expression for $g(x)$ is $x^{2}-2 x+2$.", "We use the expressions for $f(x)$ and $g(f(x))$ to construct $g(x)$.\n\nSince $f(x)$ is linear and $g(f(x))$ is quadratic, then it is likely that $g(x)$ is also quadratic.\n\nSuppose that $g(x)=a x^{2}+b x+c$ for some real numbers $a, b, c$.\n\nThen\n\n$$\n\\begin{aligned}\ng(f(x)) & =g(2 x+1) \\\\\n& =a(2 x+1)^{2}+b(2 x+1)+c \\\\\n& =a\\left(4 x^{2}+4 x+1\\right)+b(2 x+1)+c \\\\\n& =4 a x^{2}+(4 a+2 b) x+(a+b+c)\n\\end{aligned}\n$$\n\nSince we are told that $g(f(x))=4 x^{2}+1$, then we can compare coefficients to deduce that $4 a=4$ and $4 a+2 b=0$ and $a+b+c=1$.\n\nFrom the first equation, $a=1$.\n\nFrom the second equation, $b=-2 a=-2$.\n\nFrom the third equation, $c=1-a-b=2$.\n\nTherefore, an expression for $g(x)$ is $x^{2}-2 x+2$." ]

[ "$g(x)=x^2-2x+2$" ]

[ "Since the sum of the first two terms is 40 and the sum of the first three terms is 76, then the third term is $76-40=36$.\n\nSince the sum of the first three terms is 76 and the sum of the first four terms is 130, then the fourth term is $130-76=54$.\n\nSince the third term is 36 and the fourth term is 54 , then the common ratio in the geometric sequence is $\\frac{54}{36}=\\frac{3}{2}$.\n\nTherefore, the fifth term is $54 \\cdot \\frac{3}{2}=81$ and the sixth term is $81 \\cdot \\frac{3}{2}=\\frac{243}{2}$.\n\n\n\nAlso, the second term is $36 \\div \\frac{3}{2}=36 \\cdot \\frac{2}{3}=24$ and the first term is $24 \\div \\frac{3}{2}=24 \\cdot \\frac{2}{3}=16$.\n\nThus, the first six terms of the sequence are $16,24,36,54,81, \\frac{243}{2}$.\n\nSince the first term equals $2^{4}$ and the common ratio is $\\frac{3}{2}$, then the $n$th term in the sequence is $2^{4}\\left(\\frac{3}{2}\\right)^{n-1}=\\frac{3^{n-1}}{2^{n-5}}$.\n\nWhen $n \\geq 6$, this is a fraction whose numerator is odd and whose denominator is even, and so, when $n \\geq 6$, the $n$th term is not an integer. (An odd integer is never divisible by an even integer.)\n\nTherefore, there will be 5 integers in the sequence.", "Suppose that $a$ is the first term and $r$ is the common ratio between consecutive terms (so that $a r$ is the second term, $a r^{2}$ is the third term, and so on).\n\nFrom the given information, $a+a r=40$ and $a+a r+a r^{2}=76$ and $a+a r+a r^{2}+a r^{3}=130$.\n\nSubtracting the first equation from the second, we obtain $a r^{2}=36$.\n\nSubtracting the second equation from the third, we obtain $a r^{3}=54$.\n\nSince $a r^{3}=54$ and $a r^{2}=36$, then $r=\\frac{a r^{3}}{a r^{2}}=\\frac{54}{36}=\\frac{3}{2}$.\n\nSince $a r^{2}=36$ and $r=\\frac{3}{2}$, then $a\\left(\\frac{3}{2}\\right)^{2}=36$ or $\\frac{9}{4} a=36$ or $a=\\frac{4}{9} \\cdot 36=16$.\n\nSince $a=16$ and $r=\\frac{3}{2}$, then the first six terms of the sequence are 16, 24, 36, 54, 81, $\\frac{243}{2}$. Since the first term equals $2^{4}$ and the common ratio is $\\frac{3}{2}$, then the $n$th term in the sequence is $2^{4}\\left(\\frac{3}{2}\\right)^{n-1}=\\frac{3^{n-1}}{2^{n-5}}$.\n\nWhen $n \\geq 6$, this is a fraction whose numerator is odd and whose denominator is even, and so, when $n \\geq 6$, the $n$th term is not an integer. (An odd integer is never divisible by an even integer.)\n\nTherefore, there will be 5 integers in the sequence." ]

[ "5" ]

[ "Using the facts that $9=3^{2}$ and $27=3^{3}$, and the laws for manipulating exponents, we have\n\n$$\n\\begin{aligned}\n3^{x-1} 9^{\\frac{3}{2 x^{2}}} & =27 \\\\\n3^{x-1}\\left(3^{2}\\right)^{\\frac{3}{2 x^{2}}} & =3^{3} \\\\\n3^{x-1} 3^{\\frac{3}{x^{2}}} & =3^{3} \\\\\n3^{x-1+\\frac{3}{x^{2}}} & =3^{3}\n\\end{aligned}\n$$\n\nWhen two powers of 3 are equal, their exponents must be equal so\n\n$$\n\\begin{aligned}\nx-1+\\frac{3}{x^{2}} & =3 \\\\\nx^{3}-x^{2}+3 & \\left.=3 x^{2} \\quad \\text { (multiplying by } x^{2}\\right) \\\\\nx^{3}-4 x^{2}+3 & =0\n\\end{aligned}\n$$\n\nSince $x=1$ satisfies the equation, then $x-1$ is a factor of the left side. Using long division or synthetic division, we can factor this out to get $(x-1)\\left(x^{2}-3 x-3\\right)=0$.\n\nUsing the quadratic formula, the quadratic equation $x^{2}-3 x-3=0$ has roots\n\n$$\nx=\\frac{3 \\pm \\sqrt{(-3)^{2}-4(1)(-3)}}{2}=\\frac{3 \\pm \\sqrt{21}}{2}\n$$\n\nTherefore, the solutions to the original equation are $x=1$ and $x=\\frac{3 \\pm \\sqrt{21}}{2}$." ]

[ "$1$,$\\frac{3 + \\sqrt{21}}{2}$,$\\frac{3 - \\sqrt{21}}{2}$" ]

[ "To determine the points of intersection, we equate $y$ values of the two curves and obtain $\\log _{10}\\left(x^{4}\\right)=\\left(\\log _{10} x\\right)^{3}$.\n\nSince $\\log _{10}\\left(a^{b}\\right)=b \\log _{10} a$, the equation becomes $4 \\log _{10} x=\\left(\\log _{10} x\\right)^{3}$.\n\nWe set $u=\\log _{10} x$ and so the equation becomes $4 u=u^{3}$, or $u^{3}-4 u=0$.\n\nWe can factor the left side as $u^{3}-4 u=u\\left(u^{2}-4\\right)=u(u+2)(u-2)$.\n\nTherefore, $u(u+2)(u-2)=0$, and so $u=0$ or $u=-2$ or $u=2$.\n\nTherefore, $\\log _{10} x=0$ or $\\log _{10} x=-2$ or $\\log _{10} x=2$.\n\nTherefore, $x=1$ or $x=\\frac{1}{100}$ or $x=100$.\n\nFinally, we must calculate the $y$-coordinates of the points of intersection. Since one of the original curves is $y=\\left(\\log _{10} x\\right)^{3}$, we can calculate the corresponding values of $y$ by using the fact that $y=u^{3}$.\n\nThe corresponding values of $y$ are $y=0^{3}=0$ and $y=(-2)^{3}=-8$ and $y=2^{3}=8$.\n\nTherefore, the points of intersection are $(1,0),\\left(\\frac{1}{100},-8\\right)$ and $(100,8)$." ]

[ "$(1,0),(\\frac{1}{100},-8),(100,8)$" ]

[ "If Oi-Lam tosses 3 heads, then George has no coins to toss, so cannot toss exactly 1 head. If Oi-Lam tosses 2, 1 or 0 heads, then George has at least one coin to toss, so can toss exactly 1 head.\n\nTherefore, the following possibilities exist:\n\n* Oi-Lam tosses 2 heads out of 3 coins and George tosses 1 head out of 1 coin\n* Oi-Lam tosses 1 head out of 3 coins and George tosses 1 head out of 2 coins\n* Oi-Lam tosses 0 heads out of 3 coins and George tosses 1 head out of 3 coins\n\nWe calculate the various probabilities.\n\nIf 3 coins are tossed, there are 8 equally likely possibilities: $\\mathrm{HHH}, \\mathrm{HHT}, \\mathrm{HTH}, \\mathrm{THH}, \\mathrm{TTH}$, THT, HTT, TTT. Each of these possibilities has probability $\\left(\\frac{1}{2}\\right)^{3}=\\frac{1}{8}$. Therefore,\n\n\n\n* the probability of tossing 0 heads out of 3 coins is $\\frac{1}{8}$\n* the probability of tossing 1 head out of 3 coins is $\\frac{3}{8}$\n* the probability of tossing 2 heads out of 3 coins is $\\frac{3}{8}$\n* the probability of tossing 3 heads out of 3 coins is $\\frac{1}{8}$\n\nIf 2 coins are tossed, there are 4 equally likely possibilities: HH, HT, TH, TT. Each of these possibilities has probability $\\left(\\frac{1}{2}\\right)^{2}=\\frac{1}{4}$. Therefore, the probability of tossing 1 head out of 2 coins is $\\frac{2}{4}=\\frac{1}{2}$.\n\nIf 1 coin is tossed, the probability of tossing 1 head is $\\frac{1}{2}$.\n\nTo summarize, the possibilities are\n\n* Oi-Lam tosses 2 heads out of 3 coins (with probability $\\frac{3}{8}$ ) and George tosses 1 head out of 1 coin (with probability $\\frac{1}{2}$ )\n* Oi-Lam tosses 1 head out of 3 coins (with probability $\\frac{3}{8}$ ) and George tosses 1 head out of 2 coins (with probability $\\frac{1}{2}$ )\n* Oi-Lam tosses 0 heads out of 3 coins (with probability $\\frac{1}{8}$ ) and George tosses 1 head out of 3 coins (with probability $\\frac{3}{8}$ )\n\nTherefore, the overall probability is $\\frac{3}{8} \\cdot \\frac{1}{2}+\\frac{3}{8} \\cdot \\frac{1}{2}+\\frac{1}{8} \\cdot \\frac{3}{8}=\\frac{27}{64}$." ]

[ "$\\frac{27}{64}$" ]

[ "We first prove Lemma(i): If $\\theta$ is an angle whose measure is not an integer multiple of $90^{\\circ}$, then\n$$\n\\cot \\theta-\\cot 2 \\theta=\\frac{1}{\\sin 2 \\theta}\n$$\n\nProof. \n$$\n\\begin{aligned}\n\\mathrm{LS} & =\\cot \\theta-\\cot 2 \\theta \\\\\n& =\\frac{\\cos \\theta}{\\sin \\theta}-\\frac{\\cos 2 \\theta}{\\sin 2 \\theta} \\\\\n& =\\frac{\\cos \\theta}{\\sin \\theta}-\\frac{\\cos 2 \\theta}{2 \\sin \\theta \\cos \\theta} \\\\\n& =\\frac{2 \\cos ^{2} \\theta-\\cos 2 \\theta}{2 \\sin \\theta \\cos \\theta} \\\\\n& =\\frac{2 \\cos ^{2} \\theta-\\left(2 \\cos ^{2} \\theta-1\\right)}{\\sin 2 \\theta} \\\\\n& =\\frac{1}{\\sin 2 \\theta} \\\\\n& =\\mathrm{RS}\n\\end{aligned}\n$$\n\nas required.\n\nWe use (i) to note that $\\frac{1}{\\sin 8^{\\circ}}=\\cot 4^{\\circ}-\\cot 8^{\\circ}$ and $\\frac{1}{\\sin 16^{\\circ}}=\\cot 8^{\\circ}-\\cot 16^{\\circ}$ and so on. Thus,\n\n$$\n\\begin{aligned}\nS= & \\frac{1}{\\sin 8^{\\circ}}+\\frac{1}{\\sin 16^{\\circ}}+\\frac{1}{\\sin 32^{\\circ}}+\\cdots+\\frac{1}{\\sin 4096^{\\circ}}+\\frac{1}{\\sin 8192^{\\circ}} \\\\\n= & \\left(\\cot 4^{\\circ}-\\cot 8^{\\circ}\\right)+\\left(\\cot 8^{\\circ}-\\cot 16^{\\circ}\\right)+\\left(\\cot 16^{\\circ}-\\cot 32^{\\circ}\\right)+ \\\\\n& \\cdots+\\left(\\cot 2048^{\\circ}-\\cot 4096^{\\circ}\\right)+\\left(\\cot 4096^{\\circ}-\\cot 8192^{\\circ}\\right) \\\\\n= & \\cot 4^{\\circ}-\\cot 8192^{\\circ}\n\\end{aligned}\n$$\n\nsince the sum \"telescopes\".\n\nSince the cotangent function has a period of $180^{\\circ}$, and $8100^{\\circ}$ is a multiple of $180^{\\circ}$, then $\\cot 8192^{\\circ}=\\cot 92^{\\circ}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nS & =\\cot 4^{\\circ}-\\cot 92^{\\circ} \\\\\n& =\\frac{\\cos 4^{\\circ}}{\\sin 4^{\\circ}}-\\frac{\\cos 92^{\\circ}}{\\sin 92^{\\circ}} \\\\\n& =\\frac{\\cos 4^{\\circ}}{\\sin 4^{\\circ}}-\\frac{-\\sin 2^{\\circ}}{\\cos 2^{\\circ}} \\\\\n& =\\frac{\\cos 4^{\\circ}}{2 \\sin 2^{\\circ} \\cos 2^{\\circ}}+\\frac{\\sin 2^{\\circ}}{\\cos 2^{\\circ}} \\\\\n& =\\frac{\\cos 4^{\\circ}+2 \\sin ^{2} 2^{\\circ}}{2 \\sin 2^{\\circ} \\cos 2^{\\circ}} \\\\\n& =\\frac{\\left(1-2 \\sin ^{2} 2^{\\circ}\\right)+2 \\sin ^{2} 2^{\\circ}}{\\sin 4^{\\circ}} \\\\\n& =\\frac{1}{\\sin 4^{\\circ}}\n\\end{aligned}\n$$\n\nTherefore, $\\alpha=4^{\\circ}$." ]

[ "$4^{\\circ}$" ]

[ "Denote the side lengths of a triangle by $a, b$ and $c$, with $0<a \\leq b \\leq c$.\n\nIn order for these lengths to form a triangle, we need $c<a+b$ and $b<a+c$ and $a<b+c$. Since $0<a \\leq b \\leq c$, then $b<a+c$ and $a<b+c$ follow automatically, so only $c<a+b$ ever needs to be checked.\n\nInstead of directly considering triangles and sets of triangle, we can consider triples $(a, b, c)$ and sets of triples $(a, b, c)$ with the appropriate conditions.\n\nFor each positive integer $k \\geq 3$, we use the notation $S_{k}$ to denote the set of triples of positive integers $(a, b, c)$ with $0<a \\leq b \\leq c$ and $c<a+b$ and $a+b+c=k$.\n\nIn this case, $c<a+b$ and $a+b+c=k$, so $c+c<a+b+c=k$, so $2 c<k$ or $c<\\frac{1}{2} k$.\n\nAlso, if $0<a \\leq b \\leq c$ and $a+b+c=k$, then $k=a+b+c \\leq c+c+c$, so $3 c \\geq k$ or $c \\geq \\frac{1}{3} k$.\n\n\nConsider $T(10)$, which is the number of elements in $S_{10}$.\n\nWe want to find all possible triples $(a, b, c)$ of integers with $0<a \\leq b \\leq c$ and $c<a+b$ and $a+b+c=10$.\n\nWe need $c<\\frac{10}{2}=5$ and $c \\geq \\frac{10}{3}$. Thus, $c=4$.\n\nTherefore, we need $0<a \\leq b \\leq 4$ and $a+b=6$.\n\nThere are two possibilities: $(a, b, c)=(2,4,4)$ or $(a, b, c)=(3,3,4)$.\n\nTherefore, $T(10)=2$.\n\nConsider $T(11)$. We want to find all possible triples $(a, b, c)$ of integers with $0<a \\leq b \\leq c$ and $c<a+b$ and $a+b+c=11$.\n\nWe need $c<\\frac{11}{2}$ and $c \\geq \\frac{11}{3}$. Thus, $c=4$ or $c=5$.\n\nIf $c=4$, we need $0<a \\leq b \\leq 4$ and $a+b=7$.\n\nThere is only one possibility: $(a, b, c)=(3,4,4)$.\n\nIf $c=5$, we need $0<a \\leq b \\leq 5$ and $a+b=6$.\n\nThere are three possibilities: $(a, b, c)=(1,5,5)$ or $(a, b, c)=(2,4,5)$ or $(a, b, c)=(3,3,5)$.\n\nTherefore, $T(11)=4$.\n\nConsider $T(12)$. We want to find all possible triples $(a, b, c)$ of integers with $0<a \\leq b \\leq c$ and $c<a+b$ and $a+b+c=12$.\n\nWe need $c<\\frac{12}{2}$ and $c \\geq \\frac{12}{3}$. Thus, $c=4$ or $c=5$.\n\nIf $c=4$, we need $0<a \\leq b \\leq 4$ and $a+b=8$.\n\nThere is only one possibility: $(a, b, c)=(4,4,4)$.\n\n\n\nIf $c=5$, we need $0<a \\leq b \\leq 5$ and $a+b=7$.\n\nThere are two possibilities: $(a, b, c)=(2,5,5)$ or $(a, b, c)=(3,4,5)$.\n\nTherefore, $T(12)=3$." ]

[ "2,4,3" ]

[ "Denote the side lengths of a triangle by $a, b$ and $c$, with $0<a \\leq b \\leq c$.\n\nIn order for these lengths to form a triangle, we need $c<a+b$ and $b<a+c$ and $a<b+c$. Since $0<a \\leq b \\leq c$, then $b<a+c$ and $a<b+c$ follow automatically, so only $c<a+b$ ever needs to be checked.\n\nInstead of directly considering triangles and sets of triangle, we can consider triples $(a, b, c)$ and sets of triples $(a, b, c)$ with the appropriate conditions.\n\nFor each positive integer $k \\geq 3$, we use the notation $S_{k}$ to denote the set of triples of positive integers $(a, b, c)$ with $0<a \\leq b \\leq c$ and $c<a+b$ and $a+b+c=k$.\n\nIn this case, $c<a+b$ and $a+b+c=k$, so $c+c<a+b+c=k$, so $2 c<k$ or $c<\\frac{1}{2} k$.\n\nAlso, if $0<a \\leq b \\leq c$ and $a+b+c=k$, then $k=a+b+c \\leq c+c+c$, so $3 c \\geq k$ or $c \\geq \\frac{1}{3} k$.\n\n\nWe will use two important facts:\n\n(F1) $T(2 m)=T(2 m-3)$ for every positive integer $m \\geq 3$, and\n\n(F2) $T(k) \\leq T(k+2)$ for every positive integer $k \\geq 3$\n\nFirst, we prove (F1):\n\n\nWe show that $T(2 m)=T(2 m-3)$ by creating a one-to-one correspondence between the triples in $S_{2 m}$ and the triples $S_{2 m-3}$.\n\nNote that $S_{2 m}$ is the set of triples $(a, b, c)$ of positive integers with $0<a \\leq b \\leq c$, with $c<a+b$, and with $a+b+c=2 m$.\n\nAlso, $S_{2 m-3}$ is the set of triples $(A, B, C)$ of positive integers with $0<A \\leq B \\leq C$, with $C<A+B$, and with $A+B+C=2 m-3$.\n\nConsider a triple $(a, b, c)$ in $S_{2 m}$ and a corresponding triple $(a-1, b-1, c-1)$.\n\nWe show that $(a-1, b-1, c-1)$ is in $S_{2 m-3}$ :\n\n* Since $(a, b, c)$ is in $S_{2 m}$, then $c<\\frac{1}{2}(2 m)=m$. This means that $b \\leq c \\leq m-1$, so $a=2 m-b-c \\geq 2$. Therefore, $a-1, b-1$ and $c-1$ are positive integers since $a, b$ and $c$ are positive integers with $2 \\leq a \\leq b \\leq c$.\n* Since $2 \\leq a \\leq b \\leq c$, then $1 \\leq a-1 \\leq b-1 \\leq c-1$, so $0<a-1 \\leq b-1 \\leq c-1$.\n* Since $a+b+c=2 m$, then $c=2 m-(a+b)$ so $a+b$ and $c$ have the same parity.\n\nSince $c<a+b$, then $c \\leq a+b-2$. (In other words, it cannot be the case that $c=a+b-1$.) Therefore, $c-1 \\leq(a-1)+(b-1)-1$; that is, $c-1<(a-1)+(b-1)$.\n\n$*$ Since $a+b+c=2 m$, then $(a-1)+(b-1)+(c-1)=2 m-3$.\n\nTherefore, $(a-1, b-1, c-1)$ is in $S_{2 m-3}$, since it satisfies all of the conditions of $S_{2 m-3}$. Note as well that two different triples in $S_{2 m}$ correspond to two different triples in $S_{2 m-3}$. Thus, every triple in $S_{2 m}$ corresponds to a different triple in $S_{2 m-3}$.\n\nThus, $T(2 m) \\leq T(2 m-3)$.\n\nConsider a triple $(A, B, C)$ in $S_{2 m-3}$ and a corresponding triple $(A+1, B+1, C+1)$. We show that $(A+1, B+1, C+1)$ is in $S_{2 m}$ :\n\n* Since $(A, B, C)$ is in $S_{2 m-3}$, then $A, B$ and $C$ are positive integers, so $A+1, B+1$ and $C+1$ are positive integers.\n* Since $0<A \\leq B \\leq C$, then $1<A+1 \\leq B+1 \\leq C+1$, so $0<A+1 \\leq B+1 \\leq C+1$.\n* Since $C<A+B$, then $C+1<(A+1)+(B+1)-1$ so $C+1<(A+1)+(B+1)$.\n* Since $A+B+C=2 m-3$, then $(A+1)+(B+1)+(C+1)=2 m$.\n\nTherefore, $(A+1, B+1, C+1)$ is in $S_{2 m}$.\n\nNote again that two different triples in $S_{2 m-3}$ correspond to two different triples in $S_{2 m}$. Thus, every triple in $S_{2 m-3}$ corresponds to a different triple in $S_{2 m}$.\n\nTherefore, $T(2 m-3) \\leq T(2 m)$.\n\nSince $T(2 m) \\leq T(2 m-3)$ and $T(2 m-3) \\leq T(2 m)$, then $T(2 m)=T(2 m-3)$.\n\n\nNext, we prove (F2):\n\nConsider a triple $(a, b, c)$ in $S_{k}$ and a corresponding triple $(a, b+1, c+1)$. We show that the triple $(a, b+1, c+1)$ is in $S_{k+2}$ :\n\n* Since $a, b$ and $c$ are positive integers, then $a, b+1$ and $c+1$ are positive integers.\n\n\n\n* Since $0<a \\leq b \\leq c$, then $0<a \\leq b+1 \\leq c+1$.\n* Since $c<a+b$, then $c+1<a+(b+1)$.\n\n$*$ Since $a+b+c=k$, then $a+(b+1)+(c+1)=k+2$.\n\nTherefore, $(a, b+1, c+1)$ is in $S_{k+2}$. Note that, using this correspondence, different triples in $S_{k}$ correspond different triples in $S_{k+2}$. Thus, every triple in $S_{k}$ corresponds to a different triple in $S_{k+2}$. This proves that $T(k) \\leq T(k+2)$.\n\nSuppose that $n=N$ is the smallest positive integer for which $T(n)>2010$.\n\nThen $N$ must be odd:\n\nIf $N$ was even, then by (F1), $T(N-3)=T(N)>2010$ and so $n=N-3$ would be an integer smaller than $N$ with $T(n)>2010$. This contradicts the fact that $n=N$ is the smallest such integer.\n\nTherefore, we want to find the smallest odd positive integer $N$ for which $T(N)>2010$. Next, we note that if we can find an odd positive integer $n$ such that $T(n)>2010 \\geq$ $T(n-2)$, then we will have found the desired value of $n$ :\n\nThis is because $n$ and $n-2$ are both odd, and by property (F2), any smaller odd positive integer $k$ will give $T(k) \\leq T(n-2) \\leq 2010$ and any larger odd positive integer $m$ will give $T(m) \\geq T(n)>2010$.\n\nWe show that $N=309$ is the desired value of $N$ by showing that $T(309)>2010$ and $T(307) \\leq 2010$.\n\nCalculation of $T(309)$\n\nWe know that $\\frac{309}{3} \\leq c<\\frac{309}{2}$, so $103 \\leq c \\leq 154$.\n\nFor each admissible value of $c$, we need to count the number of pairs of positive integers $(a, b)$ with $a \\leq b \\leq c$ and $a+b=309-c$.\n\nFor example, if $c=154$, then we need $a \\leq b \\leq 154$ and $a+b=155$.\n\nThis gives pairs $(1,154),(2,153), \\ldots,(76,79),(77,78)$, of which there are 77 .\n\nAlso, if $c=153$, then we need $a \\leq b \\leq 153$ and $a+b=156$.\n\nThis gives pairs $(3,153), \\ldots,(77,79),(78,78)$, of which there are 76 .\n\nIn general, if $c$ is even, then the minimum possible value of $a$ occurs when $b$ is as large as possible - that is, when $b=c$, so $a \\geq 309-2 c$.\n\nAlso, the largest possible value of $a$ occurs when $a$ and $b$ are as close to equal as possible. Since $c$ is even, then $309-c$ is odd, so $a$ and $b$ cannot be equal, but they can differ by 1 . In this case, $a=154-\\frac{1}{2} c$ and $b=155-\\frac{1}{2} c$.\n\nTherefore, if $c$ is even, there are $\\left(154-\\frac{1}{2} c\\right)-(309-2 c)+1=\\frac{3}{2} c-154$ possible pairs $(a, b)$ and so $\\frac{3}{2} c-154$ possible triples.\n\nIn general, if $c$ is odd, then the minimum possible value of $a$ occurs when $b$ is as large as possible - that is, when $b=c$, so $a \\geq 309-2 c$.\n\nAlso, the largest possible value of $a$ occurs when $a$ and $b$ are as close to equal as possible.\n\nSince $c$ is odd, then $309-c$ is even, so $a$ and $b$ can be equal. In this case, $a=\\frac{1}{2}(309-c)$. Therefore, if $c$ is odd, there are $\\frac{1}{2}(309-c)-(309-2 c)+1=\\frac{3}{2} c-\\frac{307}{2}$ possible pairs $(a, b)$ and so $\\frac{3}{2} c-\\frac{307}{2}$ possible triples.\n\nThe possible even values of $c$ are 104,106,...,152,154 (there are 26 such values) and the possible odd values of $c$ are 103,105,...,151,153 (there are 26 such values).\n\n\n\nTherefore,\n\n$$\n\\begin{aligned}\nT(309)= & \\left(\\frac{3}{2}(104)-154\\right)+\\left(\\frac{3}{2}(106)-154\\right)+\\cdots+\\left(\\frac{3}{2}(154)-154\\right)+ \\\\\n& \\quad\\left(\\frac{3}{2}(103)-\\frac{307}{2}\\right)+\\left(\\frac{3}{2}(105)-\\frac{307}{2}\\right)+\\cdots+\\left(\\frac{3}{2}(153)-\\frac{307}{2}\\right) \\\\\n= & \\frac{3}{2}(104+106+\\cdots+154)-26 \\cdot 154+\\frac{3}{2}(103+105+\\cdots+153)-26 \\cdot \\frac{307}{2} \\\\\n= & \\frac{3}{2}(103+104+105+106+\\cdots+153+154)-26 \\cdot 154-26 \\cdot \\frac{307}{2} \\\\\n= & \\frac{3}{2} \\cdot \\frac{1}{2}(103+154)(52)-26 \\cdot 154-26 \\cdot \\frac{307}{2} \\\\\n= & \\frac{3}{2}(26)(257)-26 \\cdot 154-26 \\cdot \\frac{307}{2} \\\\\n= & 2028\n\\end{aligned}\n$$\n\nTherefore, $T(309)>2010$, as required.\n\nCalculation of $T(307)$\n\nWe know that $\\frac{307}{3} \\leq c<\\frac{307}{2}$, so $103 \\leq c \\leq 153$.\n\nFor each admissible value of $c$, we need to count the number of pairs of positive integers $(a, b)$ with $a \\leq b \\leq c$ and $a+b=307-c$.\n\nThis can be done in a similar way to the calculation of $T(309)$ above.\n\nIf $n$ is even, there are $\\frac{3}{2} c-153$ possible triples.\n\nIf $n$ is odd, there are $\\frac{3}{2} c-\\frac{305}{2}$ possible triples.\n\nThe possible even values of $c$ are $104,106, \\ldots, 150,152$ (there are 25 such values) and the possible odd values of $c$ are $103,105, \\ldots, 151,153$ (there are 26 such values).\n\nTherefore,\n\n$$\n\\begin{aligned}\nT(307)= & \\left(\\frac{3}{2}(104)-153\\right)+\\left(\\frac{3}{2}(106)-153\\right)+\\cdots+\\left(\\frac{3}{2}(152)-153\\right)+ \\\\\n& \\quad\\left(\\frac{3}{2}(103)-\\frac{305}{2}\\right)+\\left(\\frac{3}{2}(105)-\\frac{305}{2}\\right)+\\cdots+\\left(\\frac{3}{2}(153)-\\frac{305}{2}\\right) \\\\\n= & \\frac{3}{2}(104+106+\\cdots+152)-25 \\cdot 153+\\frac{3}{2}(103+105+\\cdots+153)-26 \\cdot \\frac{305}{2} \\\\\n= & \\frac{3}{2}(103+104+105+106+\\cdots+152+153)-25 \\cdot 153-26 \\cdot \\frac{305}{2} \\\\\n= & \\frac{3}{2} \\cdot \\frac{1}{2}(103+153)(51)-25 \\cdot 153-26 \\cdot \\frac{305}{2} \\\\\n= & \\frac{3}{2}(51)(128)-25 \\cdot 153-26 \\cdot \\frac{305}{2} \\\\\n= & 2002\n\\end{aligned}\n$$\n\nTherefore, $T(307)<2010$, as required.\n\nTherefore, the smallest positive integer $n$ such that $T(n)>2010$ is $n=309$.\n\nAs a final note, we discuss briefly how one could guess that the answer was near $N=309$.\n\nConsider the values of $T(n)$ for small odd positive integers $n$.\n\nIn (a), by considering the possible values of $c$ from smallest (roughly $\\frac{1}{3} n$ ) to largest (roughly $\\frac{1}{2} n$ ), we saw that $T(11)=1+3=4$.\n\nIf we continue to calculate $T(n)$ for a few more small odd values of $n$ we will see that:\n\n$$\n\\begin{aligned}\n& T(13)=2+3=5 \\\\\n& T(15)=1+2+4=7 \\\\\n& T(17)=1+3+4=8 \\\\\n& T(19)=2+3+5=10 \\\\\n& T(21)=1+2+4+5=12 \\\\\n& T(23)=1+3+4+6=14\n\\end{aligned}\n$$\n\n\n\nThe pattern that seems to emerge is that for $n$ odd, $T(n)$ is roughly equal to the sum of the integers from 1 to $\\frac{1}{4} n$, with one out of every three integers removed. Thus, $T(n)$ is roughly equal to $\\frac{2}{3}$ of the sum of the integers from 1 to $\\frac{1}{4} n$. Therefore, $T(n) \\approx \\frac{2}{3} \\cdot \\frac{1}{2}\\left(\\frac{1}{4} n\\right)\\left(\\frac{1}{4} n+1\\right) \\approx \\frac{2}{3} \\cdot \\frac{1}{2}\\left(\\frac{1}{4} n\\right)^{2} \\approx \\frac{1}{48} n^{2}$.\n\nIt makes sense to look for an odd positive integer $n$ with $T(n) \\approx 2010$.\n\nThus, we are looking for a value of $n$ that roughly satisfies $\\frac{1}{48} n^{2} \\approx 2010$ or $n^{2} \\approx 96480$ or $n \\approx 310$.\n\nSince $n$ is odd, then it makes sense to consider $n=309$, as in the solution above." ]

[ "309" ]

[ "Since $2 \\sin ^{2} x+\\cos ^{2} x=\\frac{25}{16}$ and $\\sin ^{2} x+\\cos ^{2} x=1\\left(\\right.$ so $\\left.\\cos ^{2} x=1-\\sin ^{2} x\\right)$, then we get\n\n$$\n\\begin{aligned}\n2 \\sin ^{2} x+\\left(1-\\sin ^{2} x\\right) & =\\frac{25}{16} \\\\\n\\sin ^{2} x & =\\frac{25}{16}-1 \\\\\n\\sin ^{2} x & =\\frac{9}{16} \\\\\n\\sin x & = \\pm \\frac{3}{4}\n\\end{aligned}\n$$\n\nso $\\sin x=\\frac{3}{4}$ since $\\sin x>0$ because $0^{\\circ}<x<90^{\\circ}$." ]

[ "$\\frac{3}{4}$" ]

[ "From the given information, the first term in the sequence is 2007 and each term starting with the second can be determined from the previous term.\n\nThe second term is $2^{3}+0^{3}+0^{3}+7^{3}=8+0+0+343=351$.\n\nThe third term is $3^{3}+5^{3}+1^{3}=27+125+1=153$.\n\nThe fourth term is $1^{3}+5^{3}+3^{3}=27+125+1=153$.\n\nSince two consecutive terms are equal, then every term thereafter will be equal, because each term depends only on the previous term and a term of 153 always makes the next term 153.\n\nThus, the 2007th term will be 153 ." ]

[ "153" ]

[ "The $n$th term of sequence $\\mathrm{A}$ is $n^{2}-10 n+70$.\n\nSince sequence B is arithmetic with first term 5 and common difference 10 , then the $n$th term of sequence $\\mathrm{B}$ is equal to $5+10(n-1)=10 n-5$. (Note that this formula agrees with the first few terms.)\n\nFor the $n$th term of sequence $\\mathrm{A}$ to be equal to the $n$th term of sequence $\\mathrm{B}$, we must have\n\n$$\n\\begin{aligned}\nn^{2}-10 n+70 & =10 n-5 \\\\\nn^{2}-20 n+75 & =0 \\\\\n(n-5)(n-15) & =0\n\\end{aligned}\n$$\n\nTherefore, $n=5$ or $n=15$. That is, 5 th and 15 th terms of sequence $\\mathrm{A}$ and sequence $\\mathrm{B}$ are equal to each other." ]

[ "5,15" ]

[ "Rearranging and then squaring both sides,\n\n$$\n\\begin{aligned}\n2+\\sqrt{x-2} & =x-2 \\\\\n\\sqrt{x-2} & =x-4 \\\\\nx-2 & =(x-4)^{2} \\\\\nx-2 & =x^{2}-8 x+16 \\\\\n0 & =x^{2}-9 x+18 \\\\\n0 & =(x-3)(x-6)\n\\end{aligned}\n$$\n\nso $x=3$ or $x=6$.\n\nWe should check both solutions, because we may have introduced extraneous solutions by squaring.\n\nIf $x=3$, the left side equals $2+\\sqrt{1}=3$ and the right side equals 1 , so $x=3$ must be rejected.\n\nIf $x=6$, the left side equals $2+\\sqrt{4}=4$ and the right side equals 4 , so $x=6$ is the only solution.", "Suppose $u=\\sqrt{x-2}$.\n\nThe equation becomes $2+u=u^{2}$ or $u^{2}-u-2=0$ or $(u-2)(u+1)=0$.\n\nTherefore, $u=2$ or $u=-1$.\n\nBut we cannot have $\\sqrt{x-2}=-1$ (as square roots are always non-negative).\n\nTherefore, $\\sqrt{x-2}=2$ or $x-2=4$ or $x=6$." ]

[ "6" ]

[ "Using rules for manipulating logarithms,\n\n$$\n\\begin{aligned}\n(\\sqrt{x})^{\\log _{10} x} & =100 \\\\\n\\log _{10}\\left((\\sqrt{x})^{\\log _{10} x}\\right) & =\\log _{10} 100 \\\\\n\\left(\\log _{10} x\\right)\\left(\\log _{10} \\sqrt{x}\\right) & =2 \\\\\n\\left(\\log _{10} x\\right)\\left(\\log _{10} x^{\\frac{1}{2}}\\right) & =2 \\\\\n\\left(\\log _{10} x\\right)\\left(\\frac{1}{2} \\log _{10} x\\right) & =2 \\\\\n\\left(\\log _{10} x\\right)^{2} & =4 \\\\\n\\log _{10} x & = \\pm 2 \\\\\nx & =10^{ \\pm 2}\n\\end{aligned}\n$$\n\nTherefore, $x=100$ or $x=\\frac{1}{100}$.\n\n(We can check by substitution that each is indeed a solution.)" ]

[ "$100,\\frac{1}{100}$" ]

[ "The quadratic function $f(x)=x^{2}+(2 n-1) x+\\left(n^{2}-22\\right)$ has no real roots exactly when its discriminant, $\\Delta$, is negative.\n\nThe discriminant of this function is\n\n$$\n\\begin{aligned}\n\\Delta & =(2 n-1)^{2}-4(1)\\left(n^{2}-22\\right) \\\\\n& =\\left(4 n^{2}-4 n+1\\right)-\\left(4 n^{2}-88\\right) \\\\\n& =-4 n+89\n\\end{aligned}\n$$\n\nWe have $\\Delta<0$ exactly when $-4 n+89<0$ or $4 n>89$.\n\nThis final inequality is equivalent to $n>\\frac{89}{4}=22 \\frac{1}{4}$.\n\nTherefore, the smallest positive integer that satisfies this inequality, and hence for which $f(x)$ has no real roots, is $n=23$." ]

[ "23" ]

[ "Each possible order in which Akshan removes the marbles corresponds to a sequence of 9 colours, 3 of which are red and 6 of which are blue.\n\nWe write these as sequences of 3 R's and 6 B's.\n\nSince are told that the first marble is red and the third is blue, we would like to consider all sequences of the form\n\n$$\nR \\_B\\_\\_\\_\\_\\_\\_\n$$\n\nThe 7 blanks must be filled with the remaining 2 R's and 5 B's.\n\nThere are $\\left(\\begin{array}{l}7 \\\\ 2\\end{array}\\right)=\\frac{7 \\cdot 6}{2}=21$ ways of doing this, because 2 of the 7 blanks must be chosen in which to place the R's. (We could count these 21 ways directly by working systematically through the possible pairs of blanks.)\n\nOf these 21 ways, some have the last two marbles being blue.\n\nThese correspond to the sequences of the form\n\n$$\nR \\_B \\_\\_\\_\\_ B B\n$$\n\nIn these sequences, the 5 blanks must be filled with the remaining $2 \\mathrm{R}$ 's and 3 B's.\n\nThere are $\\left(\\begin{array}{l}5 \\\\ 2\\end{array}\\right)=\\frac{5 \\cdot 4}{2}=10$ ways of doing this, because 2 of the 5 blanks must be chosen in which to place the R's.\n\nTherefore, 10 of the 21 possible sequences end in two B's, and so the probability that the last two marbles removed are blue is $\\frac{10}{21}$." ]

[ "$\\frac{10}{21}$" ]

[ "Factoring the first equation, we obtain\n\n$$\na c+a d+b c+b d=a(c+d)+b(c+d)=(a+b)(c+d)\n$$\n\nWe now have the equations\n\n$$\n\\begin{aligned}\n(a+b)(c+d) & =2023 \\\\\n(a+b)+(c+d) & =296\n\\end{aligned}\n$$\n\nIf we let $s=a+b$ and $t=c+d$, we obtain the equations\n\n$$\n\\begin{aligned}\ns t & =2023 \\\\\ns+t & =296\n\\end{aligned}\n$$\n\nNoting that $s$ and $t$ are integers since $a, b, c$, and $d$ are integers, we look for divisor pairs of 2023 whose sum is 296 .\n\nTo find the divisors of 2023 , we first find its prime factorization:\n\n$$\n2023=7 \\cdot 289=7 \\cdot 17^{2}\n$$\n\nTherefore, the divisors of 2023 are 1, 7, 17, 119, 289, 2023.\n\nThis means that the divisor pairs of 2023 are\n\n$$\n2023=1 \\cdot 2023=7 \\cdot 289=17 \\cdot 119\n$$\n\nThe one divisor pair with a sum of 296 is 7 and 289. (Alternatively, we could have found these by substituting $t=206-s$ into $s t=2023$ and using the quadratic formula.)\n\n\n\nSince $a<b<c<d$, then $a+b<c+d$ and so $s=a+b=7$ and $t=c+d=289$.\n\nSince $a$ and $b$ are positive integers with $a<b$ and $a+b=7$, then the possible pairs $(a, b)$ are\n\n$$\n(a, b)=(1,6),(2,5),(3,4)\n$$\n\nWe know that $c$ and $d$ are positive integers with $c<d$ and $c+d=289$, but also with $b<c<d$.\n\nWhen $(a, b)=(1,6)$, this means that the possibilities are\n\n$$\n(c, d)=(7,282),(8,281),(9,280), \\ldots,(143,146),(144,145)\n$$\n\nThere are $144-7+1=138$ such pairs.\n\nWhen $(a, b)=(2,5)$, the possibilities are\n\n$$\n(c, d)=(6,283),(7,282),(8,281),(9,280), \\ldots,(143,146),(144,145)\n$$\n\nThere are $138+1=139$ such pairs.\n\nWhen $(a, b)=(3,4)$, the possibilities are\n\n$$\n(c, d)=(5,284),(6,283),(7,282),(8,281),(9,280), \\ldots,(143,146),(144,145)\n$$\n\nThere are $139+1=140$ such pairs.\n\nIn total, there are $138+139+140=417$ possible quadruples $(a, b, c, d)$." ]

[ "417" ]

[ "Since $\\triangle A B C$ is right-angled at $B$, then\n\n$$\n\\begin{aligned}\nB C^{2} & =A C^{2}-A B^{2} \\\\\n& =((n+1)(n+4))^{2}-(n(n+1))^{2} \\\\\n& =(n+1)^{2}(n+4)^{2}-n^{2}(n+1)^{2} \\\\\n& =(n+1)^{2}\\left((n+4)^{2}-n^{2}\\right) \\\\\n& =(n+1)^{2}\\left(n^{2}+8 n+16-n^{2}\\right) \\\\\n& =(n+1)^{2}(8 n+16) \\\\\n& =4(n+1)^{2}(2 n+4)\n\\end{aligned}\n$$\n\nThe length of $B C$ is an integer exactly when $4(n+1)^{2}(2 n+4)$ is a perfect square.\n\nSince $4(n+1)^{2}$ is a perfect square, then $B C$ is an integer exactly when $2 n+4$ is a perfect square.\n\nWe note that $2 n+4 \\geq 6$ (since $n \\geq 1)$ and that $2 n+4$ is even.\n\nSince $n<100000$, then $6 \\leq 2 n+4<200004$, and so we need to count the number of even perfect squares between 6 and 200004 .\n\nThe smallest even perfect square in this range is $4^{2}=16$.\n\nSince $\\sqrt{200004} \\approx 447.2$, the largest even perfect square in this range is $446^{2}$.\n\nTherefore, the number of even perfect squares in this range is $\\frac{446}{2}-1=222$.\n\nThus, there are 222 positive integers $n$ for which the length of $B C$ is an integer." ]

[ "222" ]

[ "Let $f(x)=\\sqrt{\\log _{2} x \\cdot \\log _{2}(4 x)+1}+\\sqrt{\\log _{2} x \\cdot \\log _{2}\\left(\\frac{x}{64}\\right)+9}$.\n\nUsing logarithm laws,\n\n$$\n\\begin{aligned}\n\\log _{2} x \\cdot \\log _{2}(4 x)+1 & =\\log _{2} x\\left(\\log _{2} 4+\\log _{2} x\\right)+1 \\\\\n& =\\log _{2} x\\left(2+\\log _{2} x\\right)+1 \\quad\\left(\\text { since } 2^{2}=4\\right) \\\\\n& =\\left(\\log _{2} x\\right)^{2}+2 \\cdot \\log _{2} x+1 \\\\\n& =\\left(\\log _{2} x+1\\right)^{2}\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\n\\log _{2} x \\cdot \\log _{2}\\left(\\frac{x}{64}\\right)+9 & =\\log _{2} x\\left(\\log _{2} x-\\log _{2} 64\\right)+9 \\\\\n& =\\log _{2} x\\left(\\log _{2} x-6\\right)+9 \\quad\\left(\\text { since } 2^{6}=64\\right) \\\\\n& =\\left(\\log _{2} x\\right)^{2}-6 \\log _{2} x+9 \\\\\n& =\\left(\\log _{2} x-3\\right)^{2}\n\\end{aligned}\n$$\n\nTherefore,\n\n$f(x)=\\sqrt{\\log _{2} x \\cdot \\log _{2}(4 x)+1}+\\sqrt{\\log _{2} x \\cdot \\log _{2}\\left(\\frac{x}{64}\\right)+9}=\\sqrt{\\left(\\log _{2} x+1\\right)^{2}}+\\sqrt{\\left(\\log _{2} x-3\\right)^{2}}$\n\nBefore proceeding, we recall that if $a \\leq 0$, then $\\sqrt{a^{2}}=-a$ and if $a>0$, then $\\sqrt{a^{2}}=a$.\n\nWhen $\\log _{2} x \\leq-1$, we know that $\\log _{2} x+1 \\leq 0$ and $\\log _{2} x-3<0$, and so\n\n$$\nf(x)=-\\left(\\log _{2} x+1\\right)-\\left(\\log _{2} x-3\\right)=2-2 \\log _{2} x\n$$\n\nWhen $-1<\\log _{2} x \\leq 3$, we know that $\\log _{2} x+1>0$ and $\\log _{2} x-3 \\leq 0$, and so\n\n$$\nf(x)=\\left(\\log _{2} x+1\\right)-\\left(\\log _{2} x-3\\right)=4\n$$\n\nWhen $\\log _{2} x>3$, we know that $\\log _{2} x+1 \\geq 0$ and $\\log _{2} x-3>0$, and so\n\n$$\nf(x)=\\left(\\log _{2} x+1\\right)+\\left(\\log _{2} x-3\\right)=2 \\log _{2} x-2\n$$\n\nWe want to find all values of $x$ for which $f(x)=4$.\n\nWhen $\\log _{2} x \\leq-1, f(x)=2-2 \\log _{2} x=4$ exactly when $\\log _{2} x=-1$.\n\nWhen $-1<\\log _{2} x \\leq 3, f(x)$ is always equal to 4 .\n\nWhen $\\log _{2} x>3, f(x)=2 \\log _{2} x-2=4$ exactly when $\\log _{2} x=3$.\n\nTherefore, $f(x)=4$ exactly when $-1 \\leq \\log _{2} x \\leq 3$, which is true exactly when $\\frac{1}{2} \\leq x \\leq 8$. (It seems surprising that the solution to this equation is actually an interval of values, rather than a finite number of specific values.)" ]

[ "$[\\frac{1}{2}, 8]$" ]

[ "Since $0<\\frac{1}{3}<\\frac{2}{3}<1$, then $\\left\\lfloor\\frac{1}{3}\\right\\rfloor=\\left\\lfloor\\frac{2}{3}\\right\\rfloor=0$.\n\nSince $1 \\leq \\frac{3}{3}<\\frac{4}{3}<\\frac{5}{3}<2$, then $\\left\\lfloor\\frac{3}{3}\\right\\rfloor=\\left\\lfloor\\frac{4}{3}\\right\\rfloor=\\left\\lfloor\\frac{5}{3}\\right\\rfloor=1$.\n\nThese fractions can continue to be grouped in groups of 3 with the last full group of 3 satisfying $19 \\leq \\frac{57}{3}<\\frac{58}{3}<\\frac{59}{3}<20$, which means that $\\left\\lfloor\\frac{57}{3}\\right\\rfloor=\\left\\lfloor\\frac{58}{3}\\right\\rfloor=\\left\\lfloor\\frac{59}{3}\\right\\rfloor=19$.\n\nThe last term is $\\left\\lfloor\\frac{60}{3}\\right\\rfloor=\\lfloor 20\\rfloor=20$.\n\n\n\nIf the given sum is $S$, we obtain\n\n$$\n\\begin{aligned}\nS & =2 \\cdot 0+3 \\cdot 1+3 \\cdot 2+\\cdots+3 \\cdot 19+1 \\cdot 20 \\\\\n& =0+3(1+2+\\cdot+19)+20 \\\\\n& =3 \\cdot \\frac{1}{2} \\cdot 19 \\cdot 20+20 \\\\\n& =570+20 \\\\\n& =590\n\\end{aligned}\n$$" ]

[ "590" ]

[ "For every positive integer $m>4$, let\n\n$$\nq(m)=\\left\\lfloor\\frac{1}{3}\\right\\rfloor+\\left\\lfloor\\frac{2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3}{3}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{m-2}{3}\\right\\rfloor+\\left\\lfloor\\frac{m-1}{3}\\right\\rfloor\n$$\n\nExtending our work from (a), we know that $k-1 \\leq \\frac{3 k-3}{3}<\\frac{3 k-2}{3}<\\frac{3 k-1}{3}<k$ for each positive integer $k$, and so $\\left\\lfloor\\frac{3 k-3}{3}\\right\\rfloor=\\left\\lfloor\\frac{3 k-2}{3}\\right\\rfloor=\\left\\lfloor\\frac{3 k-1}{3}\\right\\rfloor=k-1$.\n\nEvery positive integer $m>4$ can be written as $m=3 s$ or $m=3 s+1$ or $m=3 s+2$, for some positive integer $s$, depending on its remainder when divided by 3 .\n\nWe can thus write\n\n$$\n\\begin{aligned}\nq(3 s) & =\\left\\lfloor\\frac{1}{3}\\right\\rfloor+\\left\\lfloor\\frac{2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3}{3}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{3 s-2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3 s-1}{3}\\right\\rfloor \\\\\n& =2 \\cdot 0+3(1+2+3+\\cdots+(s-1)) \\\\\n& =3 \\cdot \\frac{1}{2} \\cdot(s-1) s \\\\\n& =\\frac{3 s(s-1)}{2} \\\\\n& =\\frac{3 s(3 s-3)}{6} \\\\\nq(3 s+1) & =\\left\\lfloor\\frac{1}{3}\\right\\rfloor+\\left\\lfloor\\frac{2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3}{3}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{3 s-2}{3}\\right\\rfloor+\\left\\lfloor\\frac{3 s-1}{3}\\right\\rfloor+\\left\\lfloor\\frac{3 s}{3}\\right\\rfloor \\\\\n& =q(3 s)+s \\\\\n& =\\frac{3 s(3 s-3)}{6}+\\frac{3 s \\cdot 2}{6} \\\\\n& =\\frac{3 s(3 s-1)}{6} \\\\\nq(3 s+2) & =q(3 s+1)+\\left\\lfloor\\frac{3 s+1}{3}\\right\\rfloor \\\\\n& =\\frac{3 s(3 s-1)}{6}+s \\\\\n& =\\frac{3 s(3 s-1)}{6}+\\frac{3 s \\cdot 2}{6} \\\\\n& =\\frac{3 s(3 s+1)}{6}\n\\end{aligned}\n$$\n\nWe want to find a polynomial $p(x)$ for which $q(m)=\\lfloor p(m)\\rfloor$ for every positive integer $m>4$.\n\n\n\nIn other words, we want to find a polynomial $p(x)$ for which\n\n$$\n\\lfloor p(3 s)\\rfloor=\\frac{3 s(3 s-3)}{6} \\quad\\lfloor p(3 s+1)\\rfloor=\\frac{3 s(3 s-1)}{6} \\quad\\lfloor p(3 s+2)\\rfloor=\\frac{3 s(3 s+1)}{6}\n$$\n\nfor every positive integer $s$.\n\nWe will show that the polynomial $p(x)=\\frac{(x-1)(x-2)}{6}$ satisfies the desired conditions.\n\nIf $x=3 s+1$ for some positive integer $s$, then\n\n$$\n\\frac{(x-1)(x-2)}{6}=\\frac{(3 s+1-1)(3 s+1-2)}{6}=\\frac{3 s(3 s-1)}{6}\n$$\n\nWe note that $3 s$ is a multiple of 3 . Since $3 s$ and $3 s-1$ are consecutive integers, then one of these is a multiple of 2 . Thus, $3 s(3 s-1)$ is a multiple of 6 and so $\\frac{3 s(3 s-1)}{6}$ is an integer.\n\nThis means that $\\left\\lfloor\\frac{3 s(3 s-1)}{6}\\right\\rfloor=\\frac{3 s(3 s-1)}{6}$.\n\nTherefore, $q(3 s+1)=\\frac{3 s(3 s-1)}{6}=\\left\\lfloor\\frac{3 s(3 s-1)}{6}\\right\\rfloor=\\lfloor p(3 s+1)\\rfloor$.\n\nIf $x=3 s+2$ for some positive integer $s$, then\n\n$$\n\\frac{(x-1)(x-2)}{6}=\\frac{(3 s+2-1)(3 s+2-2)}{6}=\\frac{3 s(3 s+1)}{6}\n$$\n\nWe note that $3 s$ is a multiple of 3 . Since $3 s$ and $3 s+1$ are consecutive integers, then one of these is a multiple of 2 . Thus, $3 s(3 s+1)$ is a multiple of 6 and so $\\frac{3 s(3 s+1)}{6}$ is an integer.\n\nThis means that $\\left\\lfloor\\frac{3 s(3 s+1)}{6}\\right\\rfloor=\\frac{3 s(3 s+1)}{6}$.\n\nTherefore, $q(3 s+2)=\\frac{3 s(3 s+1)}{6}=\\left\\lfloor\\frac{3 s(3 s+1)}{6}\\right\\rfloor=\\lfloor p(3 s+2)\\rfloor$.\n\nIf $x=3 s$ for some positive integer $s$, then\n\n$$\n\\frac{(x-1)(x-2)}{6}=\\frac{(3 s-1)(3 s-2)}{6}=\\frac{9 s^{2}-9 s+2}{6}\n$$\n\nNow, $\\frac{9 s^{2}-9 s}{6}=\\frac{9 s(s-1)}{6}$ is an integer because $9 s$ is a multiple of 3 and one of $s$ and $s-1$ is even.\n\nSince $\\frac{9 s^{2}-9 s+2}{6}=\\frac{9 s^{2}-9 s}{6}+\\frac{1}{3}$, then $\\frac{9 s^{2}-9 s+2}{6}$ is $\\frac{1}{3}$ more than an integer which means that $\\left\\lfloor\\frac{9 s^{2}-9 s+2}{6}\\right\\rfloor=\\frac{9 s^{2}-9 s}{6}=\\frac{3 s(3 s-3)}{6}=q(3 s)$.\n\nTherefore, $q(3 s)=\\frac{3 s(3 s-3)}{6}=\\left\\lfloor\\frac{(3 s-1)(3 s-2)}{6}\\right\\rfloor=\\lfloor p(3 s)\\rfloor$.\n\nThis means that the polynomial $p(x)=\\frac{(x-1)(x-2)}{6}$ satisfies the required conditions." ]

[ "$p(x)=\\frac{(x-1)(x-2)}{6}$" ]

[ "Suppose that the rectangular prism has dimensions $a \\mathrm{~cm}$ by $b \\mathrm{~cm}$ by $c \\mathrm{~cm}$.\n\nSuppose further that one of the faces that is $a \\mathrm{~cm}$ by $b \\mathrm{~cm}$ is the face with area $27 \\mathrm{~cm}^{2}$ and that one of the faces that is $a \\mathrm{~cm}$ by $c \\mathrm{~cm}$ is the face with area $32 \\mathrm{~cm}^{2}$. (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.)\n\nTherefore, $a b=27$ and $a c=32$.\n\nFurther, we are told that the volume of the prism is $144 \\mathrm{~cm}^{3}$, and so $a b c=144$.\n\n\n\nThus, $b c=\\frac{a^{2} b^{2} c^{2}}{a^{2} b c}=\\frac{(a b c)^{2}}{(a b)(a c)}=\\frac{144^{2}}{(27)(32)}=24$.\n\n(We could also note that $a b c=144$ means $a^{2} b^{2} c^{2}=144^{2}$ or $(a b)(a c)(b c)=144^{2}$ and so $b c=\\frac{144^{2}}{(27)(32)}$.)\n\nIn other words, the third type of face of the prism has area $24 \\mathrm{~cm}^{2}$.\n\nThus, since the prism has two faces of each type, the surface area of the prism is equal to $2\\left(27 \\mathrm{~cm}^{2}+32 \\mathrm{~cm}^{2}+24 \\mathrm{~cm}^{2}\\right)$ or $166 \\mathrm{~cm}^{2}$.", "Suppose that the rectangular prism has dimensions $a \\mathrm{~cm}$ by $b \\mathrm{~cm}$ by $c \\mathrm{~cm}$.\n\nSuppose further that one of the faces that is $a \\mathrm{~cm}$ by $b \\mathrm{~cm}$ is the face with area $27 \\mathrm{~cm}^{2}$ and that one of the faces that is $a \\mathrm{~cm}$ by $c \\mathrm{~cm}$ is the face with area $32 \\mathrm{~cm}^{2}$. (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.)\n\nTherefore, $a b=27$ and $a c=32$.\n\nFurther, we are told that the volume of the prism is $144 \\mathrm{~cm}^{3}$, and so $a b c=144$.\n\nSince $a b c=144$ and $a b=27$, then $c=\\frac{144}{27}=\\frac{16}{3}$.\n\nSince $a b c=144$ and $a c=32$, then $b=\\frac{144}{32}=\\frac{9}{2}$.\n\nThis means that $b c=\\frac{16}{3} \\cdot \\frac{9}{2}=24$.\n\nIn $\\mathrm{cm}^{2}$, the surface area of the prism equals $2 a b+2 a c+2 b c=2(27)+2(32)+2(24)=166$. Thus, the surface area of the prism is $166 \\mathrm{~cm}^{2}$." ]

[ "$166$" ]

[ "We expand the right sides of the two equations, collecting like terms in each case:\n\n$$\n\\begin{aligned}\n& y=a(x-2)(x+4)=a\\left(x^{2}+2 x-8\\right)=a x^{2}+2 a x-8 a \\\\\n& y=2(x-h)^{2}+k=2\\left(x^{2}-2 h x+h^{2}\\right)+k=2 x^{2}-4 h x+\\left(2 h^{2}+k\\right)\n\\end{aligned}\n$$\n\nSince these two equations represent the same parabola, then the corresponding coefficients must be equal. That is, $a=2$ and $2 a=-4 h$ and $-8 a=2 h^{2}+k$.\n\nSince $a=2$ and $2 a=-4 h$, then $4=-4 h$ and so $h=-1$.\n\nSince $-8 a=2 h^{2}+k$ and $a=2$ and $h=-1$, then $-16=2+k$ and so $k=-18$.\n\nThus, $a=2, h=-1$, and $k=-18$.", "From the equation $y=a(x-2)(x+4)$, we can find the axis of symmetry by calculating the midpoint of the $x$-intercepts.\n\nSince the $x$-intercepts are 2 and -4 , the axis of symmetry is at $x=\\frac{1}{2}(2+(-4))=-1$.\n\nSince the vertex of the parabola lies on the axis of symmetry, then the $x$-coordinate of the vertex is -1 .\n\nTo find the $y$-coordinate of the vertex, we substitute $x=-1$ back into the equation $y=a(x-2)(x+4)$ to obtain $y=a(-1-2)(-1+4)=-9 a$.\n\nThus, the vertex of the parabola is $(-1,-9 a)$.\n\nSince the second equation for the same parabola is in vertex form, $y=2(x-h)^{2}+k$, we can see that the vertex is at $(h, k)$ and $a=2$.\n\nSince $a=2$, the vertex has coordinates $(-1,-18)$, which means that $h=-1$ and $k=-18$. Thus, $a=2, h=-1$ and $k=-18$." ]

[ "$2,-1,-18$" ]

[ "Let the common difference in this arithmetic sequence be $d$.\n\nSince the first term in the sequence is 5 , then the 5 terms are $5,5+d, 5+2 d, 5+3 d, 5+4 d$.\n\nFrom the given information, $5^{2}+(5+d)^{2}+(5+2 d)^{2}=(5+3 d)^{2}+(5+4 d)^{2}$.\n\nManipulating, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n5^{2}+(5+d)^{2}+(5+2 d)^{2} & =(5+3 d)^{2}+(5+4 d)^{2} \\\\\n25+\\left(25+10 d+d^{2}\\right)+\\left(25+20 d+4 d^{2}\\right) & =\\left(25+30 d+9 d^{2}\\right)+\\left(25+40 d+16 d^{2}\\right) \\\\\n75+30 d+5 d^{2} & =50+70 d+25 d^{2} \\\\\n0 & =20 d^{2}+40 d-25 \\\\\n0 & =4 d^{2}+8 d-5 \\\\\n0 & =(2 d+5)(2 d-1)\n\\end{aligned}\n$$\n\nTherefore, $d=-\\frac{5}{2}$ or $d=\\frac{1}{2}$.\n\nThese give possible fifth terms of $5+4 d=5+4\\left(-\\frac{5}{2}\\right)=-5$ and $5+4 d=5+4\\left(\\frac{1}{2}\\right)=7$.\n\n(We note that, for these two values of $d$, the sequences are $5, \\frac{5}{2}, 0,-\\frac{5}{2},-5$ and $5, \\frac{11}{2}, 6, \\frac{13}{2}, 7$.)" ]

[ "-5,7" ]

[ "First, we determine the perfect squares between 1300 and 1400 and between 1400 and 1500.\n\nSince $\\sqrt{1300} \\approx 36.06$, then the first perfect square larger than 1300 is $37^{2}=1369$.\n\nThe next perfect squares are $38^{2}=1444$ and $39^{2}=1521$.\n\nSince Dan was born between 1300 and 1400 in a year that was a perfect square, then Dan was born in 1369.\n\nSince Steve was born between 1400 and 1500 in a year that was a perfect square, then Steve was born in 1444.\n\nSuppose that on April 7 in some year, Dan was $m^{2}$ years old and Steve was $n^{2}$ years old for some positive integers $m$ and $n$. Thus, Dan was $m^{2}$ years old in the year $1369+m^{2}$ and Steve was $n^{2}$ years old in the year $1444+n^{2}$.\n\nSince these represent the same years, then $1369+m^{2}=1444+n^{2}$, or $m^{2}-n^{2}=1444-$ $1369=75$.\n\nIn other words, we want to find two perfect squares less than 110 (since their ages are less than 110) whose difference is 75.\n\nThe perfect squares less than 110 are $1,4,9,16,25,36,49,64,81,100$.\n\nThe two that differ by 75 are 100 and 25 .\n\nThus, $m^{2}=100$ and $n^{2}=25$.\n\nThis means that the year in which the age of each of Dan and Steve was a perfect square was the year $1369+100=1469$." ]

[ "1469" ]

[ "$\\triangle A B C$ is right-angled exactly when one of the following statements is true:\n\n- $A B$ is perpendicular to $B C$, or\n- $A B$ is perpendicular to $A C$, or\n- $A C$ is perpendicular to $B C$.\n\nSince $A(1,2)$ and $B(11,2)$ share a $y$-coordinate, then $A B$ is horizontal.\n\nFor $A B$ and $B C$ to be perpendicular, $B C$ must be vertical.\n\nThus, $B(11,2)$ and $C(k, 6)$ must have the same $x$-coordinate, and so $k=11$.\n\nFor $A B$ and $A C$ to be perpendicular, $A C$ must be vertical.\n\nThus, $A(1,2)$ and $C(k, 6)$ must have the same $x$-coordinate, and so $k=1$.\n\n\n\nFor $A C$ to be perpendicular to $B C$, their slopes must have a product of -1 .\n\nThe slope of $A C$ is $\\frac{6-2}{k-1}$, which equals $\\frac{4}{k-1}$.\n\nThe slope of $B C$ is $\\frac{6-2}{k-11}$, which equals $\\frac{4}{k-11}$.\n\nThus, $A C$ and $B C$ are perpendicular when $\\frac{4}{k-1} \\cdot \\frac{4}{k-11}=-1$.\n\nAssuming that $k \\neq 1$ and $k \\neq 11$, we manipulate to obtain $16=-(k-1)(k-11)$ or $16=-k^{2}+12 k-11$ or $k^{2}-12 k+27=0$.\n\nFactoring, we obtain $(k-3)(k-9)=0$ and so $A C$ and $B C$ are perpendicular when $k=3$ or $k=9$.\n\nIn summary, $\\triangle A B C$ is right-angled when $k$ equals one of $1,3,9,11$.", "$\\triangle A B C$ is right-angled exactly when its three side lengths satisfy the Pythagorean Theorem in some orientation. That is, $\\triangle A B C$ is right-angled exactly when $A B^{2}+B C^{2}=A C^{2}$ or $A B^{2}+A C^{2}=B C^{2}$ or $A C^{2}+B C^{2}=A B^{2}$.\n\nUsing $A(1,2)$ and $B(11,2)$, we obtain $A B^{2}=(11-1)^{2}+(2-2)^{2}=100$.\n\nUsing $A(1,2)$ and $C(k, 6)$, we obtain $A C^{2}=(k-1)^{2}+(6-2)^{2}=(k-1)^{2}+16$.\n\nUsing $B(11,2)$ and $C(k, 6)$, we obtain $B C^{2}=(k-11)^{2}+(6-2)^{2}=(k-11)^{2}+16$.\n\nUsing the Pythagorean relationships above, $\\triangle A B C$ is right-angled when one of the following is true:\n\n(i)\n\n$$\n\\begin{aligned}\n100+\\left((k-11)^{2}+16\\right) & =(k-1)^{2}+16 \\\\\n100+k^{2}-22 k+121+16 & =k^{2}-2 k+1+16 \\\\\n220 & =20 k \\\\\nk & =11\n\\end{aligned}\n$$\n\n(ii)\n\n$$\n\\begin{aligned}\n100+\\left((k-1)^{2}+16\\right) & =(k-11)^{2}+16 \\\\\n100+k^{2}-2 k+1+16 & =k^{2}-22 k+121+16 \\\\\n20 k & =20 \\\\\nk & =1\n\\end{aligned}\n$$\n\n(iii)\n\n$$\n\\begin{aligned}\n\\left((k-1)^{2}+16\\right)+\\left((k-11)^{2}+16\\right) & =100 \\\\\nk^{2}-2 k+1+16+k^{2}-22 k+121+16 & =100 \\\\\n2 k^{2}-24 k+54 & =0 \\\\\nk^{2}-12 k+27 & =0 \\\\\n(k-3)(k-9) & =0\n\\end{aligned}\n$$\n\nand so $k=3$ or $k=9$.\n\nIn summary, $\\triangle A B C$ is right-angled when $k$ equals one of $1,3,9,11$." ]

[ "1,3,9,11" ]

[ "Since $\\tan \\theta=\\frac{\\sin \\theta}{\\cos \\theta}$, then we assume that $\\cos \\theta \\neq 0$.\n\nTherefore, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n\\cos \\theta & =\\tan \\theta \\\\\n\\cos \\theta & =\\frac{\\sin \\theta}{\\cos \\theta} \\\\\n\\cos ^{2} \\theta & =\\sin \\theta \\\\\n1-\\sin ^{2} \\theta & =\\sin \\theta \\\\\n0 & =\\sin ^{2} \\theta+\\sin \\theta-1\n\\end{aligned}\n$$\n\nLet $u=\\sin \\theta$. This quadratic equation becomes $u^{2}+u-1=0$\n\nBy the quadratic formula, $u=\\frac{-1 \\pm \\sqrt{1^{2}-4(1)(-1)}}{2(1)}=\\frac{-1 \\pm \\sqrt{5}}{2}$.\n\nTherefore, $\\sin \\theta=\\frac{-1+\\sqrt{5}}{2} \\approx 0.62$ or $\\sin \\theta=\\frac{-1-\\sqrt{5}}{2} \\approx-1.62$.\n\nSince $-1 \\leq \\sin \\theta \\leq 1$, then the second solution is inadmissible. Thus, $\\sin \\theta=\\frac{-1+\\sqrt{5}}{2}$." ]

[ "$\\frac{-1+\\sqrt{5}}{2}$" ]

[ "Suppose that the trains are travelling at $v \\mathrm{~km} / \\mathrm{h}$.\n\nConsider two consecutive points in time at which the car is passed by a train.\n\nSince these points are 10 minutes apart, and 10 minutes equals $\\frac{1}{6}$ hour, and the car travels at $60 \\mathrm{~km} / \\mathrm{h}$, then the car travels $(60 \\mathrm{~km} / \\mathrm{h}) \\cdot\\left(\\frac{1}{6} \\mathrm{~h}\\right)=10 \\mathrm{~km}$.\n\nDuring these 10 minutes, each train travels $\\frac{1}{6} v \\mathrm{~km}$, since its speed is $v \\mathrm{~km} / \\mathrm{h}$.\n\nAt the first instance, Train A and the car are next to each other.\n\nAt this time, Train B is \" 3 minutes\" behind Train A.\n\n<img_4020>\n\nSince 3 minutes is $\\frac{1}{20}$ hour, then Train B is $\\frac{1}{20} v \\mathrm{~km}$ behind Train A and the car.\n\nTherefore, the distance from the location of Train B at the first instance to the location where it passes the car is $\\left(\\frac{1}{20} v+10\\right) \\mathrm{km}$.\n\nBut this distance also equals $\\frac{1}{6} v \\mathrm{~km}$, since Train B travels for 10 minutes.\n\nThus, $\\frac{1}{6} v=\\frac{1}{20} v+10$ or $\\frac{10}{60} v-\\frac{3}{60} v=10$ and so $\\frac{7}{60} v=10$ or $v=\\frac{600}{7}$.\n\nTherefore, the trains are travelling at $\\frac{600}{7} \\mathrm{~km} / \\mathrm{h}$.", "Suppose that the trains are travelling at $v \\mathrm{~km} / \\mathrm{h}$.\n\nConsider the following three points in time: the instant when the car and Train A are next to each other, the instant when Train B is at the same location that the car and Train A were at in the previous instant, and the instant when the car and Train B are next to each other.\n\n<img_3611>\n\nFrom the first instant to the second, Train B \"catches up\" to where Train A was, so this must take a total of 3 minutes, because the trains leave the station 3 minutes apart.\n\nSince 3 minutes equals $\\frac{3}{60}$ hour and the car travels at $60 \\mathrm{~km} / \\mathrm{h}$, then the car travels $(60 \\mathrm{~km} / \\mathrm{h}) \\cdot\\left(\\frac{3}{60} \\mathrm{~h}\\right)=3 \\mathrm{~km}$ between these two instants.\n\nFrom the first instant to the third, 10 minutes passes, since these are consecutive points at which the car is passed by trains. In 10 minutes, the car travels $10 \\mathrm{~km}$.\n\nTherefore, between the second and third instants, $10-3=7$ minutes pass. During these 7 minutes, Train B travels $10 \\mathrm{~km}$.\n\nSince 7 minutes equals $\\frac{7}{60}$ hour, then $v \\mathrm{~km} / \\mathrm{h}=\\frac{10 \\mathrm{~km}}{7 / 60 \\mathrm{~h}}=\\frac{600}{7} \\mathrm{~km} / \\mathrm{h}$, and so the trains are travelling at $\\frac{600}{7} \\mathrm{~km} / \\mathrm{h}$." ]

[ "$\\frac{600}{7}$" ]

[ "From the first equation, we note that $a \\geq 0$ and $b \\geq 0$, since the argument of a square root must be non-negative.\n\nFrom the second equation, we note that $a>0$ and $b>0$, since the argument of a logarithm must be positive.\n\nCombining these restrictions, we see that $a>0$ and $b>0$.\n\nFrom the equation $\\log _{10} a+\\log _{10} b=2$, we obtain $\\log _{10}(a b)=2$ and so $a b=10^{2}=100$. From the first equation, obtain\n\n$$\n\\begin{aligned}\n(\\sqrt{a}+\\sqrt{b})^{2} & =8^{2} \\\\\na+2 \\sqrt{a b}+b & =64 \\\\\na+2 \\sqrt{100}+b & =64 \\\\\na+b & =64-2 \\sqrt{100}=44\n\\end{aligned}\n$$\n\nSince $a+b=44$, then $b=44-a$.\n\nSince $a b=100$, then $a(44-a)=100$ or $44 a-a^{2}=100$ and so $0=a^{2}-44 a+100$.\n\nBy the quadratic formula,\n\n$$\na=\\frac{44 \\pm \\sqrt{44^{2}-4(1)(100)}}{2 \\cdot 1}=\\frac{44 \\pm \\sqrt{1536}}{2}=\\frac{44 \\pm 16 \\sqrt{6}}{2}=22 \\pm 8 \\sqrt{6}\n$$\n\nSince $b=44-a$, then $b=44-(22 \\pm 8 \\sqrt{6})=22 \\mp 8 \\sqrt{6}$.\n\nTherefore, $(a, b)=(22+8 \\sqrt{6}, 22-8 \\sqrt{6})$ or $(a, b)=(22-8 \\sqrt{6}, 22+8 \\sqrt{6})$.\n\n(We note that $22+8 \\sqrt{6}>0$ and $22-8 \\sqrt{6}>0$, so the initial restrictions on $a$ and $b$ are satisfied.)" ]

[ "$(22+8 \\sqrt{6}, 22-8 \\sqrt{6})$,$(22-8 \\sqrt{6}, 22+8 \\sqrt{6})$" ]

[ "There are 4 ! $=4 \\cdot 3 \\cdot 2 \\cdot 1=24$ permutations of $1,2,3,4$.\n\nThis is because there are 4 possible choices for $a_{1}$, and for each of these there are 3 possible choices for $a_{2}$, and for each of these there are 2 possible choices for $a_{3}$, and then 1 possible choice for $a_{4}$.\n\nConsider the permutation $a_{1}=1, a_{2}=2, a_{3}=3, a_{4}=4$. (We write this as $1,2,3,4$.)\n\nHere, $\\left|a_{1}-a_{2}\\right|+\\left|a_{3}-a_{4}\\right|=|1-2|+|3-4|=1+1=2$.\n\nThis value is the same as the value for each of $2,1,3,4$ and $1,2,4,3$ and $2,1,4,3$ and $3,4,1,2$ and 4,3,1,2 and 3,4,2,1 and 4,3,2,1.\n\nConsider the permutation $1,3,2,4$.\n\nHere, $\\left|a_{1}-a_{2}\\right|+\\left|a_{3}-a_{4}\\right|=|1-3|+|2-4|=2+2=4$.\n\nThis value is the same as the value for each of $3,1,2,4$ and $1,3,4,2$ and $3,1,4,2$ and $2,4,1,3$ and 4,2,1,3 and 2,4,3,1 and 4,2,3,1.\n\nConsider the permutation $1,4,2,3$.\n\nHere, $\\left|a_{1}-a_{2}\\right|+\\left|a_{3}-a_{4}\\right|=|1-4|+|2-3|=3+1=4$.\n\nThis value is the same as the value for each of 4,1,2,3 and 1,4,3,2 and 4,1,3,2 and 2,3,1,4 and $3,2,1,4$ and $2,3,4,1$ and $3,2,4,1$.\n\nThis accounts for all 24 permutations.\n\nTherefore, the average value is $\\frac{2 \\cdot 8+4 \\cdot 8+4 \\cdot 8}{24}=\\frac{80}{24}=\\frac{10}{3}$." ]

[ "$\\frac{10}{3}$" ]

[ "There are $7 !=7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1$ permutations of $1,2,3,4,5,6,7$, because there are 7 choices for $a_{1}$, then 6 choices for $a_{2}$, and so on.\n\nWe determine the average value of $a_{1}-a_{2}+a_{3}-a_{4}+a_{5}-a_{6}+a_{7}$ over all of these permutations by determining the sum of all 7 ! values of this expression and dividing by $7 !$.\n\nTo determine the sum of all 7 ! values, we determine the sum of the values of $a_{1}$ in each of these expressions and call this total $s_{1}$, the sum of the values of $a_{2}$ in each of these expressions and call this total $s_{2}$, and so on.\n\nThe sum of the 7 ! values of the original expression must equal $s_{1}-s_{2}+s_{3}-s_{4}+s_{5}-s_{6}+s_{7}$. This uses the fact that, when adding, the order in which we add the same set of numbers does not matter.\n\nBy symmetry, the sums of the values of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ will all be equal. That is, $s_{1}=s_{2}=s_{3}=s_{4}=s_{5}=s_{6}=s_{7}$.\n\nThis means that the desired average value equals\n\n$$\n\\frac{s_{1}-s_{2}+s_{3}-s_{4}+s_{5}-s_{6}+s_{7}}{7 !}=\\frac{\\left(s_{1}+s_{3}+s_{5}+s_{7}\\right)-\\left(s_{2}+s_{4}+s_{6}\\right)}{7 !}=\\frac{4 s_{1}-3 s_{1}}{7 !}=\\frac{s_{1}}{7 !}\n$$\n\nSo we need to determine the value of $s_{1}$.\n\nNow $a_{1}$ can equal each of $1,2,3,4,5,6,7$.\n\nIf $a_{1}=1$, there are 6 ! combinations of values for $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$, since there are still 6 choices for $a_{2}, 5$ for $a_{3}$, and so on.\n\nSimilarly, there are 6 ! combinations with $a_{1}$ equal to each of $2,3,4,5,6,7$.\n\nThus, $s_{1}=1 \\cdot 6 !+2 \\cdot 6 !+3 \\cdot 6 !+4 \\cdot 6 !+5 \\cdot 6 !+6 \\cdot 6 !+7 \\cdot 6 !=6 !(1+2+3+4+5+6+7)=28(6 !)$.\n\nTherefore, the average value of the expression is $\\frac{28(6 !)}{7 !}=\\frac{28(6 !)}{7(6 !)}=\\frac{28}{7}=4$." ]

[ "4" ]

[ "There are 200! permutations of $1,2,3, \\ldots, 198,199,200$.\n\nWe determine the average value of\n\n$$\n\\left|a_{1}-a_{2}\\right|+\\left|a_{3}-a_{4}\\right|+\\cdots+\\left|a_{197}-a_{198}\\right|+\\left|a_{199}-a_{200}\\right|\n$$\n\nover all of these permutations by determining the sum of all 200! values of this expression and dividing by $200 !$.\n\nThen, we let $s_{1}$ be the sum of the values of $\\left|a_{1}-a_{2}\\right|$ in each of these expressions, $s_{2}$ be the sum of the values of $\\left|a_{3}-a_{4}\\right|$, and so on.\n\nThe sum of the 200 ! values of $(*)$ equals $s_{1}+s_{2}+\\cdots+s_{99}+s_{100}$.\n\nBy symmetry, $s_{1}=s_{2}=\\cdots=s_{99}=s_{100}$.\n\nTherefore, the average value of $(*)$ equals $\\frac{100 s_{1}}{200 !}$. So we need to determine the value of $s_{1}$.\n\nSuppose that $a_{1}=i$ and $a_{2}=j$ for some integers $i$ and $j$ between 1 and 200, inclusive.\n\nThere are 198! permutations with $a_{1}=i$ and $a_{2}=j$ because there are still 198 choices for $a_{3}, 197$ choices for $a_{4}$, and so on.\n\nSimilarly, there are 198! permutations with $a_{1}=j$ and $a_{2}=i$.\n\nSince $|i-j|=|j-i|$, then there are 2(198!) permutations with $\\left|a_{1}-a_{2}\\right|=|i-j|$ that come from $a_{1}$ and $a_{2}$ equalling $i$ and $j$ in some order.\n\nTherefore, we may assume that $i>j$ and note that $s_{1}$ equals 2(198!) times the sum of $i-j$ over all possible pairs $i>j$.\n\n(Note that there are $\\left(\\begin{array}{c}200 \\\\ 2\\end{array}\\right)=\\frac{200(199)}{2}$ choices for the pair of integers $(i, j)$ with $i>j$. For each of these choices, there are 2(198!) choices for the remaining entries in the permutation, which gives $\\frac{200(199)}{2} \\cdot 2(198 !)=200(199)(198 !)=200$ ! permutations, as expected.)\n\nSo to determine $s_{1}$, we need to determine the sum of the values of $i-j$.\n\nWe calculate this sum, which we call $D$, by letting $j=1,2,3, \\ldots, 198,199$ and for each of these, we let $i$ be the possible integers with $j<i \\leq 200$ :\n\n$$\n\\begin{aligned}\nD & =(2-1)+(3-1)+(4-1)+\\cdots+(197-1)+(198-1)+(199-1)+(200-1) \\\\\n& +(3-2)+(4-2)+(5-2)+\\cdots+(198-2)+(199-2)+(200-2) \\\\\n& +(4-3)+(5-3)+(6-3)+\\cdots+(199-3)+(200-3) \\\\\n& \\vdots \\\\\n& +(199-198)+(200-198) \\\\\n& +(200-199) \\\\\n& =199(1)+198(2)+197(3)+\\cdots+2(198)+1(199) \\quad \\quad \\quad \\text { grouping by columns }) \\\\\n& =199(200-199)+198(200-198)+197(200-197)+\\cdots+2(200-2)+1(200-1) \\\\\n& =200(199+198+197+\\cdots+3+2+1)-\\left(199^{2}+198^{2}+197^{2}+\\cdots+3^{2}+2^{2}+1^{2}\\right) \\\\\n& =200 \\cdot \\frac{1}{2}(199)(200)-\\frac{1}{6}(199)(199+1)(2(199)+1) \\\\\n& =100(199)(200)-\\frac{1}{6}(199)(200)(399) \\\\\n& =199(200)\\left(100-\\frac{133}{2}\\right) \\\\\n& =199(200) \\frac{67}{2}\n\\end{aligned}\n$$\n\nTherefore, $s_{1}=2(198 !) D=2(198 !) \\cdot \\frac{199(200)(67)}{2}=67(198 !)(199)(200)=67(200 !)$.\n\nFinally, this means that the average value of $(*)$ is $\\frac{100 s_{1}}{200 !}=\\frac{100(67)(200 !)}{200 !}=6700$.\n\n\n\nWe note that we have used the facts that, if $n$ is a positive integer, then\n\n- $1+2+\\cdots+(n-1)+n=\\frac{1}{2} n(n+1)$\n- $1^{2}+2^{2}+\\cdots+(n-1)^{2}+n^{2}=\\frac{1}{6} n(n+1)(2 n+1)$\n\nUsing sigma notation, we could have calculated $D$ as follows:\n\n$$\n\\begin{aligned}\nD & =\\sum_{i=2}^{200} \\sum_{j=1}^{i-1}(i-j) \\\\\n& =\\left(\\sum_{i=2}^{200} \\sum_{j=1}^{i-1} i\\right)-\\left(\\sum_{i=2}^{200} \\sum_{j=1}^{i-1} j\\right) \\\\\n& =\\left(\\sum_{i=2}^{200} i(i-1)\\right)-\\left(\\sum_{i=2}^{200} \\frac{1}{2}(i-1) i\\right) \\\\\n& =\\left(\\sum_{i=2}^{200} i(i-1)\\right)-\\frac{1}{2}\\left(\\sum_{i=2}^{200}(i-1) i\\right) \\\\\n& =\\frac{1}{2}\\left(\\sum_{i=2}^{200}(i-1) i\\right) \\\\\n& =\\frac{1}{2}\\left(\\sum_{i=1}^{200}(i-1) i\\right) \\\\\n& =\\frac{1}{2}\\left(\\sum_{i=1}^{200}\\left(i^{2}-i\\right)\\right) \\\\\n& =\\frac{1}{2}\\left(\\sum_{i=1}^{200} i^{2}-\\sum_{i=1}^{200} i\\right) \\\\\n& =\\frac{1}{2}\\left(\\frac{1}{6}(200)(200+1)(2(200)+1)-\\frac{1}{2}(200)(200+1)\\right) \\\\\n& =\\frac{1}{2}(200)(201)\\left(\\frac{1}{6}(401)-\\frac{1}{2}\\right) \\\\\n& =100(201) \\cdot \\frac{398}{6} \\\\\n& =100(201) \\cdot \\frac{199}{3} \\\\\n& =100(67)(199)\n\\end{aligned}\n$$\n\nwhich equals $199(200) \\frac{67}{2}$, as expected." ]

[ "6700" ]

[ "Rearranging the equation,\n\n$$\n\\begin{aligned}\n3 \\sin (x) & =\\cos \\left(15^{\\circ}\\right) \\\\\n\\sin (x) & =\\frac{1}{3} \\cos \\left(15^{\\circ}\\right) \\\\\n\\sin (x) & \\approx 0.3220\n\\end{aligned}\n$$\n\nUsing a calculator, $x \\approx 18.78^{\\circ}$. To the nearest tenth of a degree, $x=18.8^{\\circ}$." ]

[ "$18.8^{\\circ}$" ]

[ "Since we are looking for the value of $f(9)$, then it makes sense to use the given equation and to set $x=3$ in order to obtain $f(9)=2 f(3)+3$.\n\nSo we need to determine the value of $f(3)$. We use the equation again and set $x=0$ since we will then get $f(3)$ on the left side and $f(0)$ (whose value we already know) on the right side, ie.\n\n$$\nf(3)=2 f(0)+3=2(6)+3=15\n$$\n\nThus, $f(9)=2(15)+3=33$." ]

[ "33" ]

[ "We solve the system of equations for $f(x)$ and $g(x)$.\n\nDividing out the common factor of 2 from the second equation, we get\n\n$f(x)+2 g(x)=x^{2}+2$.\n\nSubtracting from the first equation, we get $g(x)=x+4$.\n\nThus, $f(x)=x^{2}+2-2 g(x)=x^{2}+2-2(x+4)=x^{2}-2 x-6$.\n\nEquating $f(x)$ and $g(x)$, we obtain\n\n$$\n\\begin{aligned}\nx^{2}-2 x-6 & =x+4 \\\\\nx^{2}-3 x-10 & =0 \\\\\n(x-5)(x+2) & =0\n\\end{aligned}\n$$\n\nTherefore, $x=5$ or $x=-2$.", "Instead of considering the equation $f(x)=g(x)$, we consider the equation $f(x)-g(x)=0$, and we try to obtain an expression for $f(x)-g(x)$ by manipulating the two given equations.\n\nIn fact, after some experimentation, we can see that\n\n$$\n\\begin{aligned}\nf(x)-g(x) & =2(2 f(x)+4 g(x))-3(f(x)+3 g(x)) \\\\\n& =2\\left(2 x^{2}+4\\right)-3\\left(x^{2}+x+6\\right) \\\\\n& =x^{2}-3 x-10\n\\end{aligned}\n$$\n\nSo to solve $f(x)-g(x)=0$, we solve $x^{2}-3 x-10=0$ or $(x-5)(x+2)=0$. Therefore, $x=5$ or $x=-2$." ]

[ "$5,-2$" ]

[ "We label the 5 skaters A, B, C, D, and E, where D and E are the two Canadians.\n\nThere are then $5 !=5 \\times 4 \\times 3 \\times 2 \\times 1=120$ ways of arranging these skaters in their order of finish (for example, $\\mathrm{ADBCE}$ indicates that A finished first, $\\mathrm{D}$ second, etc.), because there are 5 choices for the winner, 4 choices for the second place finisher, 3 choices for the third place finisher, etc.\n\n\n\nIf the two Canadians finish without winning medals, then they must finish fourth and fifth. So the $\\mathrm{D}$ and $\\mathrm{E}$ are in the final two positions, and $\\mathrm{A}, \\mathrm{B}$ and $\\mathrm{C}$ in the first three. There are $3 !=6$ ways of arranging the $\\mathrm{A}, \\mathrm{B}$ and $\\mathrm{C}$, and $2 !=2$ ways to arrange the $\\mathrm{D}$ and E. Thus, there are $6 \\times 2=12$ ways or arranging the skaters so that neither Canadian wins a medal.\n\nTherefore, the probability that neither Canadian wins a medal is\n\n$$\n\\frac{\\# \\text { of ways where Canadians don't win medals }}{\\text { Total } \\# \\text { of arrangements }}=\\frac{12}{120}=\\frac{1}{10}\n$$", "We label the 5 skaters as A, B, C, D, and E, where D and E are the two Canadians. In any race, two of the skaters finish fourth and fifth. Also, any pair of skaters are equally as likely to finish fourth and fifth, since the probability of every skater is equally likely to finish in a given position.\n\nHow many pairs of 2 skaters can we form from the 5 skaters? There are ten such pairs:\n\n$$\n\\{A, B\\},\\{A, C\\},\\{A, D\\},\\{A, E\\},\\{B, C\\},\\{B, D\\},\\{B, E\\},\\{C, D\\},\\{C, E\\},\\{D, E\\}\n$$\n\nOnly one of these ten pairs is made up of the two Canadians. Therefore, the probability is $\\frac{1}{10}$, since one out of ten choices gives the desired result." ]

[ "$\\frac{1}{10}$" ]

[ "Since the least common multiple of $3,5,10$ and 15 is 30 , then we can count the number of positive integers less than or equal to 30 satisfying these conditions, and multiply the total by 10 to obtain the number less than 300. (This is because each group of 30 consecutive integers starting with 1 more than a multiple of 30 will have the same number of integers having these properties, because we can subtract 30 from each one and not change these properties.)\n\nSo from 1 to 30, we have:\n\n$$\n3,5,6,9,12,18,21,24,25,27\n$$\n\nThus there are 10 less than or equal to 30 , and so 100 such positive integers less than or equal to 300 .", "We proceed by doing a (careful!) count.\n\nThe number of positive multiples of 3 less than or equal to 300 is 100.\n\nThe number of positive multiples of 5 less than or equal to 300 is 60 .\n\nThus, we have 160 candidates, but have included multiples of 15 twice (since 15 is a multiple of each of 3 and 5), and have also included multiples of 10.\n\nThe number of multiples of 15 less than or equal to 300 is 20 , so to remove the multiples of 15 , we must remove 40 from 160 to get 120 positive integers less than or equal to 300 which are multiples of 3 or 5 but not of 15 .\n\n\n\nThis total still included some multiples of 10 that are less or equal to 300 (but not all, since we have already removed 30 , for instance).\n\nIn fact, there are 30 multiples of 10 less than or equal 300,10 of which are multiples of 15 as well (that is, the multiples of 30). So we must remove 20 from the total of 120. We then obtain that there are 100 positive integers less than or equal to 300 which are multiples of 3 or 5 , but not of 10 or 15 ." ]

[ "100" ]

[ "Since the signs alternate every three terms, it makes sense to look at the terms in groups of 6 .\n\nThe sum of the first 6 terms is $1+3+5-7-9-11=-18$.\n\nThe sum of the next 6 terms is $13+15+17-19-21-23=-18$.\n\nIn fact, the sum of each group of 6 terms will be the same, since in each group, 12 has been added to the numerical value of each term when compared to the previous group of 6 , so overall 12 has been added three times and subtracted three times.\n\nSince we are looking for the sum of the first 300 terms, then we are looking at 50 groups of 6 terms, so the sum must be $50(-18)=-900$." ]

[ "-900" ]

[ "Let the two digit integer have tens digit $a$ and units digit $b$. Then the given information tells us\n\n$$\n\\begin{aligned}\na^{2}+10 b & =b^{2}+10 a \\\\\na^{2}-b^{2}-10 a+10 b & =0 \\\\\n(a+b)(a-b)-10(a-b) & =0 \\\\\n(a-b)(a+b-10) & =0\n\\end{aligned}\n$$\n\nand so $a=b$ or $a+b=10$.\n\nSo the possibilities for the integer are 11, 22, 33, 44, 55, 66, 77, 88, 99, 19, 28, 37, 46, 55, $64,73,82,91$. We now must determine which integers in this list are prime.\n\nWe can quickly reject all multiples of 11 bigger than 11 and all of the even integers, to reduce the list to $11,19,37,73,91$.\n\nAll of these are prime, except for $91=13 \\times 7$.\n\nTherefore, the required integers are 11, 19, 37, and 73 ." ]

[ "11,19,37,73" ]

[ "In 24 minutes, the number of atoms of isotope $\\mathrm{A}$ has halved 4 times, so the initial number of atoms is $2^{4}=16$ times the number of atoms of isotope $\\mathrm{A}$ at time 24 minutes.\n\nBut there were initially half as many atoms of isotope B as of isotope B, so there was 8 times the final number of atoms. Therefore, the number of atoms of isotope B halves 3 times in the 24 minutes, so it takes 8 minutes for the number of atoms of isotope B to halve.", "Initially, there is twice as many atoms of isotope A as of isotope B, so let the original numbers of atoms of each be $2 x$ and $x$, respectively.\n\nConsidering isotope A, after 24 minutes, if it loses half of its atoms every 6 minutes, there will be $2 x\\left(\\frac{1}{2}\\right)^{\\frac{24}{6}}$ atoms remaining.\n\nSimilarly for isotope B, after 24 minutes, there will be $x\\left(\\frac{1}{2}\\right)^{\\frac{24}{T}}$ atoms remaining, where $T$ is the length of time (in minutes) that it takes for the number of atoms to halve.\n\nFrom the given information,\n\n$$\n\\begin{aligned}\n2 x\\left(\\frac{1}{2}\\right)^{\\frac{24}{6}} & =x\\left(\\frac{1}{2}\\right)^{\\frac{24}{T}} \\\\\n2\\left(\\frac{1}{2}\\right)^{4} & =\\left(\\frac{1}{2}\\right)^{\\frac{24}{T}} \\\\\n\\left(\\frac{1}{2}\\right)^{3} & =\\left(\\frac{1}{2}\\right)^{\\frac{24}{T}} \\\\\n\\frac{24}{T} & =3 \\\\\nT & =8\n\\end{aligned}\n$$\n\nTherefore, it takes 8 minutes for the number of atoms of isotope B to halve." ]

[ "8" ]

[ "Using the facts that $\\log _{10} A+\\log _{10} B=\\log _{10} A B$ and that $\\log _{10} A-\\log _{10} B=\\log _{10} \\frac{A}{B}$, then we can convert the two equations to\n\n$$\n\\begin{aligned}\n\\log _{10}\\left(x^{3} y^{2}\\right) & =11 \\\\\n\\log _{10}\\left(\\frac{x^{2}}{y^{3}}\\right) & =3\n\\end{aligned}\n$$\n\nRaising both sides to the power of 10 , we obtain\n\n$$\n\\begin{aligned}\nx^{3} y^{2} & =10^{11} \\\\\n\\frac{x^{2}}{y^{3}} & =10^{3}\n\\end{aligned}\n$$\n\nTo eliminate the $y$ 's, we raise the first equation to the power 3 and the second to the power 2 to obtain\n\n$$\n\\begin{aligned}\nx^{9} y^{6} & =10^{33} \\\\\n\\frac{x^{4}}{y^{6}} & =10^{6}\n\\end{aligned}\n$$\n\nand multiply to obtain $x^{9} x^{4}=x^{13}=10^{39}=10^{33} 10^{6}$.\n\nTherefore, since $x^{13}=10^{39}$, then $x=10^{3}$.\n\n\n\nSubstituting back into $x^{3} y^{2}=10^{11}$, we get $y^{2}=10^{2}$, and so $y= \\pm 10$. However, substituting into $\\frac{x^{2}}{y^{3}}=10^{3}$ we see that $y$ must be positive, so $y=10$.\n\nTherefore, the solution to the system of equation is $x=10^{3}$ and $y=10$.", "Since the domain of the logarithm is the positive real numbers, then the quantities $\\log _{10}\\left(x^{3}\\right)$ and $\\log _{10}\\left(y^{3}\\right)$ tell us that $x$ and $y$ are positive.\n\nUsing the fact that $\\log _{10}\\left(a^{b}\\right)=b \\log _{10}(a)$, we rewrite the equations as\n\n$$\n\\begin{aligned}\n& 3 \\log _{10} x+2 \\log _{10} y=11 \\\\\n& 2 \\log _{10} x-3 \\log _{10} y=3\n\\end{aligned}\n$$\n\nWe solve the system of equations for $\\log _{10} x$ and $\\log _{10} y$ by multiplying the first equation by 3 and adding two times the second equation in order to eliminate $\\log _{10} y$. Thus we obtain $13 \\log _{10} x=39$ or $\\log _{10} x=3$.\n\nSubstituting back into the first equation, we obtain $\\log _{10} y=1$.\n\nTherefore, $x=10^{3}$ and $y=10$." ]

[ "$10^{3},10$" ]

[ "First, we prove lemma (b): if $n$ is an even savage integer, then $\\frac{n+4}{12}$ is an integer.\n\nProof of lemma (b):\nWe use the strategy of putting all of the multiples of 3 between 1 and $n$ in the set $C$, all of the remaining even numbers in the set $B$, and all of the remaining numbers in the set $A$. The sums of these sets will not likely all be equal, but we then try to adjust the sums to by moving elements out of $A$ and $B$ into $C$ to try to make these sums equal. (Notice that we can't move elements either into $A$ or $B$, or out of $C$.) We will use the notation $|C|$ to denote the sum of the elements of $C$.\n\nSince we are considering the case of $n$ even and we want to examine multiples of 3 less than or equal to $n$, it makes sense to consider $n$ as having one of the three forms $6 k$, $6 k+2$ or $6 k+4$. (These forms allow us to quickly tell what the greatest multiple of 3 less than $n$ is.)\n\nCase 1: $n=6 k$\n\nIn this case, $C$ contains at least the integers $3,6,9, \\ldots, 6 k$, and so the sum of $C$ is greater than one-third of the sum of the integers from 1 to $n$, since if we divide the integers from 1 to $n=6 k$ into groups of 3 consecutive integers starting with 1,2, 3 , then the set $C$ will always contain the largest of the 3 .\n\n\n\nCase 2: $n=6 k+4$\n\nHere, the sum of the integers from 1 to $n=6 k+4$ is $\\frac{1}{2}(6 k+4)(6 k+5)=18 k^{2}+27 k+10=3\\left(6 k^{2}+9 k+3\\right)+1$, which is never divisible by 3 . Therefore, $n$ cannot be savage in this case because the integers from 1 to $n$ cannot be partitioned into 3 sets with equal sums.\n\nCase 3: $n=6 k+2$\n\nHere, the sum of the integers from 1 to $n=6 k+2$ is\n\n$\\frac{1}{2}(6 k+2)(6 k+3)=18 k^{2}+15 k+3$, so the sum of the elements of each of the sets $A, B$ and $C$ should be $6 k^{2}+5 k+1$, so that the sums are equal.\n\nIn this case $C$, contains at least the integers $3,6,9, \\ldots, 6 k$, and so $|C| \\geq 3+6+9+\\cdots 6 k=3(1+2+3+\\cdots+2 k)=3\\left(\\frac{1}{2}(2 k)(2 k+1)\\right)=6 k^{2}+3 k$\n\nThe set $A$ contains at most the integers $1,3,5,7, \\ldots, 6 k+1$, but does not contain the odd multiples of 3 less than $n$, ie. the integers $3,9,15, \\ldots, 6 k-3$. Therefore, $|A| \\leq(1+3+5+\\cdots+6 k+1)-(3+9+\\cdots+6 k-3)$\n\n$=\\frac{1}{2}(3 k+1)[1+6 k+1]-\\frac{1}{2}(k)[3+6 k-3]$\n\n$=(3 k+1)(3 k+1)-k(3 k)$\n\n$=6 k^{2}+6 k+1$\n\n(To compute the sum of each of these arithmetic sequences, we use the fact that the sum of an arithmetic sequence is equal to half of the number of terms times the sum of the first and last terms.)\n\nThe set $B$ contains at most the integers $2,4,6,8, \\ldots, 6 k+2$, but does not contain the even multiples of 3 less than $n$, ie. the integers $6,12, \\ldots, 6 k$. Therefore, $|B| \\leq(2+4+6+\\cdots+6 k+2)-(6+12+\\cdots+6 k)$\n\n$=\\frac{1}{2}(3 k+1)[2+6 k+2]-\\frac{1}{2}(k)[6+6 k]$\n\n$=(3 k+1)(3 k+2)-k(3 k+3)$\n\n$=6 k^{2}+6 k+2$\n\nThus, the set $C$ is $2 k+1$ short of the desired sum, while the set $A$ has a sum that is $k$ too big and the set $B$ has a sum that is $k+1$ too big.\n\nSo in order to correct this, we would like to move elements from $A$ adding to $k$, and elements from $B$ which add to $k+1$ all to set $C$.\n\n\n\nSince we are assuming that $n$ is savage, then this is possible, which means that $k+1$ must be even since every element in $B$ is even, so the sum of any number of elements of $B$ is even.\n\nTherefore, $k$ is odd, and so $k=2 l+1$ for some integer $l$, and so\n\n$n=6(2 l+1)+2=12 l+8$, ie. $\\frac{n+4}{12}$ is an integer.\n\nHaving examined all cases, we see that if $n$ is an even savage integer, then $\\frac{n+4}{12}$ is an integer.\n\n\nFrom the proof of (b) above, the only possible even savage integers less than 100 are those satisfying the condition that $\\frac{n+4}{12}$ is an integer, ie. $8,20,32,44,56,68,80,92$. We already know that 8 is savage, so we examine the remaining 7 possibilities.\n\nWe make a table of the possibilities, using the notation from the proof of (b):\n\n| $n$ | $k$ | Sum of elements <br> to remove from $A$ | Sum of elements <br> to remove from $B$ | Possible? |\n| :---: | :---: | :---: | :---: | :---: |\n| 20 | 3 | 3 | 4 | No - cannot remove a sum of 3 from <br> A. |\n| 32 | 5 | 5 | 6 | Yes - remove 5 from $A, 2$ and 4 <br> from $B$ |\n| 44 | 7 | 7 | 8 | Yes - remove 7 from $A, 8$ from $B$ |\n| 56 | 9 | 9 | 10 | No - cannot remove a sum of 9 from <br> A. |\n| 68 | 11 | 11 | 12 | Yes - remove 11 from $A, 4$ and 8 <br> from $B$ |\n| 80 | 13 | 13 | 14 | Yes - remove 13 from $A, 14$ from $B$ |\n| 92 | 15 | 15 | 16 | No - cannot remove a sum of 15 <br> from $A$ (since could only use $1,5,7$, <br> 11,13 ) |\n\nTherefore, the only even savage integers less than 100 are 8, 32, 44, 68 and 80." ]

[ "8,32,44,68,80" ]

[ "We make a table of the 36 possible combinations of rolls and the resulting sums:\n\n| | 2 | 3 | 5 | 7 | 11 | 13 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 2 | 4 | 5 | 7 | 9 | 13 | 15 |\n| 3 | 5 | 6 | 8 | 10 | 14 | 16 |\n| 5 | 7 | 8 | 10 | 12 | 16 | 18 |\n| 7 | 9 | 10 | 12 | 14 | 18 | 20 |\n| 11 | 13 | 14 | 16 | 18 | 22 | 24 |\n| 13 | 15 | 16 | 18 | 20 | 24 | 26 |\n\nOf the 36 entries in the table, 6 are prime numbers (two entries each of 5, 7 and 13).\n\nTherefore, the probability that the sum is a prime number is $\\frac{6}{36}$ or $\\frac{1}{6}$.\n\n(Note that each sum is at least 4 and so must be odd to be prime. Since odd plus odd equals even, then the only possibilities that really need to be checked are even plus odd and odd plus even (that is, the first row and first column of the table).)" ]

[ "$\\frac{1}{6}$" ]

[ "Beginning with the given equation, we have\n\n$$\n\\begin{aligned}\n\\frac{1}{\\cos x}-\\tan x & =3 \\\\\n\\frac{1}{\\cos x}-\\frac{\\sin x}{\\cos x} & =3 \\\\\n1-\\sin x & =3 \\cos x \\quad(\\text { since } \\cos x \\neq 0) \\\\\n(1-\\sin x)^{2} & =9 \\cos ^{2} x \\quad \\text { (squaring both sides) } \\\\\n1-2 \\sin x+\\sin ^{2} x & =9\\left(1-\\sin ^{2} x\\right) \\\\\n10 \\sin ^{2} x-2 \\sin x-8 & =0 \\\\\n5 \\sin ^{2} x-\\sin x-4 & =0 \\\\\n(5 \\sin x+4)(\\sin x-1) & =0\n\\end{aligned}\n$$\n\nTherefore, $\\sin x=-\\frac{4}{5}$ or $\\sin x=1$.\n\nIf $\\sin x=1$, then $\\cos x=0$ and $\\tan x$ is undefined, which is inadmissible in the original equation.\n\nTherefore, $\\sin x=-\\frac{4}{5}$.\n\n(We can check that if $\\sin x=-\\frac{4}{5}$, then $\\cos x= \\pm \\frac{3}{5}$ and the possibility that $\\cos x=\\frac{3}{5}$ satisfies the original equation, since in this case $\\frac{1}{\\cos x}=\\frac{5}{3}$ and $\\tan x=-\\frac{4}{3}$ and the difference between these fractions is 3 .)" ]

[ "$-\\frac{4}{5}$" ]

[ "Since $f(x)=a x+b$, we can determine an expression for $g(x)=f^{-1}(x)$ by letting $y=f(x)$ to obtain $y=a x+b$. We then interchange $x$ and $y$ to obtain $x=a y+b$ which we solve for $y$ to obtain $a y=x-b$ or $y=\\frac{x}{a}-\\frac{b}{a}$.\n\nTherefore, $f^{-1}(x)=\\frac{x}{a}-\\frac{b}{a}$.\n\nNote that $a \\neq 0$. (This makes sense since the function $f(x)=b$ has a graph which is a horizontal line, and so cannot be invertible.)\n\nTherefore, the equation $f(x)-g(x)=44$ becomes $(a x+b)-\\left(\\frac{x}{a}-\\frac{b}{a}\\right)=44$ or $\\left(a-\\frac{1}{a}\\right) x+\\left(b+\\frac{b}{a}\\right)=44=0 x+44$, and this equation is true for all $x$.\n\nWe can proceed in two ways.\n\nMethod \\#1: Comparing coefficients\n\nSince the equation\n\n$$\n\\left(a-\\frac{1}{a}\\right) x+\\left(b+\\frac{b}{a}\\right)=0 x+44\n$$\n\nis true for all $x$, then the coefficients of the linear expression on the left side must match the coefficients of the linear expression on the right side.\n\nTherefore, $a-\\frac{1}{a}=0$ and $b+\\frac{b}{a}=44$.\n\nFrom the first of these equations, we obtain $a=\\frac{1}{a}$ or $a^{2}=1$, which gives $a=1$ or $a=-1$. If $a=1$, the equation $b+\\frac{b}{a}=44$ becomes $b+b=44$, which gives $b=22$.\n\n\n\nIf $a=-1$, the equation $b+\\frac{b}{a}=44$ becomes $b-b=44$, which is not possible.\n\nTherefore, we must have $a=1$ and $b=22$, and so $f(x)=x+22$.\n\nMethod \\#2: Trying specific values for $x$\n\nSince the equation\n\n$$\n\\left(a-\\frac{1}{a}\\right) x+\\left(b+\\frac{b}{a}\\right)=0 x+44\n$$\n\nis true for all values of $x$, then it must be true for any specific values of $x$ that we choose.\n\nChoosing $x=0$, we obtain $0+\\left(b+\\frac{b}{a}\\right)=44$ or $b+\\frac{b}{a}=44$.\n\nChoosing $x=b$, we obtain $\\left(a-\\frac{1}{a}\\right) b+\\left(b+\\frac{b}{a}\\right)=44$ or $a b+b=44$.\n\nWe can rearrange the first of these equations to get $\\frac{a b+b}{a}=44$.\n\nUsing the second equation, we obtain $\\frac{44}{a}=44$ or $a=1$.\n\nSince $a=1$, then $a b+b=44$ gives $2 b=44$ or $b=22$.\n\nThus, $f(x)=x+22$.\n\nIn summary, the only linear function $f$ for which the given equation is true for all $x$ is $f(x)=x+22$." ]

[ "$f(x)=x+22$" ]

[ "First, we factor the left side of the given equation to obtain $a\\left(a^{2}+2 b\\right)=2013$.\n\nNext, we factor the integer 2013 as $2013=3 \\times 671=3 \\times 11 \\times 61$. Note that each of 3,11 and 61 is prime, so we can factor 2013 no further. (We can find the factors of 3 and 11 using tests for divisibility by 3 and 11, or by systematic trial and error.)\n\nSince $2013=3 \\times 11 \\times 61$, then the positive divisors of 2013 are\n\n$$\n1,3,11,33,61,183,671,2013\n$$\n\nSince $a$ and $b$ are positive integers, then $a$ and $a^{2}+2 b$ are both positive integers.\n\nSince $a$ and $b$ are positive integers, then $a^{2} \\geq a$ and $2 b>0$, so $a^{2}+2 b>a$.\n\nSince $a\\left(a^{2}+2 b\\right)=2013$, then $a$ and $a^{2}+2 b$ must be a divisor pair of 2013 (that is, a pair of positive integers whose product is 2013) with $a<a^{2}+2 b$.\n\nWe make a table of the possibilities:\n\n| $a$ | $a^{2}+2 b$ | $2 b$ | $b$ |\n| :---: | :---: | :---: | :---: |\n| 1 | 2013 | 2012 | 1006 |\n| 3 | 671 | 662 | 331 |\n| 11 | 183 | 62 | 31 |\n| 33 | 61 | -1028 | N/A |\n\nNote that the last case is not possible, since $b$ must be positive.\n\nTherefore, the three pairs of positive integers that satisfy the equation are $(1,1006)$, $(3,331),(11,31)$.\n\n(We can verify by substitution that each is a solution of the original equation.)" ]

[ "$(1,1006),(3,331),(11,31)$" ]

[ "We successively manipulate the given equation to produce equivalent equations:\n\n$$\n\\begin{aligned}\n\\log _{2}\\left(2^{x-1}+3^{x+1}\\right) & =2 x-\\log _{2}\\left(3^{x}\\right) \\\\\n\\log _{2}\\left(2^{x-1}+3^{x+1}\\right)+\\log _{2}\\left(3^{x}\\right) & =2 x \\\\\n\\log _{2}\\left(\\left(2^{x-1}+3^{x+1}\\right) 3^{x}\\right) & =2 x \\quad\\left(\\text { using } \\log _{2} A+\\log _{2} B=\\log _{2} A B\\right) \\\\\n\\left(2^{x-1}+3^{x+1}\\right) 3^{x} & =2^{2 x} \\quad \\text { (exponentiating both sides) } \\\\\n2^{-1} 2^{x} 3^{x}+3^{1} 3^{x} 3^{x} & =2^{2 x} \\\\\n\\frac{1}{2} \\cdot 2^{x} 3^{x}+3 \\cdot 3^{2 x} & =2^{2 x} \\\\\n2^{x} 3^{x}+6 \\cdot 3^{2 x} & \\left.=2 \\cdot 2^{2 x} \\quad \\text { (multiplying by } 2\\right) \\\\\n2^{x} 3^{x}+6 \\cdot\\left(3^{x}\\right)^{2} & =2 \\cdot\\left(2^{x}\\right)^{2}\n\\end{aligned}\n$$\n\nNext, we make the substitution $a=2^{x}$ and $b=3^{x}$.\n\nThis gives $a b+6 b^{2}=2 a^{2}$ or $2 a^{2}-a b-6 b^{2}=0$.\n\nFactoring, we obtain $(a-2 b)(2 a+3 b)=0$.\n\nTherefore, $a=2 b$ or $2 a=-3 b$.\n\nSince $a>0$ and $b>0$, then $a=2 b$ which gives $2^{x}=2 \\cdot 3^{x}$.\n\nTaking $\\log$ of both sides, we obtain $x \\log 2=\\log 2+x \\log 3$ and so $x(\\log 2-\\log 3)=\\log 2$ or $x=\\frac{\\log 2}{\\log 2-\\log 3}$.", "We successively manipulate the given equation to produce equivalent equations:\n\n$$\n\\begin{aligned}\n\\log _{2}\\left(2^{x-1}+3^{x+1}\\right) & =2 x-\\log _{2}\\left(3^{x}\\right) \\\\\n\\log _{2}\\left(2^{x-1}+3^{x+1}\\right)+\\log _{2}\\left(3^{x}\\right) & =2 x \\\\\n\\log _{2}\\left(\\left(2^{x-1}+3^{x+1}\\right) 3^{x}\\right) & =2 x \\quad\\left(\\text { using } \\log _{2} A+\\log _{2} B=\\log _{2} A B\\right) \\\\\n\\left(2^{x-1}+3^{x+1}\\right) 3^{x} & =2^{2 x} \\quad \\text { (exponentiating both sides) } \\\\\n2^{-1} 2^{x} 3^{x}+3^{1} 3^{x} 3^{x} & =2^{2 x} \\\\\n\\frac{1}{2} \\cdot 2^{x} 3^{x}+3 \\cdot 3^{2 x} & =2^{2 x} \\\\\n2^{x} 3^{x}+6 \\cdot 3^{2 x} & \\left.=2 \\cdot 2^{2 x} \\quad \\text { (multiplying by } 2\\right) \\\\\n2^{x} 3^{x} 2^{-2 x}+6 \\cdot 3^{2 x} 2^{-2 x} & \\left.=2 \\quad \\text { (dividing both sides by } 2^{2 x} \\neq 0\\right) \\\\\n2^{-x} 3^{x}+6 \\cdot 3^{2 x} 2^{-2 x} & =2 \\\\\n\\left(\\frac{3}{2}\\right)^{x}+6\\left(\\frac{3}{2}\\right)^{2 x} & =2\n\\end{aligned}\n$$\n\nNext, we make the substitution $t=\\left(\\frac{3}{2}\\right)^{x}$, noting that $\\left(\\frac{3}{2}\\right)^{2 x}=\\left(\\left(\\frac{3}{2}\\right)^{x}\\right)^{2}=t^{2}$.\n\nThus, we obtain the equivalent equations\n\n$$\n\\begin{aligned}\nt+6 t^{2} & =2 \\\\\n6 t^{2}+t-2 & =0 \\\\\n(3 t+2)(2 t-1) & =0\n\\end{aligned}\n$$\n\nTherefore, $t=-\\frac{2}{3}$ or $t=\\frac{1}{2}$.\n\nSince $t=\\left(\\frac{3}{2}\\right)^{x}>0$, then we must have $t=\\left(\\frac{3}{2}\\right)^{x}=\\frac{1}{2}$.\n\nThus,\n\n$$\nx=\\log _{3 / 2}(1 / 2)=\\frac{\\log (1 / 2)}{\\log (3 / 2)}=\\frac{\\log 1-\\log 2}{\\log 3-\\log 2}=\\frac{-\\log 2}{\\log 3-\\log 2}=\\frac{\\log 2}{\\log 2-\\log 3}\n$$" ]

[ "$\\frac{\\log 2}{\\log 2-\\log 3}$" ]

[ "We determine the multiplicative partitions of 64 by considering the number of parts in the various partitions. Note that 64 is a power of 2 so any divisor of 64 is also a power of 2 . In each partition, since the order of parts is not important, we list the parts in increasing order to make it easier to systematically find all of these.\n\n* One part. There is one possibility: 64.\n* Two parts. There are three possibilities: $64=2 \\times 32=4 \\times 16=8 \\times 8$.\n* Three parts. We start with the smallest possible first and second parts. We keep the first part fixed while adjusting the second and third parts. We then increase the first part and repeat.\n\nWe get: $64=2 \\times 2 \\times 16=2 \\times 4 \\times 8=4 \\times 4 \\times 4$.\n\n* Four parts. A partition of 64 with four parts must include at least two $2 \\mathrm{~s}$, since if it didn't, it would include at least three parts that are at least 4 , and so would be too large. With two $2 \\mathrm{~s}$, the remaining two parts have a product of 16 .\n\nWe get: $64=2 \\times 2 \\times 2 \\times 8=2 \\times 2 \\times 4 \\times 4$.\n\n* Five parts. A partition of 64 with five parts must include at least three $2 \\mathrm{~s}$, since if it didn't, it would include at least three parts that are at least 4 , and so would be too large. With three $2 \\mathrm{~s}$, the remaining two parts have a product of 8 .\n\nWe get: $64=2 \\times 2 \\times 2 \\times 2 \\times 4$.\n\n$*$ Six parts. Since $64=2^{6}$, there is only one possibility: $64=2 \\times 2 \\times 2 \\times 2 \\times 2 \\times 2$.\n\nTherefore, $P(64)=1+3+3+2+1+1=11$." ]

[ "11" ]

[ "First, we note that $1000=10^{3}=(2 \\cdot 5)^{3}=2^{3} 5^{3}$.\n\nWe calculate the value of $P\\left(p^{3} q^{3}\\right)$ for two distinct prime numbers $p$ and $q$. It will turn out that this value does not depend on $p$ and $q$. This value will be the value of $P(1000)$, since 1000 has this form of prime factorization.\n\nLet $n=p^{3} q^{3}$ for distinct prime numbers $p$ and $q$.\n\nThe integer $n$ has three prime factors equal to $p$.\n\nIn a given partition, these can be all together in one part (as $p^{3}$ ), can be split between two different parts (as $p$ and $p^{2}$ ), or can be split between three different parts (as $p, p$ and $p)$. There are no other ways to divide up three divisors of $p$.\n\nSimilarly, $n$ has three prime factors equal to $q$ which can be divided in similar ways.\n\nWe determine $P\\left(p^{3} q^{3}\\right)$ by considering the possible combination of the number of parts divisible by $p$ and the number of parts divisible by $q$ and counting partitions in each case. In other words, we complete the following table:\n\n<img_3502>\n\nWe note that the table is symmetric, since the factors of $p$ and $q$ are interchangeable.\n\nWe proceed to consider cases, considering only those on the top left to bottom right diagonal and and those below this diagonal in the table.\n\n\n\nCase 1: One part divisible by $p$, one part divisible by $q$\n\nThe partition must be $p^{3} q^{3}$ ( $n$ itself) or $p^{3} \\times q^{3}$.\n\nThere are two partitions in this case.\n\nCase 2: One part divisible by $p$, two parts divisible by $q$\n\nThe three factors of $p$ occur together as $p^{3}$. The three factors of $q$ occur as $q$ and $q^{2}$.\n\nThe $p^{3}$ can occur in one of the parts divisible by $q$ or not.\n\nThis gives partitions $p^{3} \\times q \\times q^{2}$ and $p^{3} q \\times q^{2}$ and $q \\times p^{3} q^{2}$.\n\nThere are three partitions in this case. Similarly, there are three partitions with one part divisible by $q$ and two parts divisible by $p$.\n\nCase 3: One part divisible by $p$, three parts divisible by $q$\n\nThe three factors of $p$ occur together as $p^{3}$. The three factors of $q$ occur as $q, q$ and $q$.\n\nThe $p^{3}$ can occur in one of the parts divisible by $q$ or not.\n\nThis gives partitions $p^{3} \\times q \\times q \\times q$ and $p^{3} q \\times q \\times q$.\n\n(Note that the three divisors of $q$ are interchangeable so $p^{3}$ only needs to be placed with one of them.)\n\nThere are two partitions in this case. Similarly, there are two partitions with one part divisible by $q$ and three parts divisible by $p$.\n\nCase 4: Two parts divisible by $p$, two parts divisible by $q$\n\nThe three factors of $p$ occur as $p$ and $p^{2}$. The three factors of $q$ occur as $q$ and $q^{2}$.\n\nEach of $p$ and $p^{2}$ can occur in one of the parts divisible by $q$ or not.\n\nIf no part is a multiple of both $p$ and $q$, we have one partition: $p \\times p^{2} \\times q \\times q^{2}$.\n\nIf one part is a multiple of both $p$ and $q$, there are two choices for which power of $p$ to include in this part and two choices for which power of $q$ to include. (There is no choice for the remaining parts.) Thus, there are $2 \\times 2=4$ such partitions:\n\n$$\np^{2} q^{2} \\times p \\times q \\quad p q^{2} \\times p^{2} \\times q \\quad p^{2} q \\times p \\times q^{2} \\quad p q \\times p^{2} \\times q^{2}\n$$\n\nIf two parts are a multiple of both $p$ and $q$, there are two ways to choose the power of $p$ in the part containing just $q$, so there are two such partitions: $p q \\times p^{2} q^{2}$ and $p^{2} q \\times p q^{2}$. There are seven partitions in this case.\n\nCase 5: Two parts divisible by $p$, three parts divisible by $q$\n\nThe three factors of $p$ occur as $p$ and $p^{2}$. The three factors of $q$ occur as $q, q$ and $q$.\n\nEach of $p$ and $p^{2}$ can occur in one of the parts divisible by $q$ or not.\n\nIf no part is a multiple of both $p$ and $q$, we have one partition: $p \\times p^{2} \\times q \\times q \\times q$.\n\nIf one part is a multiple of both $p$ and $q$, there are two choices for which power of $p$ to include in this part (since all powers of $q$ are identical).\n\nThus, there are 2 such partitions: $p^{2} q \\times p \\times q \\times q$ and $p q \\times p^{2} \\times q \\times q$.\n\nIf two parts are a multiple of both $p$ and $q$, there is one partition, since all of the powers of $q$ are identical: $p q \\times p^{2} q \\times q$.\n\nThere are four partitions in this case. Similarly, there are four partitions with two parts divisible by $q$ and three parts divisible by $p$.\n\nCase 6: Three parts divisible by $p$, three parts divisible by $q$\n\nThe three factors of $p$ as $p, p$ and $p$. The three factors of $q$ appear as $q, q$ and $q$.\n\nHere, the number of parts in the partition that are multiples of both $p$ and $q$ can be 0 , 1,2 or 3 . Since all of the powers of $p$ and $q$ are identical, the partitions are completely determined by this and are\n\n$$\np \\times p \\times p \\times q \\times q \\times q \\quad p \\times p \\times p q \\times q \\times q \\quad p \\times p q \\times p q \\times q \\quad p q \\times p q \\times p q\n$$\n\nThere are four partitions in this case.\n\n\n\nFinally, we complete the table:\n\nNumber of parts divisible by $p$ (Column)\n\nNumber of parts divisible by $q$ (Row)\n\n| | 1 | 2 | 3 |\n| :--- | :--- | :--- | :--- |\n| 1 | 2 | 3 | 2 |\n| 2 | 3 | 7 | 4 |\n| 3 | 2 | 4 | 4 |\n\nAdding the entries in the table, we obtain $P\\left(p^{3} q^{3}\\right)=31$.\n\nThus, $P(1000)=31$." ]

[ "31" ]

[ "Combining the logarithms,\n\n$$\n\\begin{aligned}\n\\log _{5}(x+3)+\\log _{5}(x-1) & =1 \\\\\n\\log _{5}((x+3)(x-1)) & =1 \\\\\n\\log _{5}\\left(x^{2}+2 x-3\\right) & =1 \\\\\nx^{2}+2 x-3 & =5 \\\\\nx^{2}+2 x-8 & =0 \\\\\n(x+4)(x-2) & =0\n\\end{aligned}\n$$\n\nTherefore, $x=-4$ or $x=2$. Substituting the two values for $x$ back into the original equation, we see that $x=2$ works, but that $x=-4$ does not, since we cannot take the logarithm of a negative number." ]

[ "2" ]

[ "From the table we have two pieces of information, so we substitute both of these into the given formula.\n\n$$\n\\begin{aligned}\n& 2.75=a(3.00)^{b} \\\\\n& 3.75=a(6.00)^{b}\n\\end{aligned}\n$$\n\nWe can now proceed in either of two ways to solve for $b$.\n\nMethod 1 to find $b$\n\nDividing the second equation by the first, we obtain\n\n$$\n\\frac{3.75}{2.75}=\\frac{a(6.00)^{b}}{a(3.00)^{b}}=\\frac{(6.00)^{b}}{(3.00)^{b}}=\\left(\\frac{6.00}{3.00}\\right)^{b}=2^{b}\n$$\n\nor\n\n$$\n2^{b} \\approx 1.363636\n$$\n\nTaking logarithms of both sides,\n\n\n\n$$\n\\begin{aligned}\n\\log \\left(2^{b}\\right) & \\approx \\log (1.363636) \\\\\nb \\log (2) & \\approx \\log (1.363636) \\\\\nb & \\approx \\frac{\\log (1.363636)}{\\log (2)} \\\\\nb & \\approx 0.4475\n\\end{aligned}\n$$\n\nMethod 2 to find $b$ \n\nTaking logarithms of both sides of the above equations, we obtain\n\n$$\n\\begin{aligned}\n\\log (2.75) & =\\log \\left(a(3.00)^{b}\\right) \\\\\n& =\\log (a)+\\log \\left((3.00)^{b}\\right) \\\\\n& =\\log (a)+b \\log (3.00)\n\\end{aligned}\n$$\n\nSimilarly,\n\n$$\n\\log (3.75)=\\log (a)+b \\log (6.00)\n$$\n\nSubtracting the first equation from the second, we obtain\n\n$$\n\\begin{aligned}\n\\log (3.75)-\\log (2.75) & =b(\\log (6.00)-\\log (3.00)) \\\\\nb & =\\frac{\\log (3.75)-\\log (2.75)}{\\log (6.00)-\\log (3.00)} \\\\\nb & \\approx 0.4475\n\\end{aligned}\n$$\n\nWe now continue in the same way for both methods.\n\nSubstituting this value for $b$ back into the first equation above,\n\n$$\n\\begin{aligned}\n2.75 & \\approx a(3.00)^{0.4475} \\\\\na & \\approx \\frac{2.75}{(3.00)^{0.4475}} \\\\\na & \\approx 1.6820\n\\end{aligned}\n$$\n\nTherefore, to two decimal places, $a=1.68$ and $b=0.45$." ]

[ "$1.68,0.45$" ]

[ "We first determine the three points through which the circle passes.\n\nThe first point is the origin $(0,0)$.\n\nThe second and third points are found by determining the points of intersection of the two parabolas $y=x^{2}-3$ and $y=-x^{2}-2 x+9$. We do this by setting the $y$ values equal.\n\n$$\nx^{2}-3=-x^{2}-2 x+9\n$$\n\n$2 x^{2}+2 x-12=0$\n\n$x^{2}+x-6=0$\n\n$(x+3)(x-2)=0$\n\nso $x=-3$ or $x=2$.\n\n\n\nWe determine the points of intersection by substituting into the first parabola.\n\nIf $x=2, y=2^{2}-3=1$, so the point of intersection is $(2,1)$.\n\nIf $x=-3, y=(-3)^{2}-3=6$, so the point of intersection is $(-3,6)$.\n\nTherefore, the circle passes through the three points $A(0,0), B(2,1)$ and $C(-3,6)$.\n\nLet the centre of the circle be the point $Q(a, b)$.\n\n<img_4046>\n\nFinding the centre of the circle can be done in a variety of ways.\n\nWe use the fact $Q$ is of equal distance from each of the points $A, B$ and $C$. In particular $Q A^{2}=Q B^{2}=Q C^{2}$ or $x^{2}+y^{2}=(x-2)^{2}+(y-1)^{2}=(x+3)^{2}+(y-6)^{2}$\n\nFrom the first equality,\n\n$$\n\\begin{aligned}\n& x^{2}+y^{2}=(x-2)^{2}+(y-1)^{2} \\\\\n& 4 x+2 y=5\n\\end{aligned}\n$$\n\n<img_4000>\n\n\n\nFrom the second equality,\n\n$$\n\\begin{aligned}\n(x-2)^{2}+(y-1)^{2} & =(x+3)^{2}+(y-6)^{2} \\\\\n-10 x+10 y & =40 \\\\\ny & =x+4\n\\end{aligned}\n$$\n\nSubstituting the equation above into into $4 x+2 y=5$, we obtain $4 x+2(x+4)=5$ or $6 x=-3$ or $x=-\\frac{1}{2}$. Thus, $y=-\\frac{1}{2}+4=\\frac{7}{2}$, and so the centre of the circle is $\\left(-\\frac{1}{2}, \\frac{7}{2}\\right)$." ]

[ "$(-\\frac{1}{2}, \\frac{7}{2})$" ]

[ "In total, there are $\\frac{1}{2} \\times 5 \\times 20=50$ games played, since each of 5 teams plays 20 games (we divide by 2 since each game is double-counted).\n\nIn each game, there is either a loss or a tie.\n\nThe number of games with a loss is $44+y$ from the second column, and the number of games with a tie is $\\frac{1}{2}(11+z)$ (since any game ending in a tie has 2 ties).\n\n\n\nSo\n\n$$\n\\begin{aligned}\n50 & =44+y+\\frac{1}{2}(11+z) \\\\\n100 & =88+2 y+11+z \\\\\n1 & =2 y+z\n\\end{aligned}\n$$\n\nSince $y$ and $z$ are non-negative integers, $z=1$ and $y=0$. So $x=19$ since Team E plays 20 games.", "In any game played, the final result is either both teams earning a tie, or one team earning a win, and the other getting a loss. Therefore, the total number of wins among all teams equals the total number of losses, ie.\n\n$$\n\\begin{aligned}\n25+x & =44+y \\\\\nx-y & =19\n\\end{aligned}\n$$\n\nAlso, since team E plays 20 games, then\n\n$$\nx+y+z=20\n$$\n\nSo from (1), $x$ must be at least 19, and from (2), $x$ can be at most 20.\n\nLastly, we know that the total of all of the teams numbers of ties must be even, ie. $11+z$ is even, ie. $z$ is odd.\n\nSince $x$ is at least 19, then $z$ can be at most 1 by (2).\n\nTherefore, $z=1$. Thus, $x=19$ and $y=0$.", "In any game played, the final result is either both teams earning a tie, or one team earning a win, and the other getting a loss. Therefore, the total number of wins among all teams equals the total number of losses, ie.\n\n$$\n\\begin{aligned}\n25+x & =44+y \\\\\nx-y & =19\n\\end{aligned}\n\\tag{1}\n$$\n\nAlso, since team E plays 20 games, then\n\n$$\nx+y+z=20\n\\tag{2}\n$$\n\nSo from (1), $x$ must be at least 19, and from (2), $x$ can be at most 20.\n\nConsider the possibility that $x=20$. From (2), then $y=z=0$, which does not agree with (1).\n\nThus, the only possibility is $x=19$. From (1), $y=0$, and so $z=1$ from (2). (These three values agree with both equations (1) and (2).)" ]

[ "19,0,1" ]

[ "Consider the cross-section of the sphere in the plane defined by the triangle. This crosssection will be a circle, since any cross-section of a sphere is a circle. This circle will be tangent to the three sides of the triangle, ie. will be the inscribed circle (or incircle) of the triangle. Let the centre of this circle be $O$, and its radius be $r$. We calculate the value of $r$.\n\n<img_3449>\n\nJoin $O$ to the three points of tangency, $P, Q, R$, and to the three vertices $A, B, C$. Then $O P, O Q$ and $O R$ (radii) will form right angles with the three sides of the triangle. Consider the three triangles $\\triangle A O B$, $\\triangle B O C$ and $\\triangle C O A$. Each of these triangles has a height of $r$ and they have bases 15, 9 and 12, respectively. Since the area of $\\triangle A B C$ is equal to the sum of the areas of $\\triangle A O B, \\triangle B O C$, and $\\triangle C O A$, So comparing areas,\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(9)(12) & =\\frac{1}{2}(9)(r)+\\frac{1}{2}(12)(r)+\\frac{1}{2}(15)(r) \\\\\n54 & =\\frac{1}{2} r(9+12+15) \\\\\nr & =3\n\\end{aligned}\n$$\n\n<img_3572>\n\nNow join the centre of the cross-sectional circle to the centre of the sphere and let this distance be $h$. Now, the line joining the centre of the circle to the centre of the sphere will be perpendicular to the plane of the triangle, so we can form a right-angled triangle by joining the centre of the sphere to any point on the circumference of the cross-sectional circle. By Pythagoras,\n\n$$\n\\begin{aligned}\nh^{2}+r^{2} & =25 \\\\\nh & =4\n\\end{aligned}\n$$\n\nThis tells us that the top of the sphere is 9 units above the plane of the triangle, since the top of the sphere is 5 units above the centre of the sphere." ]

[ "5" ]

[ "First, we calculate the side lengths of $\\triangle A B C$ :\n\n$$\n\\begin{aligned}\n& A B=\\sqrt{(0-3)^{2}+(5-0)^{2}}=\\sqrt{34} \\\\\n& B C=\\sqrt{(3-8)^{2}+(0-3)^{2}}=\\sqrt{34} \\\\\n& A C=\\sqrt{(0-8)^{2}+(5-3)^{2}}=\\sqrt{68}\n\\end{aligned}\n$$\n\nSince $A B=B C$ and $A C=\\sqrt{2} A B=\\sqrt{2} B C$, then $\\triangle A B C$ is an isosceles right-angled triangle, with the\n\n<img_3758>\nright angle at $B$.\n\nTherefore, $\\angle A C B=45^{\\circ}$.", "First, we calculate the side lengths of $\\triangle A B C$ :\n\n$$\n\\begin{aligned}\n& A B=\\sqrt{(0-3)^{2}+(5-0)^{2}}=\\sqrt{34} \\\\\n& B C=\\sqrt{(3-8)^{2}+(0-3)^{2}}=\\sqrt{34} \\\\\n& A C=\\sqrt{(0-8)^{2}+(5-3)^{2}}=\\sqrt{68}\n\\end{aligned}\n$$\n\nLine segment $A B$ has slope $\\frac{5-0}{0-3}=-\\frac{5}{3}$.\n\nLine segment $B C$ has slope $\\frac{0-3}{3-8}=\\frac{3}{5}$.\n\nSince the product of these two slopes is -1 , then $A B$ and $B C$ are perpendicular.\n\nTherefore, $\\triangle A B C$ is right-angled at $B$.\n\nSince $A B=B C$, then $\\triangle A B C$ is an isosceles right-angled triangle, so $\\angle A C B=45^{\\circ}$.", "First, we calculate the side lengths of $\\triangle A B C$ :\n\n$$\n\\begin{aligned}\n& A B=\\sqrt{(0-3)^{2}+(5-0)^{2}}=\\sqrt{34} \\\\\n& B C=\\sqrt{(3-8)^{2}+(0-3)^{2}}=\\sqrt{34} \\\\\n& A C=\\sqrt{(0-8)^{2}+(5-3)^{2}}=\\sqrt{68}\n\\end{aligned}\n$$\n\nUsing the cosine law,\n\n$$\n\\begin{aligned}\nA B^{2} & =A C^{2}+B C^{2}-2(A C)(B C) \\cos (\\angle A C B) \\\\\n34 & =68+34-2(\\sqrt{68})(\\sqrt{34}) \\cos (\\angle A C B) \\\\\n0 & =68-2(\\sqrt{2} \\sqrt{34})(\\sqrt{34}) \\cos (\\angle A C B) \\\\\n0 & =68-68 \\sqrt{2} \\cos (\\angle A C B) \\\\\n68 \\sqrt{2} \\cos (\\angle A C B) & =68 \\\\\n\\cos (\\angle A C B) & =\\frac{1}{\\sqrt{2}}\n\\end{aligned}\n$$\n\nSince $\\cos (\\angle A C B)=\\frac{1}{\\sqrt{2}}$ and $0^{\\circ}<\\angle A C B<180^{\\circ}$, then $\\angle A C B=45^{\\circ}$." ]

[ "$45^{\\circ}$" ]

[ "There are two possibilities: either each player wins three games or one player wins more games than the other.\n\nSince the probability that each player wins three games is $\\frac{5}{16}$, then the probability that any one player wins more games than the other is $1-\\frac{5}{16}=\\frac{11}{16}$.\n\nSince each of Blaise and Pierre is equally likely to win any given game, then each must be equally likely to win more games than the other.\n\nTherefore, the probability that Blaise wins more games than Pierre is $\\frac{1}{2} \\times \\frac{11}{16}=\\frac{11}{32}$.", "We consider the results of the 6 games as a sequence of 6 Bs or Ps, with each letter a B if Blaise wins the corresponding game or $\\mathrm{P}$ if Pierre wins.\n\nSince the two players are equally skilled, then the probability that each wins a given game is $\\frac{1}{2}$. This means that the probability of each letter being a $B$ is $\\frac{1}{2}$ and the probability of each letter being a $\\mathrm{P}$ is also $\\frac{1}{2}$.\n\nSince each sequence consists of 6 letters, then the probability of a particular sequence occurring is $\\left(\\frac{1}{2}\\right)^{6}=\\frac{1}{64}$, because each of the letters is specified.\n\nSince they play 6 games in total, then the probability that Blaise wins more games than Pierre is the sum of the probabilities that Blaise wins 4 games, that Blaise wins 5 games, and that Blaise wins 6 games.\n\nIf Blaise wins 6 games, then the sequence consists of 6 Bs. The probability of this is $\\frac{1}{64}$, since there is only one way to arrange $6 \\mathrm{Bs}$.\n\nIf Blaise wins 5 games, then the sequence consists of $5 \\mathrm{Bs}$ and $1 \\mathrm{P}$. The probability of this is $6 \\times \\frac{1}{64}=\\frac{6}{64}$, since there are 6 possible positions in the list for the $1 \\mathrm{P}$ (eg. PBBBBB,BPBBBB, BBPBBB, BBBPBB, BBBBPB, BBBBBP).\n\nThe probability that Blaise wins 4 games is $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right) \\times \\frac{1}{64}=\\frac{15}{64}$, since there are $\\left(\\begin{array}{l}6 \\\\ 2\\end{array}\\right)=15$ ways for 4 Bs and 2 Ps to be arranged.\n\nTherefore, the probability that Blaise wins more games than Pierre is $\\frac{1}{64}+\\frac{6}{64}+\\frac{15}{64}=\\frac{22}{64}=\\frac{11}{32}$." ]

[ "$\\frac{11}{32}$" ]

[ "Using exponent rules and arithmetic, we manipulate the given equation:\n\n$$\n\\begin{aligned}\n3^{x+2}+2^{x+2}+2^{x} & =2^{x+5}+3^{x} \\\\\n3^{x} 3^{2}+2^{x} 2^{2}+2^{x} & =2^{x} 2^{5}+3^{x} \\\\\n9\\left(3^{x}\\right)+4\\left(2^{x}\\right)+2^{x} & =32\\left(2^{x}\\right)+3^{x} \\\\\n8\\left(3^{x}\\right) & =27\\left(2^{x}\\right) \\\\\n\\frac{3^{x}}{2^{x}} & =\\frac{27}{8} \\\\\n\\left(\\frac{3}{2}\\right)^{x} & =\\left(\\frac{3}{2}\\right)^{3}\n\\end{aligned}\n$$\n\nSince the two expressions are equal and the bases are equal, then the exponents must be equal, so $x=3$." ]

[ "3" ]

[ "We manipulate the given equation into a sequence of equivalent equations:\n\n$$\n\\begin{array}{rll}\n\\log _{5 x+9}\\left(x^{2}+6 x+9\\right)+\\log _{x+3}\\left(5 x^{2}+24 x+27\\right) & =4 & \\\\\n\\frac{\\log \\left(x^{2}+6 x+9\\right)}{\\log (5 x+9)}+\\frac{\\log \\left(5 x^{2}+24 x+27\\right)}{\\log (x+3)} & =4 & \\text { (using the \"change of base\" formula) } \\\\\n\\frac{\\log \\left((x+3)^{2}\\right)}{\\log (5 x+9)}+\\frac{\\log ((5 x+9)(x+3))}{\\log (x+3)} & =4 & \\text { (factoring) } \\\\\n\\frac{2 \\log (x+3)}{\\log (5 x+9)}+\\frac{\\log (5 x+9)+\\log (x+3)}{\\log (x+3)} & =4 & \\text { (using logarithm rules) } \\\\\n2\\left(\\frac{\\log (x+3)}{\\log (5 x+9)}\\right)+\\frac{\\log (5 x+9)}{\\log (x+3)}+\\frac{\\log (x+3)}{\\log (x+3)} & =4 & \\text { (rearranging fractions) }\n\\end{array}\n$$\n\n\n\nMaking the substitution $t=\\frac{\\log (x+3)}{\\log (5 x+9)}$, we obtain successively\n\n$$\n\\begin{aligned}\n2 t+\\frac{1}{t}+1 & =4 \\\\\n2 t^{2}+1+t & =4 t \\\\\n2 t^{2}-3 t+1 & =0 \\\\\n(2 t-1)(t-1) & =0\n\\end{aligned}\n$$\n\nTherefore, $t=1$ or $t=\\frac{1}{2}$.\n\nIf $\\frac{\\log (x+3)}{\\log (5 x+9)}=1$, then $\\log (x+3)=\\log (5 x+9)$ or $x+3=5 x+9$, which gives $4 x=-6$ or $x=-\\frac{3}{2}$.\n\nIf $\\frac{\\log (x+3)}{\\log (5 x+9)}=\\frac{1}{2}$, then $2 \\log (x+3)=\\log (5 x+9)$ or $\\log \\left((x+3)^{2}\\right)=\\log (5 x+9)$ or $(x+3)^{2}=5 x+9$.\n\nHere, $x^{2}+6 x+9=5 x+9$ or $x^{2}+x=0$ or $x(x+1)=0$, and so $x=0$ or $x=-1$.\n\nTherefore, there are three possible values for $x: x=0, x=-1$ and $x=-\\frac{3}{2}$.\n\nWe should check each of these in the original equation.\n\nIf $x=0$, the left side of the original equation is $\\log _{9} 9+\\log _{3} 27=1+3=4$.\n\nIf $x=-1$, the left side of the original equation is $\\log _{4} 4+\\log _{2} 8=1+3=4$.\n\nIf $x=-\\frac{3}{2}$, the left side of the original equation is $\\log _{3 / 2}(9 / 4)+\\log _{3 / 2}(9 / 4)=2+2=4$.\n\nTherefore, the solutions are $x=0,-1,-\\frac{3}{2}$." ]

[ "$0,-1,-\\frac{3}{2}$" ]

[ "The Eden sequences from $\\{1,2,3,4,5\\}$ are\n\n$$\n135 \\quad 5 \\quad 1,2 \\quad 1,4 \\quad 3,4 \\quad 1,2,3 \\quad 1,2,5 \\quad 1,4,5 \\quad 3,4,5 \\quad 1,2,3,4 \\quad 1,2,3,4,5\n$$\n\nThere are 12 such sequences.\n\nWe present a brief justification of why these are all of the sequences.\n\n* An Eden sequence of length 1 consists of a single odd integer. The possible choices are 1 and 3 and 5 .\n* An Eden sequence of length 2 consists of an odd integer followed by a larger even integer. Since the only possible even integers here are 2 and 4 , then the possible sequences are 1, 2 and 1, 4 and 3,4 .\n* An Eden sequence of length 3 starts with an Eden sequence of length 2 and appends (that is, adds to the end) a larger odd integer. Starting with 1,2, we form 1,2,3 and $1,2,5$. Starting with 1,4 , we form $1,4,5$. Starting with 3,4 , we form $3,4,5$.\n* An Eden sequence of length 4 starts with an Eden sequence of length 3 and appends a larger even integer. Since 2 and 4 are the only possible even integers, then the only possible sequence here is $1,2,3,4$.\n* An Eden sequence of length 5 from $\\{1,2,3,4,5\\}$ must include all 5 elements, so is $1,2,3,4,5$." ]

[ "12" ]

[ "We will prove that, for all positive integers $n \\geq 3$, we have $e(n)=e(n-1)+e(n-2)+1$. Thus, if $e(18)=m$, then $e(19)=e(18)+e(17)+1=m+4181$ and\n\n$$\ne(20)=e(19)+e(18)+1=(m+4181)+m+1\n$$\n\nSince $e(20)=17710$, then $17710=2 m+4182$ or $2 m=13528$ and so $m=6764$.\n\nTherefore, $e(18)=6764$ and $e(19)=6764+4181=10945$.\n\nSo we must prove that, for all positive integers $n \\geq 3$, we have $e(n)=e(n-1)+e(n-2)+1$.\n\nTo simplify the reading, we use a number of abbreviations:\n\n* ES means \"Eden sequence\"\n\n$* \\operatorname{ES}(m)$ means \"Eden sequence from $\\{1,2,3, \\ldots, m\\}$\n\n* ESE and ESO mean \"Eden sequence of even length\" and \"Eden sequence of odd length\", respectively\n* $\\operatorname{ESE}(m)$ and $\\operatorname{ESO}(m)$ mean \"Eden sequence of even length from $\\{1,2,3, \\ldots, m\\}$ \" and \"Eden sequence of odd length from $\\{1,2,3, \\ldots, m\\}$ \", respectively\n\n\nMethod 1 \n\nFor each positive integer $n$, let $A(n)$ be the number of $\\operatorname{ESE}(n)$, and let $B(n)$ be the number of $\\operatorname{ESO}(n)$.\n\nThen $e(n)=A(n)+B(n)$ for each positive integer $n$.\n\nNote also that for each positive integer $n \\geq 2$, we have $e(n) \\geq e(n-1)$ and $A(n) \\geq A(n-1)$ and $B(n) \\geq B(n-1)$. This is because every $\\operatorname{ES}(n-1)$ is also an $\\operatorname{ES}(n)$ because it satisfies the three required conditions. So there are at least as many $\\operatorname{ES}(n)$ as there are $\\operatorname{ES}(n-1)$. (The same argument works to show that there are at least as many $\\operatorname{ESE}(n)$ as there are $\\operatorname{ESE}(n-1)$, and at least as many $\\operatorname{ESO}(n)$ as there are $\\operatorname{ESO}(n-1)$.\n\nNote that if $k$ is a positive integer, then $2 k+1$ is odd and $2 k$ is even.\n\nThe following four facts are true for every positive integer $k \\geq 1$ :\n\n\n\n(i) $A(2 k+1)=A(2 k)$\n\n(ii) $B(2 k)=B(2 k-1)$\n\n(iii) $A(2 k)=A(2 k-1)+B(2 k-1)$\n\n(iv) $B(2 k+1)=A(2 k)+B(2 k)+1$\n\nHere are justifications for these facts:\n\n(i) An ESE must end with an even integer. Thus, an $\\operatorname{ESE}(2 k+1)$ cannot include $2 k+1$, since it would then have to include a larger even positive integer, which it cannot. Therefore, an $\\operatorname{ESE}(2 k+1)$ has largest term at most $2 k$ and so is an $\\operatorname{ES}(2 k)$.\n\nThus, $A(2 k+1) \\leq A(2 k)$.\n\nBut from above, $A(2 k+1) \\geq A(2 k)$, and so $A(2 k+1)=A(2 k)$.\n\n(ii) An ESO must end with an odd integer. Thus, an $\\operatorname{ESO}(2 k)$ cannot include $2 k$, since it would then have to include a larger odd positive integer, which it cannot. Therefore, an $\\operatorname{ESO}(2 k)$ has largest term at most $2 k-1$ and so is an $\\operatorname{ESO}(2 k-1)$. Thus, $B(2 k) \\leq B(2 k-1)$.\n\nBut from above, $B(2 k) \\geq B(2 k-1)$, and so $B(2 k)=B(2 k-1)$.\n\n(iii) An $\\operatorname{ESE}(2 k)$ either includes $2 k$ or does not include $2 k$.\n\nIf such a sequence includes $2 k$, then removing the $2 k$ produces an $\\operatorname{ESO}(2 k-1)$. Also, every $\\operatorname{ESO}(2 k-1)$ can be produced in this way.\n\nTherefore, the number of sequences in this case is $B(2 k-1)$.\n\nIf such a sequence does not include $2 k$, then the sequence can be thought of as an $\\operatorname{ESE}(2 k-1)$. Note that every $\\operatorname{ESE}(2 k-1)$ is an $\\operatorname{ESE}(2 k)$.\n\nTherefore, the number of sequences in this case is $A(2 k-1)$.\n\nThus, $A(2 k)=A(2 k-1)+B(2 k-1)$.\n\n(iv) $\\operatorname{An} \\operatorname{ESO}(2 k+1)$ is either the one term sequence $2 k+1$, or includes $2 k+1$ and more terms, or does not include $2 k+1$.\n\nThere is 1 sequence of the first kind.\n\nAs in (iii), there are $A(2 k)$ sequences of the second kind and $B(2 k)$ sequences of the third kind.\n\nThus, $B(2 k+1)=1+A(2 k)+B(2 k)$.\n\nCombining these facts, for each positive integer $k$, we obtain\n\n$$\n\\begin{aligned}\ne(2 k+1) & =A(2 k+1)+B(2 k+1) \\\\\n& =A(2 k)+(A(2 k)+B(2 k)+1) \\\\\n& =(A(2 k)+B(2 k))+A(2 k)+1 \\\\\n& =e(2 k)+(A(2 k-1)+B(2 k-1))+1 \\\\\n& =e(2 k)+e(2 k-1)+1\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\ne(2 k) & =A(2 k)+B(2 k) \\\\\n& =(A(2 k-1)+B(2 k-1))+B(2 k-1) \\\\\n& =e(2 k-1)+(A(2 k-2)+B(2 k-2)+1) \\\\\n& =e(2 k-1)+e(2 k-2)+1\n\\end{aligned}\n$$\n\nTherefore, for all positive integers $n \\geq 3$, we have $e(n)=e(n-1)+e(n-2)+1$, as required.\n\n\n\nMethod 2 \n\nLet $n$ be a positive integer with $n \\geq 3$, and consider the $\\operatorname{ES}(n)$.\n\nWe divide the sequences into three sets:\n\n(i) The sequence 1 (there is 1 such sequence)\n\n(ii) The sequences which begin with 1 and have more than 1 term\n\n(iii) The sequences which do not begin with 1\n\nWe show that in case (ii) there are $e(n-1)$ sequences and in case (iii) there are $e(n-2)$ sequences. This will show that $e(n)=1+e(n-1)+e(n-2)$, as required.\n\n(ii) Consider the set of $\\operatorname{ES}(n)$ that begin with 1 . We call this set of sequences $P$.\n\nWe remove the 1 from each of these and consider the set of resulting sequences. We call this set $Q$. Note that the number of sequences in $P$ and in $Q$ is the same.\n\nEach of the sequences in $Q$ includes numbers from the set $\\{2,3, \\ldots, n\\}$, is increasing, and has even terms in odd positions and odd terms in even positions (since each term has been shifted one position to the left).\n\nThe sequences in $Q$ are in a one-to-one correspondence with the $\\operatorname{ES}(n-1)$ (we call this set of sequences $R$ ) and so there are exactly $e(n-1)$ of them (and so $e(n-1)$ sequences in $P$ ).\n\nWe can show that this one-to-one correspondence exists by subtracting 1 from each term of each sequence in $Q$, to form a set of sequences $S$. Each of the resulting sequences is distinct, includes numbers from the set $\\{1,2,3, \\ldots, n-1\\}$, is increasing, and has odd terms in odd positions and even terms in even positions (since each term has been reduced by 1). Also, each sequence in $R$ can be obtained in this way (since adding 1 to each term in one of these ES gives a distinct sequence in $Q$ ).\n\nTherefore, the number of sequences in this case is $e(n-1)$.\n\n(iii) Consider the set of $\\operatorname{ES}(n)$ that do not begin with 1 . We call this set of sequences $T$. Since each sequence in $T$ does not begin with 1 , then the minimum number in each sequence is 3 .\n\nThus, each of the sequences in $T$ includes numbers from the set $\\{3,4, \\ldots, n\\}$, is increasing, and has odd terms in odd positions and even terms in even positions.\n\nThe sequences in $T$ are in a one-to-one correspondence with the $\\mathrm{ES}(n-2)$ (we call this set of sequences $U$ ) and so there are exactly $e(n-2)$ of them.\n\nWe can show that this one-to-one correspondence exists by subtracting 2 from each term of each sequence in $T$, to form a set of sequences $V$. Each of the resulting sequences is distinct, includes numbers from the set $\\{1,2,3, \\ldots, n-2\\}$, is increasing, and has odd terms in odd positions and even terms in even positions (since each term has been reduced by 2). Also, each sequence in $U$ can be obtained in this way (since adding 2 to each term in one of these $\\mathrm{ES}$ gives a distinct sequence in $U$ ).\n\nTherefore, the number of sequences in this case is $e(n-2)$.\n\nThis concludes our proof and shows that $e(n)=1+e(n-1)+e(n-2)$, as required." ]

[ "$6764$,$10945$" ]

[ "Since there are 5 choices for $a$ and 3 choices for $b$, there are fifteen possible ways of choosing $a$ and $b$.\n\nIf $a$ is even, $a^{b}$ is even; if $a$ is odd, $a^{b}$ is odd.\n\nSo the choices of $a$ and $b$ which give an even value for $a^{b}$ are those where $a$ is even, or 6 of the choices (since there are two even choices for $a$ and three ways of choosing $b$ for each of these). (Notice that in fact the value of $b$ does not affect whether $a^{b}$ is even or odd, so the probability depends only on the choice of $a$.)\n\nThus, the probability is $\\frac{6}{15}=\\frac{2}{5}$." ]

[ "$\\frac{2}{5}$" ]

[ "Starting with 4 blue hats and 2 green hats, the probability that Julia removes a blue hat is $\\frac{4}{6}=\\frac{2}{3}$. The result would be 3 blue hats and 3 green hats, since a blue hat is replaced with a green hat.\n\nIn order to return to 4 blue hats and 2 green hats from 3 blue and 3 green, Julia would need remove a green hat (which would be replaced by a blue hat). The probability of her\n\n\n\nremoving a green hat from 3 blue and 3 green is $\\frac{3}{6}=\\frac{1}{2}$.\n\nSummarizing, the probability of choosing a blue hat and then a green hat is $\\frac{2}{3} \\times \\frac{1}{2}=\\frac{1}{3}$.\n\nStarting with 4 blue hats and 2 green hats, the probability that Julia removes a green hat is $\\frac{2}{6}=\\frac{1}{3}$. The result would be 5 blue hats and 1 green hat, since a green hat is replaced with a blue hat.\n\nIn order to return to 4 blue hats and 2 green hats from 5 blue and 1 green, Julia would need remove a blue hat (which would be replaced by a green hat). The probability of her removing a green hat from 5 blue and 1 green is $\\frac{5}{6}$.\n\nSummarizing, the probability of choosing a green hat and then a blue hat is $\\frac{1}{3} \\times \\frac{5}{6}=\\frac{5}{18}$.\n\nThese are the only two ways to return to 4 blue hats and 2 green hats after two turns removing a blue hat then a green, or removing a green then a blue.\n\nTherefore, the total probability of returning to 4 blue hats and 2 green hats after two turns is $\\frac{1}{3}+\\frac{5}{18}=\\frac{11}{18}$." ]

[ "$\\frac{11}{18}$" ]

[ "Adding the two equations, we obtain\n\n$$\n\\begin{aligned}\n\\sin ^{2} x+\\cos ^{2} x+\\sin ^{2} y+\\cos ^{2} y & =\\frac{3}{2} a+\\frac{1}{2} a^{2} \\\\\n2 & =\\frac{3}{2} a+\\frac{1}{2} a^{2} \\\\\n4 & =3 a+a^{2} \\\\\n0 & =a^{2}+3 a-4 \\\\\n0 & =(a+4)(a-1)\n\\end{aligned}\n$$\n\nand so $a=-4$ or $a=1$.\n\nHowever, $a=-4$ is impossible, since this would give $\\sin ^{2} x+\\cos ^{2} y=-6$, whose left side is non-negative and whose right side is negative.\n\nTherefore, the only possible value for $a$ is $a=1$.\n\n(We can check that angles $x=90^{\\circ}$ and $y=45^{\\circ}$ give $\\sin ^{2} x+\\cos ^{2} y=\\frac{3}{2}$ and $\\cos ^{2} x+\\sin ^{2} y=$ $\\frac{1}{2}$, so $a=1$ is indeed possible.)" ]

[ "1" ]

[ "We calculate the first 15 terms, writing each as an integer times a power of 10:\n\n$$\n\\begin{gathered}\n2,5,10,5 \\times 10,5 \\times 10^{2}, 5^{2} \\times 10^{3}, 5^{3} \\times 10^{5}, 5^{5} \\times 10^{8}, 5^{8} \\times 10^{13}, 5^{13} \\times 10^{21}, 5^{21} \\times 10^{34} \\\\\n5^{34} \\times 10^{55}, 5^{55} \\times 10^{89}, 5^{89} \\times 10^{144}, 5^{144} \\times 10^{233}\n\\end{gathered}\n$$\n\nSince the 15 th term equals an odd integer times $10^{233}$, then the 15 th term ends with 233 zeroes.", "To obtain the 6 th term, we calculate $50 \\times 500=25 \\times 1000$.\n\nEach of the 4th and 5th terms equals an odd integer followed by a number of zeroes, so the 6th term also equals an odd integer followed by a number of zeroes, where the number of zeroes is the sum of the numbers of zeroes at the ends of the 4th and 5th terms.\n\nThis pattern will continue. Thus, starting with the 6th term, the number of zeroes at the end of the term will be the sum of the number of zeroes at the ends of the two previous terms.\n\nThis tells us that, starting with the 4th term, the number of zeroes at the ends of the terms is\n\n$$\n1,2,3,5,8,13,21,34,55,89,144,233\n$$\n\nTherefore, the 15 th term ends with 233 zeroes." ]

[ "233" ]

[ "We use logarithm rules to rearrange the equation to solve for $y$ :\n\n$$\n\\begin{aligned}\n\\log _{2} x-2 \\log _{2} y & =2 \\\\\n\\log _{2} x-\\log _{2}\\left(y^{2}\\right) & =2 \\\\\n\\log _{2}\\left(\\frac{x}{y^{2}}\\right) & =2 \\\\\n\\frac{x}{y^{2}} & =2^{2} \\\\\n\\frac{1}{4} x & =y^{2} \\\\\ny & = \\pm \\frac{1}{2} \\sqrt{x}\n\\end{aligned}\n$$\n\nBut since the domain of the $\\log _{2}$ function is all positive real numbers, we must have $x>0$ and $y>0$, so we can reject the negative square root to obtain\n\n$$\ny=\\frac{1}{2} \\sqrt{x}, \\quad x>0\n$$" ]

[ "$\\frac{1}{2},\\sqrt{x}$" ]

[ "Since $\\sin ^{2} x+\\cos ^{2} x=1$, then $\\cos ^{2} x=1-\\sin ^{2} x$, so\n\n$$\n\\begin{aligned}\nf(x) & =\\sin ^{6} x+\\left(1-\\sin ^{2} x\\right)^{3}+k\\left(\\sin ^{4} x+\\left(1-\\sin ^{2} x\\right)^{2}\\right) \\\\\n& =\\sin ^{6} x+1-3 \\sin ^{2} x+3 \\sin ^{4} x-\\sin ^{6} x+k\\left(\\sin ^{4} x+1-2 \\sin ^{2} x+\\sin ^{4} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x+(3+2 k) \\sin ^{4} x\n\\end{aligned}\n$$\n\nTherefore, if $3+2 k=0$ or $k=-\\frac{3}{2}$, then $f(x)=1+k=-\\frac{1}{2}$ for all $x$ and so is constant. (If $k \\neq-\\frac{3}{2}$, then we get\n\n$$\n\\begin{aligned}\nf(0) & =1+k \\\\\nf\\left(\\frac{1}{4} \\pi\\right) & =(1+k)-(3+2 k)\\left(\\frac{1}{2}\\right)+(3+2 k)\\left(\\frac{1}{4}\\right)=\\frac{1}{4}+\\frac{1}{2} k \\\\\nf\\left(\\frac{1}{6} \\pi\\right) & =(1+k)-(3+2 k)\\left(\\frac{1}{4}\\right)+(3+2 k)\\left(\\frac{1}{16}\\right)=\\frac{7}{16}+\\frac{5}{8} k\n\\end{aligned}\n$$\n\nwhich cannot be all equal for any single value of $k$, so $f(x)$ is not constant if $k \\neq-\\frac{3}{2}$.)", "Since $\\sin ^{2} x+\\cos ^{2} x=1$, then\n\n$$\n\\begin{aligned}\nf(x) & =\\left(\\sin ^{2} x+\\cos ^{2} x\\right)\\left(\\sin ^{4} x-\\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x\\right)+k\\left(\\sin ^{4} x+\\cos ^{4} x\\right) \\\\\n& =\\left(\\sin ^{4}+2 \\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x-3 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& \\quad k\\left(\\sin ^{4} x+2 \\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =\\left(\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{2}-3 \\sin ^{2} x \\cos ^{2} x\\right)+k\\left(\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{2}-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =1-3 \\sin ^{2} x \\cos ^{2} x+k\\left(1-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x \\cos ^{2} x\n\\end{aligned}\n$$\n\nTherefore, if $3+2 k=0$ or $k=-\\frac{3}{2}$, then $f(x)=1+k=-\\frac{1}{2}$ for all $x$ and so is constant.", "For $f(x)$ to be constant, we need $f^{\\prime}(x)=0$ for all values of $x$.\n\nCalculating using the Chain Rule,\n\n$$\n\\begin{aligned}\nf^{\\prime}(x) & =6 \\sin ^{5} x \\cos x-6 \\cos ^{5} x \\sin x+k\\left(4 \\sin ^{3} x \\cos x-4 \\cos ^{3} x \\sin x\\right) \\\\\n& =2 \\sin x \\cos x\\left(3\\left(\\sin ^{4} x-\\cos ^{4} x\\right)+2 k\\left(\\sin ^{2} x-\\cos ^{2} x\\right)\\right) \\\\\n& =2 \\sin x \\cos x\\left(\\sin ^{2} x-\\cos ^{2} x\\right)\\left(3\\left(\\sin ^{2} x+\\cos ^{2} x\\right)+2 k\\right) \\\\\n& =2 \\sin x \\cos x\\left(\\sin ^{2} x-\\cos ^{2} x\\right)(3+2 k)\n\\end{aligned}\n$$\n\nIf $3+2 k=0$ or $k=-\\frac{3}{2}$, then $f^{\\prime}(x)=0$ for all $x$, so $f(x)$ is constant.\n\n(If $3+2 k \\neq 0$, then choosing $x=\\frac{1}{6} \\pi$ for example gives $f^{\\prime}(x) \\neq 0$ so $f(x)$ is not constant.)" ]

[ "$-\\frac{3}{2}$" ]

[ "Since $\\sin ^{2} x+\\cos ^{2} x=1$, then $\\cos ^{2} x=1-\\sin ^{2} x$, so\n\n$$\n\\begin{aligned}\nf(x) & =\\sin ^{6} x+\\left(1-\\sin ^{2} x\\right)^{3}+k\\left(\\sin ^{4} x+\\left(1-\\sin ^{2} x\\right)^{2}\\right) \\\\\n& =\\sin ^{6} x+1-3 \\sin ^{2} x+3 \\sin ^{4} x-\\sin ^{6} x+k\\left(\\sin ^{4} x+1-2 \\sin ^{2} x+\\sin ^{4} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x+(3+2 k) \\sin ^{4} x\n\\end{aligned}\n$$\n\nNow, we have\n\n$$\nf(x)=(1+k)-(3+2 k) \\sin ^{2} x+(3+2 k) \\sin ^{4} x\n$$\n\nand so we want to solve\n\n$$\n\\begin{array}{r}\n0.3-(1.6) \\sin ^{2} x+(1.6) \\sin ^{4} x=0 \\\\\n16 \\sin ^{4} x-16 \\sin ^{2} x+3=0 \\\\\n\\left(4 \\sin ^{2} x-3\\right)\\left(4 \\sin ^{2} x-1\\right)=0\n\\end{array}\n$$\n\nTherefore, $\\sin ^{2} x=\\frac{1}{4}, \\frac{3}{4}$, and so $\\sin x= \\pm \\frac{1}{2}, \\pm \\frac{\\sqrt{3}}{2}$.\n\nTherefore,\n\n$$\nx=\\frac{1}{6} \\pi+2 \\pi k, \\frac{5}{6} \\pi+2 \\pi k, \\frac{7}{6} \\pi+2 \\pi k, \\frac{11}{6} \\pi+2 \\pi k, \\frac{1}{3} \\pi+2 \\pi k, \\frac{2}{3} \\pi+2 \\pi k, \\frac{4}{3} \\pi+2 \\pi k, \\frac{5}{3} \\pi+2 \\pi k\n$$\n\nfor $k \\in \\mathbb{Z}$.", "Since $\\sin ^{2} x+\\cos ^{2} x=1$, then\n\n$$\n\\begin{aligned}\nf(x) & =\\left(\\sin ^{2} x+\\cos ^{2} x\\right)\\left(\\sin ^{4} x-\\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x\\right)+k\\left(\\sin ^{4} x+\\cos ^{4} x\\right) \\\\\n& =\\left(\\sin ^{4}+2 \\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x-3 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& \\quad k\\left(\\sin ^{4} x+2 \\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =\\left(\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{2}-3 \\sin ^{2} x \\cos ^{2} x\\right)+k\\left(\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{2}-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =1-3 \\sin ^{2} x \\cos ^{2} x+k\\left(1-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x \\cos ^{2} x\n\\end{aligned}\n$$\n\nNow, we have\n\n$$\nf(x)=(1+k)-(3+2 k) \\sin ^{2} x \\cos ^{2} x\n$$\n\nUsing the fact that $\\sin 2 x=2 \\sin x \\cos x$, we can further simplify $f(x)$ to\n\n$$\nf(x)=(1+k)-\\frac{1}{4}(3+2 k) \\sin ^{2} 2 x\n$$\n\n\n\nand so we want to solve\n\n$$\n\\begin{aligned}\n0.3-\\frac{1}{4}(1.6) \\sin ^{2} 2 x & =0 \\\\\n4 \\sin ^{2} 2 x & =3 \\\\\n\\sin ^{2} 2 x & =\\frac{3}{4}\n\\end{aligned}\n$$\n\nand so $\\sin 2 x= \\pm \\frac{\\sqrt{3}}{2}$.\n\nTherefore,\n\n$$\n2 x=\\frac{1}{3} \\pi+2 \\pi k, \\frac{2}{3} \\pi+2 \\pi k, \\frac{4}{3} \\pi+2 \\pi k, \\frac{5}{3} \\pi+2 \\pi k\n$$\n\nfor $k \\in \\mathbb{Z}$, and so\n\n$$\nx=\\frac{1}{6} \\pi+\\pi k, \\frac{1}{3} \\pi+\\pi k, \\frac{2}{3} \\pi+\\pi k, \\frac{5}{6} \\pi+\\pi k\n$$\n\nfor $k \\in \\mathbb{Z}$." ]

[ "$x=\\frac{1}{6} \\pi+\\pi k, \\frac{1}{3} \\pi+\\pi k, \\frac{2}{3} \\pi+\\pi k, \\frac{5}{6} \\pi+\\pi k$" ]

[ "Since $\\sin ^{2} x+\\cos ^{2} x=1$, then $\\cos ^{2} x=1-\\sin ^{2} x$, so\n\n$$\n\\begin{aligned}\nf(x) & =\\sin ^{6} x+\\left(1-\\sin ^{2} x\\right)^{3}+k\\left(\\sin ^{4} x+\\left(1-\\sin ^{2} x\\right)^{2}\\right) \\\\\n& =\\sin ^{6} x+1-3 \\sin ^{2} x+3 \\sin ^{4} x-\\sin ^{6} x+k\\left(\\sin ^{4} x+1-2 \\sin ^{2} x+\\sin ^{4} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x+(3+2 k) \\sin ^{4} x\n\\end{aligned}\n$$\n\nNow, we have\n\n$$\nf(x)=(1+k)-(3+2 k) \\sin ^{2} x+(3+2 k) \\sin ^{4} x\n$$\n\nWe want to determine the values of $k$ for which there is an $a$ such that $f(a)=0$.\n\nFrom (a), if $k=-\\frac{3}{2}, f(x)$ is constant and equal to $-\\frac{1}{2}$, so has no roots.\n\nLet $u=\\sin ^{2} x$.\n\nThen $u$ takes all values between 0 and 1 as $\\sin x$ takes all values between -1 and 1 . Then we want to determine for which $k$ the equation\n\n$$\n(3+2 k) u^{2}-(3+2 k) u+(1+k)=0\n$$\n\nhas a solution for $u$ with $0 \\leq u \\leq 1$.\n\nFirst, we must ensure that the equation $(*)$ has real solutions, ie.\n\n$$\n\\begin{aligned}\n(3+2 k)^{2}-4(3+2 k)(1+k) & \\geq 0 \\\\\n(3+2 k)(3+2 k-4(1+k)) & \\geq 0 \\\\\n(3+2 k)(-1-2 k) & \\geq 0 \\\\\n(3+2 k)(1+2 k) & \\leq 0\n\\end{aligned}\n$$\n\nThis is true if and only if $-\\frac{3}{2}<k \\leq-\\frac{1}{2}$. (We omit $k=-\\frac{3}{2}$ because of the earlier comment.)\n\nNext, we have to check for which values of $k$ the equation $(*)$ has a solution $u$ with $0 \\leq u \\leq 1$. We may assume that $-\\frac{3}{2}<k \\leq-\\frac{1}{2}$.\n\nTo do this, we solve the equation $(*)$ using the quadratic formula to obtain\n\n$$\nu=\\frac{(3+2 k) \\pm \\sqrt{(3+2 k)^{2}-4(3+2 k)(1+k)}}{2(3+2 k)}\n$$\n\nor\n\n$$\nu=\\frac{(3+2 k) \\pm \\sqrt{-(3+2 k)(1+2 k)}}{2(3+2 k)}=\\frac{1}{2} \\pm \\frac{1}{2} \\sqrt{-\\frac{1+2 k}{3+2 k}}\n$$\n\n\n\nSince $k>-\\frac{3}{2}$ then $3+2 k>0$.\n\nFor $u$ to be between 0 and 1, we need to have\n\n$$\n0 \\leq \\sqrt{-\\frac{1+2 k}{3+2 k}} \\leq 1\n$$\n\nThus\n\n$$\n0 \\leq-\\frac{1+2 k}{3+2 k} \\leq 1\n$$\n\nSince $-\\frac{3}{2}<k \\leq-\\frac{1}{2}$ then $3+2 k>0$ and $1+2 k \\leq 0$, so the left inequality is true.\n\nTherefore, we need $-\\frac{1+2 k}{3+2 k} \\leq 1$ or $-(1+2 k) \\leq(3+2 k)$ (we can multiply by $(3+2 k)$ since it is positive), and so $-4 \\leq 4 k$ or $k \\geq-1$.\n\nCombining with $-\\frac{3}{2}<k \\leq-\\frac{1}{2}$ gives $-1 \\leq k \\leq-\\frac{1}{2}$.", "Since $\\sin ^{2} x+\\cos ^{2} x=1$, then\n\n$$\n\\begin{aligned}\nf(x) & =\\left(\\sin ^{2} x+\\cos ^{2} x\\right)\\left(\\sin ^{4} x-\\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x\\right)+k\\left(\\sin ^{4} x+\\cos ^{4} x\\right) \\\\\n& =\\left(\\sin ^{4}+2 \\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x-3 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& \\quad k\\left(\\sin ^{4} x+2 \\sin ^{2} x \\cos ^{2} x+\\cos ^{4} x-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =\\left(\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{2}-3 \\sin ^{2} x \\cos ^{2} x\\right)+k\\left(\\left(\\sin ^{2} x+\\cos ^{2} x\\right)^{2}-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =1-3 \\sin ^{2} x \\cos ^{2} x+k\\left(1-2 \\sin ^{2} x \\cos ^{2} x\\right) \\\\\n& =(1+k)-(3+2 k) \\sin ^{2} x \\cos ^{2} x\n\\end{aligned}\n$$\n\nNow, we have\n\n$$\nf(x)=(1+k)-(3+2 k) \\sin ^{2} x \\cos ^{2} x\n$$\n\nUsing the fact that $\\sin 2 x=2 \\sin x \\cos x$, we can further simplify $f(x)$ to\n\n$$\nf(x)=(1+k)-\\frac{1}{4}(3+2 k) \\sin ^{2} 2 x\n$$\n\nNow, we have\n\n$$\nf(x)=(1+k)-\\frac{1}{4}(3+2 k) \\sin ^{2} 2 x\n$$\n\nIf we tried to solve $f(x)=0$, we would obtain\n\n$$\n(1+k)-\\frac{1}{4}(3+2 k) \\sin ^{2} 2 x=0\n$$\n\nor\n\n$$\n\\sin ^{2} 2 x=\\frac{4(1+k)}{3+2 k}\n$$\n\n(From (a), if $k=-\\frac{3}{2}, f(x)$ is constant and equal to $-\\frac{1}{2}$, so has no roots.)\n\nIn order to be able to solve this (first for $\\sin 2 x$, then for $2 x$ then for $x$ ), we therefore need\n\n$$\n0 \\leq \\frac{4(1+k)}{3+2 k} \\leq 1\n$$\n\nIf $3+2 k>0$, we can multiply the inequality by $3+2 k$ to obtain\n\n$$\n0 \\leq 4(1+k) \\leq 3+2 k\n$$\n\nand so we get $k \\geq-1$ from the left inequality and $k \\leq-\\frac{1}{2}$ from the right inequality.\n\nCombining these with $-\\frac{3}{2}<k$, we obtain $-1 \\leq k \\leq-\\frac{1}{2}$.\n\nIf $3+2 k<0$, we would obtain $0 \\geq 4(1+k) \\geq 3+2 k$ which would give $k \\leq-1$ and $k \\geq-\\frac{1}{2}$, which are inconsistent.\n\nTherefore, $-1 \\leq k \\leq-\\frac{1}{2}$." ]

[ "$[-1,-\\frac{1}{2}]$" ]

[ "Let $P$ be the point with coordinates $(7,0)$ and let $Q$ be the point with coordinates $(0,5)$.\n\n<img_4025>\n\nThen $A P D Q$ is a rectangle with width 7 and height 5 , and so it has area $7 \\cdot 5=35$.\n\nHexagon $A B C D E F$ is formed by removing two triangles from rectangle $A P D Q$, namely $\\triangle B P C$ and $\\triangle E Q F$.\n\nEach of $\\triangle B P C$ and $\\triangle E Q F$ is right-angled, because each shares an angle with rectangle $A P D Q$.\n\nEach of $\\triangle B P C$ and $\\triangle E Q F$ has a base of length 3 and a height of 2.\n\nThus, their combined area is $2 \\cdot \\frac{1}{2} \\cdot 3 \\cdot 2=6$.\n\nThis means that the area of hexagon $A B C D E F$ is $35-6=29$." ]

[ "29" ]

[ "If $r$ is a term in the sequence and $s$ is the next term, then $s=1+\\frac{1}{1+r}$.\n\nThis means that $s-1=\\frac{1}{1+r}$ and so $\\frac{1}{s-1}=1+r$ which gives $r=\\frac{1}{s-1}-1$.\n\nTherefore, since $a_{3}=\\frac{41}{29}$, then\n\n$$\na_{2}=\\frac{1}{a_{3}-1}-1=\\frac{1}{(41 / 29)-1}-1=\\frac{1}{12 / 29}-1=\\frac{29}{12}-1=\\frac{17}{12}\n$$\n\nFurther, since $a_{2}=\\frac{17}{12}$, then\n\n$$\na_{1}=\\frac{1}{a_{2}-1}-1=\\frac{1}{(17 / 12)-1}-1=\\frac{1}{5 / 12}-1=\\frac{12}{5}-1=\\frac{7}{5}\n$$" ]

[ "$\\frac{7}{5}$" ]

[ "Initially, the water in the hollow tube forms a cylinder with radius $10 \\mathrm{~mm}$ and height $h \\mathrm{~mm}$. Thus, the volume of the water is $\\pi(10 \\mathrm{~mm})^{2}(h \\mathrm{~mm})=100 \\pi h \\mathrm{~mm}^{3}$.\n\nAfter the rod is inserted, the level of the water rises to $64 \\mathrm{~mm}$. Note that this does not overflow the tube, since the tube's height is $100 \\mathrm{~mm}$.\n\nUp to the height of the water, the tube is a cylinder with radius $10 \\mathrm{~mm}$ and height 64 mm.\n\nThus, the volume of the tube up to the height of the water is\n\n$$\n\\pi(10 \\mathrm{~mm})^{2}(64 \\mathrm{~mm})=6400 \\pi \\mathrm{mm}^{3}\n$$\n\nThis volume consists of the water that is in the tube (whose volume, which has not changed, is $100 \\pi h \\mathrm{~mm}^{3}$ ) and the rod up to a height of $64 \\mathrm{~mm}$.\n<img_3180>\n\nSince the radius of the rod is $2.5 \\mathrm{~mm}$, the volume of the rod up to a height of $64 \\mathrm{~mm}$ is $\\pi(2.5 \\mathrm{~mm})^{2}(64 \\mathrm{~mm})=400 \\pi \\mathrm{mm}^{3}$.\n\nComparing volumes, $6400 \\pi \\mathrm{mm}^{3}=100 \\pi h \\mathrm{~mm}^{3}+400 \\pi \\mathrm{mm}^{3}$ and so $100 h=6000$ which gives $h=60$." ]

[ "60" ]

[ "We note that $\\frac{2 x+1}{x}=\\frac{2 x}{x}+\\frac{1}{x}=2+\\frac{1}{x}$.\n\nTherefore, $\\frac{2 x+1}{x}=4$ exactly when $2+\\frac{1}{x}=4$ or $\\frac{1}{x}=2$ and so $x=\\frac{1}{2}$.\n\nAlternatively, we could solve $\\frac{2 x+1}{x}=4$ directly to obtain $2 x+1=4 x$, which gives $2 x=1$ and so $x=\\frac{1}{2}$.\n\nThus, to determine the value of $f(4)$, we substitute $x=\\frac{1}{2}$ into the given equation $f\\left(\\frac{2 x+1}{x}\\right)=x+6$ and obtain $f(4)=\\frac{1}{2}+6=\\frac{13}{2}$." ]

[ "$\\frac{13}{2}$" ]

[ "Since the graph passes through $(3,5),(5,4)$ and $(11,3)$, we can substitute these three points and obtain the following three equations:\n\n$$\n\\begin{aligned}\n& 5=\\log _{a}(3+b)+c \\\\\n& 4=\\log _{a}(5+b)+c \\\\\n& 3=\\log _{a}(11+b)+c\n\\end{aligned}\n$$\n\nSubtracting the second equation from the first and the third equation from the second, we obtain:\n\n$$\n\\begin{aligned}\n& 1=\\log _{a}(3+b)-\\log _{a}(5+b) \\\\\n& 1=\\log _{a}(5+b)-\\log _{a}(11+b)\n\\end{aligned}\n$$\n\nEquating right sides and manipulating, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n\\log _{a}(5+b)-\\log _{a}(11+b) & =\\log _{a}(3+b)-\\log _{a}(5+b) \\\\\n2 \\log _{a}(5+b) & =\\log _{a}(3+b)+\\log _{a}(11+b) \\\\\n\\log _{a}\\left((5+b)^{2}\\right) & =\\log _{a}((3+b)(11+b)) \\quad(\\text { using log laws }) \\\\\n(5+b)^{2} & =(3+b)(11+b) \\quad \\text { (raising both sides to the power of } a) \\\\\n25+10 b+b^{2} & =33+14 b+b^{2} \\quad \\\\\n-8 & =4 b \\\\\nb & =-2\n\\end{aligned}\n$$\n\nSince $b=-2$, the equation $1=\\log _{a}(3+b)-\\log _{a}(5+b)$ becomes $1=\\log _{a} 1-\\log _{a} 3$.\n\nSince $\\log _{a} 1=0$ for every admissible value of $a$, then $\\log _{a} 3=-1$ which gives $a=3^{-1}=\\frac{1}{3}$.\n\nFinally, the equation $5=\\log _{a}(3+b)+c$ becomes $5=\\log _{1 / 3}(1)+c$ and so $c=5$.\n\nTherefore, $a=\\frac{1}{3}, b=-2$, and $c=5$, which gives $y=\\log _{1 / 3}(x-2)+5$.\n\nChecking:\n\n- When $x=3$, we obtain $y=\\log _{1 / 3}(3-2)+5=\\log _{1 / 3} 1+5=0+5=5$.\n- When $x=5$, we obtain $y=\\log _{1 / 3}(5-2)+5=\\log _{1 / 3} 3+5=-1+5=4$.\n- When $x=11$, we obtain $y=\\log _{1 / 3}(11-2)+5=\\log _{1 / 3} 9+5=-2+5=3$." ]

[ "$\\frac{1}{3},-2,5$" ]

[ "The probability that the integer $n$ is chosen is $\\log _{100}\\left(1+\\frac{1}{n}\\right)$.\n\nThe probability that an integer between 81 and 99 , inclusive, is chosen equals the sum of the probabilities that the integers $81,82, \\ldots, 98,99$ are selected, which equals\n\n$$\n\\log _{100}\\left(1+\\frac{1}{81}\\right)+\\log _{100}\\left(1+\\frac{1}{82}\\right)+\\cdots+\\log _{100}\\left(1+\\frac{1}{98}\\right)+\\log _{100}\\left(1+\\frac{1}{99}\\right)\n$$\n\nSince the second probability equals 2 times the first probability, the following equations are equivalent:\n\n$$\n\\begin{aligned}\n\\log _{100}\\left(1+\\frac{1}{81}\\right)+\\log _{100}\\left(1+\\frac{1}{82}\\right)+\\cdots+\\log _{100}\\left(1+\\frac{1}{98}\\right)+\\log _{100}\\left(1+\\frac{1}{99}\\right) & =2 \\log _{100}\\left(1+\\frac{1}{n}\\right) \\\\\n\\log _{100}\\left(\\frac{82}{81}\\right)+\\log _{100}\\left(\\frac{83}{82}\\right)+\\cdots+\\log _{100}\\left(\\frac{99}{98}\\right)+\\log _{100}\\left(\\frac{100}{99}\\right) & =2 \\log _{100}\\left(1+\\frac{1}{n}\\right)\n\\end{aligned}\n$$\n\nUsing logarithm laws, these equations are further equivalent to\n\n$$\n\\begin{aligned}\n\\log _{100}\\left(\\frac{82}{81} \\cdot \\frac{83}{82} \\cdots \\cdot \\frac{99}{98} \\cdot \\frac{100}{99}\\right) & =\\log _{100}\\left(1+\\frac{1}{n}\\right)^{2} \\\\\n\\log _{100}\\left(\\frac{100}{81}\\right) & =\\log _{100}\\left(1+\\frac{1}{n}\\right)^{2}\n\\end{aligned}\n$$\n\nSince logarithm functions are invertible, we obtain $\\frac{100}{81}=\\left(1+\\frac{1}{n}\\right)^{2}$.\n\nSince $n>0$, then $1+\\frac{1}{n}=\\sqrt{\\frac{100}{81}}=\\frac{10}{9}$, and so $\\frac{1}{n}=\\frac{1}{9}$, which gives $n=9$." ]

[ "9" ]

[ "Since $10^{y} \\neq 0$, the equation $\\frac{1}{32}=\\frac{x}{10^{y}}$ is equivalent to $10^{y}=32 x$.\n\nSo the given question is equivalent to asking for the smallest positive integer $x$ for which $32 x$ equals a positive integer power of 10 .\n\nNow $32=2^{5}$ and so $32 x=2^{5} x$.\n\nFor $32 x$ to equal a power of 10, each factor of 2 must be matched with a factor of 5 .\n\nTherefore, $x$ must be divisible by $5^{5}$ (that is, $x$ must include at least 5 powers of 5 ), and so $x \\geq 5^{5}=3125$.\n\nBut $32\\left(5^{5}\\right)=2^{5} 5^{5}=10^{5}$, and so if $x=5^{5}=3125$, then $32 x$ is indeed a power of 10 , namely $10^{5}$.\n\nThis tells us that the smallest positive integer $x$ for which $\\frac{1}{32}=\\frac{x}{10^{y}}$ for some positive integer $y$ is $x=5^{5}=3125$." ]

[ "3125" ]

[ "Since the three side lengths of a right-angled triangle form an arithemetic sequence and must include 60 , then the three side lengths are $60,60+d, 60+2 d$ or $60-d, 60,60+d$ or $60-2 d, 60-d, 60$, for some $d \\geq 0$.\n\nFor a triangle with sides of length $60,60+d, 60+2 d$ to be right-angled, by the Pythagorean Theorem, the following equivalent equations must be true:\n\n$$\n\\begin{aligned}\n60^{2}+(60+d)^{2} & =(60+2 d)^{2} \\\\\n3600+3600+120 d+d^{2} & =3600+240 d+4 d^{2} \\\\\n0 & =3 d^{2}+120 d-3600 \\\\\n0 & =d^{2}+40 d-1200 \\\\\n0 & =(d+60)(d-20)\n\\end{aligned}\n$$\n\n(Note that, since $d \\geq 0$, then $60+2 d$ must be the hypotenuse of the triangle.) Since $d \\geq 0$, then $d=20$, which gives the triangle with side lengths $60,80,100$.\n\nThe longest side length is the hypotenuse and the shorter two sides meet at right angles, giving an area of $\\frac{1}{2}(60)(80)=2400$.\n\n\n\nFor a triangle with sides of length $60-d, 60,60+d$ to be right-angled, by the Pythagorean Theorem, the following equivalent equations must be true:\n\n$$\n\\begin{aligned}\n(60-d)^{2}+60^{2} & =(60+d)^{2} \\\\\n3600-120 d+d^{2}+3600 & =3600+120 d+d^{2} \\\\\n3600 & =240 d \\\\\nd & =15\n\\end{aligned}\n$$\n\nSince $d \\geq 0$, then $d=15$ is admissible, which gives the triangle with side lengths 45, 60,75. Using a similar analysis, the area of this triangle is $\\frac{1}{2}(45)(60)=1350$.\n\nFor a triangle with sides of length $60-2 d, 60-d, 60$ to be right-angled, by the Pythagorean Theorem, the following equivalent equations must be true:\n\n$$\n\\begin{aligned}\n(60-2 d)^{2}+(60-d)^{2} & =60^{2} \\\\\n3600-240 d+4 d^{2}+3600-120 d+d^{2} & =3600 \\\\\n5 d^{2}-360 d+3600 & =0 \\\\\nd^{2}-72 d+720 & =0 \\\\\n(d-60)(d-12) & =0\n\\end{aligned}\n$$\n\nSince $d \\geq 0$, then $d=60$ or $d=12$, which give possible side lengths of $-60,0,60$ (which do not form a triangle) and 36,48,60 (which do form a triangle).\n\nUsing a similar analysis, the area of this triangle is $\\frac{1}{2}(36)(48)=864$.\n\nTherefore, the possible values for the area of such a triangle are 2400, 1350, and 864.", "Suppose that a triangle has side lengths in arithemetic sequence.\n\nThen the side lengths can be written as $a-d, a, a+d$ for some $a>0$ and $d \\geq 0$.\n\nNote that $a-d \\leq a \\leq a+d$.\n\nFor such a triangle to be right-angled, by the Pythagorean Theorem, the following equivalent equations are true:\n\n$$\n\\begin{aligned}\n(a-d)^{2}+a^{2} & =(a+d)^{2} \\\\\na^{2}-2 a d+d^{2}+a^{2} & =a^{2}+2 a d+d^{2} \\\\\na^{2} & =4 a d\n\\end{aligned}\n$$\n\nSince $a>0$, then $a=4 d$, and so the side lengths of the triangle are $a-d=3 d, a=4 d$, and $a+d=5 d$ for some $d \\geq 0$.\n\n(Note that such triangles are all similar to the 3-4-5 triangle.)\n\nIf such a triangle has 60 as a side length, then there are three possibilities:\n\n(i) $3 d=60$ : This gives $d=20$ and side lengths $60,80,100$.\n\nSince the triangle is right-angled and its hypotenuse has length 100, then its area will equal $\\frac{1}{2}(60)(80)=2400$.\n\n(ii) $4 d=60$ : This gives $d=15$ and side lengths $45,60,75$.\n\nIn a similar way to case (i), its area will equal $\\frac{1}{2}(45)(60)=1350$.\n\n(iii) $5 d=60$ : This gives $d=12$ and side lengths $36,48,60$.\n\nIn a similar way to case (i), its area will equal $\\frac{1}{2}(36)(48)=864$.\n\nTherefore, the possible values for the area of such a triangle are 2400, 1350, and 864 ." ]

[ "2400, 1350, 864" ]

[ "Suppose that Amrita paddles the kayak for $p \\mathrm{~km}$ and swims for $s \\mathrm{~km}$.\n\nSince Amrita leaves the kayak in the lake and it does not move, then Zhang swims $p \\mathrm{~km}$ and paddles the kayak for $s \\mathrm{~km}$.\n\nNote that each paddles at $7 \\mathrm{~km} / \\mathrm{h}$ and each swims at $2 \\mathrm{~km} / \\mathrm{h}$ and each takes exactly 90 minutes (or 1.5 hours) to complete the trip.\n\nIf $s<p$, then Amrita would paddle farther and swim less distance than Zhang and so would reach the other side in less time than Zhang.\n\nIf $s>p$, then Zhang would paddle farther and swim less distance than Amrita and so would reach the other side in less time than Amrita.\n\nSince they each take 90 minutes, then we must have $s=p$.\n\nAlternatively, since each paddles at $7 \\mathrm{~km} / \\mathrm{h}$ and each swims at $2 \\mathrm{~km} / \\mathrm{h}$ and each takes exactly 90 minutes (or 1.5 hours) to complete the trip, then we obtain the two equations\n\n$$\n\\frac{p}{7}+\\frac{s}{2}=1.5 \\quad \\frac{p}{2}+\\frac{s}{7}=1.5\n$$\n\nUsing the fact that the right sides of these equations are equal, we obtain\n\n$$\n\\begin{aligned}\n\\frac{p}{7}+\\frac{s}{2} & =\\frac{s}{7}+\\frac{p}{2} \\\\\n\\frac{s}{2}-\\frac{s}{7} & =\\frac{p}{2}-\\frac{p}{7} \\\\\ns\\left(\\frac{1}{2}-\\frac{1}{7}\\right) & =p\\left(\\frac{1}{2}-\\frac{1}{7}\\right) \\\\\ns & =p\n\\end{aligned}\n$$\n\nTherefore, $\\frac{p}{7}+\\frac{p}{2}=1.5$ or $\\frac{9}{14} p=1.5=\\frac{3}{2}$ and so $p=\\frac{7}{3}$.\n\nFor Amrita to paddle these $\\frac{7}{3} \\mathrm{~km}$ at $7 \\mathrm{~km} / \\mathrm{h}$, it takes $\\frac{7 / 3}{7}=\\frac{1}{3}$ hour, or 20 minutes.\n\nFor Zhang to swim these $\\frac{7}{3} \\mathrm{~km}$ at $2 \\mathrm{~km} / \\mathrm{h}$, it takes $\\frac{7 / 3}{2}=\\frac{7}{6}$ hour, or 70 minutes.\n\nThe kayak is not being paddled for the period of time from when Amrita stops paddling to the time when Zhang stops swimming, which is a period of $70-20=50$ minutes.", "Let $t_{1}$ hours be the length of time during which Amrita paddles and Zhang swims.\n\nLet $t_{2}$ hours be the length of time during which Amrita swims and Zhang swims; the kayak is not moving during this time.\n\nLet $t_{3}$ hours be the length of time during which Amrita swims and Zhang paddles.\n\nLet $d \\mathrm{~km}$ be the total distance across the lake.\n\nSince Amrita paddles at $7 \\mathrm{~km} / \\mathrm{h}$ and swims at $2 \\mathrm{~km} / \\mathrm{h}$, then $7 t_{1}+2 t_{2}+2 t_{3}=d$.\n\nSince Zhang paddles at $7 \\mathrm{~km} / \\mathrm{h}$ and swims at $2 \\mathrm{~km} / \\mathrm{h}$, then $2 t_{1}+2 t_{2}+7 t_{3}=d$.\n\nSince the kayak travels at $7 \\mathrm{~km} / \\mathrm{h}$ and does not move while both Amrita and Zhang are swimming, then $7 t_{1}+0 t_{2}+7 t_{3}=d$.\n\nSince Amrita and Zhang each take 90 minutes ( $\\frac{3}{2}$ hours) to cross the lake, then the total time gives $t_{1}+t_{2}+t_{3}=\\frac{3}{2}$.\n\nFrom $7 t_{1}+2 t_{2}+2 t_{3}=d$ and $2 t_{1}+2 t_{2}+7 t_{3}=d$, we obtain $7 t_{1}+2 t_{2}+2 t_{3}=2 t_{1}+2 t_{2}+7 t_{3}$ or $5 t_{1}=5 t_{3}$ and so $t_{1}=t_{3}$.\n\nSince $7 t_{1}+2 t_{2}+2 t_{3}=d$ and $7 t_{1}+0 t_{2}+7 t_{3}=d$ and $t_{1}=t_{3}$, then $7 t_{1}+2 t_{2}+2 t_{1}=7 t_{1}+7 t_{1}$ or $2 t_{2}=5 t_{1}$ or $t_{2}=\\frac{5}{2} t_{1}$.\n\nSince $t_{1}+t_{2}+t_{3}=\\frac{3}{2}$, then $t_{1}+\\frac{5}{2} t_{1}+t_{1}=\\frac{3}{2}$ or $\\frac{9}{2} t_{1}=\\frac{3}{2}$ and so $t_{1}=\\frac{1}{3}$.\n\nThus, $t_{2}=\\frac{5}{2} \\cdot \\frac{1}{3}=\\frac{5}{6}$ hours (or 50 minutes) is the period of time that the kayak is not moving." ]

[ "50" ]

[ "From the first equation, $x\\left(\\frac{1}{2}+y-2 x^{2}\\right)=0$, we obtain $x=0$ or $\\frac{1}{2}+y-2 x^{2}=0$.\n\nFrom the second equation, $y\\left(\\frac{5}{2}+x-y\\right)=0$, we obtain $y=0$ or $\\frac{5}{2}+x-y=0$.\n\nIf $x=0$, the first equation is satisified.\n\nFor the second equation to be true in this case, we need $y=0$ (giving the solution $(0,0)$ ) or $\\frac{5}{2}+0-y=0$. The second equation gives $y=\\frac{5}{2}$ (giving the solution $\\left(0, \\frac{5}{2}\\right)$ ).\n\nIf $y=0$, the second equation is satisified.\n\nFor the first equation to be true in this case, we need $x=0$ (giving the solution $(0,0)$ ) or $\\frac{1}{2}+0-2 x^{2}=0$. The second equation gives $x^{2}=\\frac{1}{4}$ or $x= \\pm \\frac{1}{2}$ (giving the solutions $\\left(\\frac{1}{2}, 0\\right)$ and $\\left.\\left(-\\frac{1}{2}, 0\\right)\\right)$.\n\nSo far, we have accounted for all solutions with $x=0$ or $y=0$.\n\nIf $x \\neq 0$ and $y \\neq 0$, then for both equations to be true, we need $\\frac{1}{2}+y-2 x^{2}=0$ (or $1+2 y-4 x^{2}=0$ ) and $\\frac{5}{2}+x-y=0$ ( or $5+2 x-2 y=0$ ).\n\nAdding these two equations, we obtain $6+2 x-4 x^{2}=0$.\n\nThis is equivalent to $2 x^{2}-x-3=0$ or $(2 x-3)(x+1)=0$, whose solutions are $x=\\frac{3}{2}$ and $x=-1$.\n\nThe equation $\\frac{5}{2}+x-y=0$ tells us that $y=x+\\frac{5}{2}$.\n\nIf $x=\\frac{3}{2}$, then $y=4$; if $x=-1$, then $y=\\frac{3}{2}$.\n\nTherefore, the complete list of pairs that satisfy the given system of equations is\n\n$$\n(x, y)=(0,0),\\left(0, \\frac{5}{2}\\right),\\left(\\frac{1}{2}, 0\\right),\\left(-\\frac{1}{2}, 0\\right),\\left(\\frac{3}{2}, 4\\right),\\left(-1, \\frac{3}{2}\\right)\n$$" ]

[ "$(0,0),(0, \\frac{5}{2}),(\\frac{1}{2}, 0),(-\\frac{1}{2}, 0),(\\frac{3}{2}, 4),(-1, \\frac{3}{2})$" ]

[ "Note that $x \\neq 1$ since 1 cannot be the base of a logarithm. This tells us that $\\log x \\neq 0$. Using the fact that $\\log _{a} b=\\frac{\\log b}{\\log a}$ and then using other logarithm laws, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n\\log _{4} x-\\log _{x} 16 & =\\frac{7}{6}-\\log _{x} 8 \\\\\n\\frac{\\log x}{\\log 4}-\\frac{\\log 16}{\\log x} & =\\frac{7}{6}-\\frac{\\log 8}{\\log x} \\quad(\\text { note that } x \\neq 1, \\text { so } \\log x \\neq 0) \\\\\n\\frac{\\log x}{\\log 4} & =\\frac{7}{6}+\\frac{\\log 16-\\log 8}{\\log x} \\\\\n\\frac{\\log x}{\\log \\left(2^{2}\\right)} & =\\frac{7}{6}+\\frac{\\log \\left(\\frac{16}{8}\\right)}{\\log x} \\\\\n\\frac{\\log x}{2 \\log 2} & =\\frac{7}{6}+\\frac{\\log 2}{\\log x} \\\\\n\\frac{1}{2}\\left(\\frac{\\log x}{\\log 2}\\right) & =\\frac{7}{6}+\\frac{\\log 2}{\\log x}\n\\end{aligned}\n$$\n\nLetting $t=\\frac{\\log x}{\\log 2}=\\log _{2} x$ and noting that $t \\neq 0$ since $x \\neq 1$, we obtain the following equations equivalent to the previous ones:\n\n$$\n\\begin{aligned}\n\\frac{t}{2} & =\\frac{7}{6}+\\frac{1}{t} \\\\\n3 t^{2} & =7 t+6 \\quad(\\text { multiplying both sides by } 6 t) \\\\\n3 t^{2}-7 t-6 & =0 \\\\\n(3 t+2)(t-3) & =0\n\\end{aligned}\n$$\n\nTherefore, the original equation is equivalent to $t=-\\frac{2}{3}$ or $t=3$.\n\nConverting back to the variable $x$, we obtain $\\log _{2} x=-\\frac{2}{3}$ or $\\log _{2} x=3$, which gives $x=2^{-2 / 3}$ or $x=2^{3}=8$." ]

[ "$2^{-2 / 3}$, $8$" ]

[ "There are $2^{10}=1024$ strings of ten letters, each of which is $A$ or $B$, because there are 2 choices for each of the 10 positions in the string.\n\nWe determine the number of these strings that do not include the \"substring\" $A B B A$ (that is, that do not include consecutive letters $A B B A$ ) by counting the number of strings that do include the substring $A B B A$ and subtracting this total from 1024.\n\nIf a string includes the substring $A B B A$, there are 7 possible positions in which this substring could start ( $A B B A x x x x x x, x A B B A x x x x x, \\ldots, \\operatorname{xxxxxxABBA).}$\n\nThere are 2 choices for each of the remaining 6 letters in such a string, so there are $7 \\cdot 2^{6}=448$ occurrences of the substring $A B B A$ among the 1024 strings.\n\nThis does not mean that there are 448 strings that contain the substring $A B B A$. Since $A B B A$ can appear multiple times in a single string, this total will count some strings more than once. (For example, the string $A B B A A A A B B A$ is included in both the first and seventh of these categories, so this string is counted twice.)\n\nSo we must \"correct\" this total of 448 by accounting for the strings in which $A B B A$ occurs more than once.\n\nWe note that, since two substrings of $A B B A$ can overlap in 0 letters (for example, $A B B A A B B A x x$ ) or in 1 letter (for example, $A B B A B B A x x x$ ), then the maximum number of times that the substring $A B B A$ can appear is 3 , and there is only one such string: $A B B A B B A B B A$.\n\nIf a string contains two copies of $A B B A$ that overlap, then it must be of one of the following forms:\n\n$A B B A B B A x x \\quad x A B B A B B A x x \\quad x x A B B A B B A x \\quad x x x A B B A B B A$\n\nSince there are 4 choices for the starting position of $A B B A B B A$ and 2 choices for each of the three unknown letters, then there are $4 \\cdot 2^{3}=32$ occurrences of $A B B A B B A$ among all of these strings.\n\nBut the string $A B B A B B A B B A$ is counted in each of the first and last categories, so we subtract 2 occurrences from this total to obtain 30 , the total number of strings of ten letters that included exactly two overlapping copies of $A B B A$. (We'll count the string $A B B A B B A B B A$ later.)\n\nIf a string contains exactly two substrings of $A B B A$ and these do not overlap, then it must be of one of the following forms:\n\n$$\n\\begin{array}{lll}\nA B B A A B B A x x & A B B A x A B B A x & A B B A x x A B B A \\\\\nx A B B A A B B A x & x A B B A x A B B A & x x A B B A A B B A\n\\end{array}\n$$\n\nSince there are 6 such forms and 2 choices for each of the 2 unknown letters in each case, then there appear to be $6 \\cdot 2^{2}=24$ such strings.\n\nThis total includes the string $A B B A B B A B B A$ in the third category, so we subtract 1 from this total to obtain 23 , the total number of strings of ten letters that include exactly two copies of $A B B A$ which do not overlap.\n\nSo there are 30 strings that contain exactly two overlapping substrings $A B B A, 23$ strings that contain exactly two non-overlapping substrings $A B B A$, and 1 string that contains exactly three substrings $A B B A$.\n\nTo get the total number of strings with one or more substrings $A B B A$ we take the total number of occurrences of $A B B A$ (448), subtract the number of strings with exactly two substrings $A B B A$ (since these were included twice in the original count), and subtract two times the number of strings with exactly three substrings $A B B A$ (since these were included three times in the original count).\n\nTherefore, there are $448-23-30-2 \\cdot 1=393$ strings that include at least one substring\n$A B B A$, and so there are $1024-393=631$ strings of ten letters that do not include the substring $A B B A$." ]

[ "631" ]

[ "Here, $k=10$ and so there are 10 balls in each bag.\n\nSince there are 10 balls in each bag, there are $10 \\cdot 10=100$ pairs of balls that can be chosen.\n\nLet $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen. To determine $P(10)$, we count the number of pairs $(a, b)$ for which $a b$ is divisible by 10 .\n\nIf the number of pairs is $m$, then $P(10)=\\frac{m}{100}$.\n\nFor $a b$ to be divisible by 10, at least one of $a$ and $b$ must be a multiple of 5 and at least one of $a$ and $b$ must be even.\n\nIf $a=10$ or $b=10$, then the pair $(a, b)$ gives a product $a b$ divisible by 10 .\n\nIn this case, we obtain the 19 pairs\n\n$$\n(a, b)=(1,10),(2,10), \\ldots,(9,10),(10,10),(10,9), \\ldots,(10,2),(10,1)\n$$\n\nIf neither $a$ nor $b$ equals 10 , then either $a=5$ or $b=5$ in order for $a$ or $b$ to be divisible by 5 .\n\nIn this case, the other of $a$ and $b$ must be even and not equal to 10. (We have already counted the pairs where $a=10$ or $b=10$.)\n\nIn this case, we obtain the 8 pairs\n\n$$\n(a, b)=(5,2),(5,4),(5,6),(5,8),(2,5),(4,5),(6,5),(8,5)\n$$\n\nFrom our work above, there are no additional pairs for which $a b$ is divisible by 10 .\n\nThus, there are $19+8=27$ pairs $(a, b)$ for which $a b$ is divisible by 10 , which means that $P(10)=\\frac{27}{100}$.\n\n(We note that we could have made a 10 by 10 table that listed all possible combinations of $a$ and $b$ and their product, from which we could obtain $P(10)$.)" ]

[ "$\\frac{27}{100}$" ]

[ "The sum of the terms in an arithmetic sequence is equal to the average of the first and last terms times the number of terms.\n\nIf $n$ is the number of terms in the sequence, then $\\frac{1}{2}(1+19) n=70$ or $10 n=70$ and so $n=7$.", "Let $n$ be the number of terms in the sequence and $d$ the common difference.\n\nSince the first term is 1 and the $n$th term equals 19 , then $1+(n-1) d=19$ and so $(n-1) d=18$.\n\nSince the sum of the terms in the sequence is 70 , then $\\frac{1}{2} n(1+1+(n-1) d)=70$.\n\nThus, $\\frac{1}{2} n(2+18)=70$ or $10 n=70$ and so $n=7$." ]

[ "7" ]

[ "Since the given equation is true for all values of $x$, then it is true for any particular value of $x$ that we try.\n\nIf $x=-3$, the equation becomes $a(-3+b(0))=2(3)$ or $-3 a=6$ and so $a=-2$.\n\nIf $x=0$, the equation becomes $-2(0+b(3))=2(6)$ or $-6 b=12$ and so $b=-2$.\n\nTherefore, $a=-2$ and $b=-2$.", "We expand both sides of the equation:\n\n$$\n\\begin{aligned}\na(x+b(x+3)) & =2(x+6) \\\\\na(x+b x+3 b) & =2 x+12 \\\\\na x+a b x+3 a b & =2 x+12 \\\\\n(a+a b) x+3 a b & =2 x+12\n\\end{aligned}\n$$\n\nSince this equation is true for all values of $x$, then the coefficients on the left side and right side must be equal, so $a+a b=2$ and $3 a b=12$.\n\nFrom the second equation, $a b=4$ so the first equation becomes $a+4=2$ or $a=-2$.\n\nSince $a b=4$, then $-2 b=4$ and so $b=-2$.\n\nThus, $a=b=-2$." ]

[ "-2,-2" ]

[ "The number of integers between 100 and 999 inclusive is $999-100+1=900$.\n\nAn integer $n$ in this range has three digits, say $a, b$ and $c$, with the hundreds digit equal to $a$.\n\nNote that $0 \\leq b \\leq 9$ and $0 \\leq c \\leq 9$ and $1 \\leq a \\leq 9$.\n\nTo have $a+b+c=24$, then the possible triples for $a, b, c$ in some order are $9,9,6 ; 9,8,7$; $8,8,8$. (There cannot be three 9's. If there are two 9's, the the other digit equals 6 . If there is one 9 , the second and third digits add to 15 but are both less than 9 , so must equal 8 and 7 . If there are zero 9's, the maximum for each digit is 8 , and so each digt must be 8 in order for the sum of all three to equal 24.)\n\nIf the digits are 9, 9 and 6, there are 3 arrangements: 996, 969, 699.\n\n\n\nIf the digits are 9, 8 and 7, there are 6 arrangements: 987, 978, 897, 879, 798, 789.\n\nIf the digits are 8,8 and 8 , there is only 1 arrangement: 888 .\n\nTherefore, there are $3+6+1=10$ integers $n$ in the range 100 to 999 with the sum of the digits of $n$ equal to 24 .\n\nThe required probability equals the number of possible values of $n$ with the sum of digits equal to 24 divided by the total number of integers in the range, or $\\frac{10}{900}=\\frac{1}{90}$." ]

[ "$\\frac{1}{90}$" ]

[ "Completing the square on the original parabola, we obtain\n\n$$\ny=x^{2}-2 x+4=x^{2}-2 x+1-1+4=(x-1)^{2}+3\n$$\n\nTherefore, the vertex of the original parabola is $(1,3)$.\n\nSince the new parabola is a translation of the original parabola and has $x$-intercepts 3 and 5 , then its equation is $y=1(x-3)(x-5)=x^{2}-8 x+15$.\n\nCompleting the square here, we obtain\n\n$$\ny=x^{2}-8 x+15=x^{2}-8 x+16-16+15=(x-4)^{2}-1\n$$\n\nTherefore, the vertex of the new parabola is $(4,-1)$.\n\nThus, the point $(1,3)$ is translated $p$ units to the right and $q$ units down to reach $(4,-1)$, so $p=3$ and $q=4$." ]

[ "3,4" ]

[ "First, we convert each of the logarithms to a logarithm with base 2:\n\n$$\n\\begin{aligned}\n1+\\log _{4} x & =1+\\frac{\\log _{2} x}{\\log _{2} 4}=1+\\frac{\\log _{2} x}{2}=1+\\frac{1}{2} \\log _{2} x \\\\\n\\log _{8} 4 x & =\\frac{\\log _{2} 4 x}{\\log _{2} 8}=\\frac{\\log _{2} 4+\\log _{2} x}{3}=\\frac{2}{3}+\\frac{1}{3} \\log _{2} x\n\\end{aligned}\n$$\n\nLet $y=\\log _{2} x$. Then the three terms are $y, 1+\\frac{1}{2} y$, and $\\frac{2}{3}+\\frac{1}{3} y$. Since these three are in geometric sequence, then\n\n$$\n\\begin{aligned}\n\\frac{y}{1+\\frac{1}{2} y} & =\\frac{1+\\frac{1}{2} y}{\\frac{2}{3}+\\frac{1}{3} y} \\\\\ny\\left(\\frac{2}{3}+\\frac{1}{3} y\\right) & =\\left(1+\\frac{1}{2} y\\right)^{2} \\\\\n\\frac{2}{3} y+\\frac{1}{3} y^{2} & =1+y+\\frac{1}{4} y^{2} \\\\\n8 y+4 y^{2} & =12+12 y+3 y^{2} \\\\\ny^{2}-4 y-12 & =0 \\\\\n(y-6)(y+2) & =0\n\\end{aligned}\n$$\n\nTherefore, $y=\\log _{2} x=6$ or $y=\\log _{2} x=-2$, which gives $x=2^{6}=64$ or $x=2^{-2}=\\frac{1}{4}$." ]

[ "$64,\\frac{1}{4}$" ]

[ "First, we note that $\\sqrt{50}=5 \\sqrt{2}$.\n\nNext, we note that $\\sqrt{2}+4 \\sqrt{2}=5 \\sqrt{2}$ and $2 \\sqrt{2}+3 \\sqrt{2}=5 \\sqrt{2}$.\n\nFrom the first of these, we obtain $\\sqrt{2}+\\sqrt{32}=\\sqrt{50}$.\n\nFrom the second of these, we obtain $\\sqrt{8}+\\sqrt{18}=\\sqrt{50}$.\n\nThus, $(a, b)=(2,32)$ and $(a, b)=(8,18)$ are solutions to the original equation.\n\n(We are not asked to justify why these are the only two solutions.)" ]

[ "(2,32), (8,18)" ]

[ "From the second equation, we note that $d \\neq 0$.\n\nRearranging this second equation, we obtain $c=k d$.\n\nSubstituting into the first equation, we obtain $k d+d=2000$ or $(k+1) d=2000$.\n\nSince $k \\geq 0$, note that $k+1 \\geq 1$.\n\nThis means that if $(c, d)$ is a solution, then $k+1$ is a divisor of 2000 .\n\nAlso, if $k+1$ is a divisor of 2000 , then the equation $(k+1) d=2000$ gives us an integer value of $d$ (which is non-zero) from which we can find an integer value of $c$ using the first equation.\n\nTherefore, the values of $k$ that we want to count correspond to the positive divisors of 2000.\n\nSince $2000=10 \\cdot 10 \\cdot 20=2^{4} \\cdot 5^{3}$, then 2000 has $(4+1)(3+1)=20$ positive divisors.\n\nThis comes from the fact that if $p$ and $q$ are distinct prime numbers then the positive integer $p^{a} \\cdot q^{b}$ has $(a+1)(b+1)$ positive divisors.\n\nWe could list these divisors as\n\n$$\n1,2,4,5,8,10,16,20,25,40,50,80,100,125,200,250,400,500,1000,2000\n$$\n\n\n\nif we did not know the earlier formula.\n\nSince 2000 has 20 positive divisors, then there are 20 values of $k$ for which the system of equations has at least one integer solution.\n\nFor example, if $k+1=8$, then $k=7$. This gives the system $c+d=2000$ and $\\frac{c}{d}=7$ which has solution $(c, d)=(1750,250)$." ]

[ "20" ]

[ "Using logarithm and exponent laws, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n2 \\log _{2}(x-1) & =1-\\log _{2}(x+2) \\\\\n2 \\log _{2}(x-1)+\\log _{2}(x+2) & =1 \\\\\n\\log _{2}\\left((x-1)^{2}\\right)+\\log _{2}(x+2) & =1 \\\\\n\\log _{2}\\left((x-1)^{2}(x+2)\\right) & =1 \\\\\n(x-1)^{2}(x+2) & =2^{1} \\\\\n\\left(x^{2}-2 x+1\\right)(x+2) & =2 \\\\\nx^{3}-3 x+2 & =2 \\\\\nx^{3}-3 x & =0 \\\\\nx\\left(x^{2}-3\\right) & =0\n\\end{aligned}\n$$\n\nand so $x=0$ or $x=\\sqrt{3}$ or $x=-\\sqrt{3}$.\n\nNote that if $x=0$, then $x-1=-1<0$ and so $\\log _{2}(x-1)$ is not defined. Thus, $x \\neq 0$. Note that if $x=-\\sqrt{3}$, then $x-1=-\\sqrt{3}-1<0$ and so $\\log _{2}(x-1)$ is not defined. Thus, $x \\neq-\\sqrt{3}$.\n\nIf $x=\\sqrt{3}$, we can verify that both logarithms in the original equation are defined and that the original equation is true. We could convince ourselves of this with a calculator or we could algebraically verify that raising 2 to the power of both sides gives the same number, so the expressions must actually be equal.\n\nTherefore, $x=\\sqrt{3}$ is the only solution." ]

[ "$\\sqrt{3}$" ]

[ "Let $a=f(f(x))$.\n\nThus, the equation $f(f(f(x)))=3$ is equivalent to $f(a)=3$.\n\nSince $f(a)=a^{2}-2 a$, then we obtain the equation $a^{2}-2 a=3$ which gives $a^{2}-2 a-3=0$ and $(a-3)(a+1)=0$.\n\nThus, $a=3$ or $a=-1$ which means that $f(f(x))=3$ or $f(f(x))=-1$.\n\nLet $b=f(x)$.\n\nThus, the equations $f(f(x))=3$ and $f(f(x))=-1$ become $f(b)=3$ and $f(b)=-1$.\n\nIf $f(b)=3$, then $b=f(x)=3$ or $b=f(x)=-1$ using similar reasoning to above when $f(a)=3$.\n\nIf $f(b)=-1$, then $b^{2}-2 b=-1$ and so $b^{2}-2 b+1=0$ or $(b-1)^{2}=0$ which means that $b=f(x)=1$.\n\nThus, $f(x)=3$ or $f(x)=-1$ or $f(x)=1$.\n\nIf $f(x)=3$, then $x=3$ or $x=-1$ as above.\n\nIf $f(x)=-1$, then $x=1$ as above.\n\nIf $f(x)=1$, then $x^{2}-2 x=1$ and so $x^{2}-2 x-1=0$.\n\nBy the quadratic formula,\n\n$$\nx=\\frac{-(-2) \\pm \\sqrt{(-2)^{2}-4(1)(-1)}}{2(1)}=\\frac{2 \\pm \\sqrt{8}}{2}=1 \\pm \\sqrt{2}\n$$\n\nTherefore, the solutions to the equation $f(f(f(x)))=3$ are $x=3,1,-1,1+\\sqrt{2}, 1-\\sqrt{2}$." ]

[ "$3,1,-1,1+\\sqrt{2}, 1-\\sqrt{2}$" ]

[ "Since $0<x<\\frac{\\pi}{2}$, then $0<\\cos x<1$ and $0<\\sin x<1$.\n\nThis means that $0<\\frac{3}{2} \\cos x<\\frac{3}{2}$ and $0<\\frac{3}{2} \\sin x<\\frac{3}{2}$. Since $3<\\pi$, then $0<\\frac{3}{2} \\cos x<\\frac{\\pi}{2}$ and $0<\\frac{3}{2} \\sin x<\\frac{\\pi}{2}$.\n\nIf $Y$ and $Z$ are angles with $0<Y<\\frac{\\pi}{2}$ and $0<Z<\\frac{\\pi}{2}$, then $\\cos Y=\\sin Z$ exactly when $Y+Z=\\frac{\\pi}{2}$. To see this, we could picture points $R$ and $S$ on the unit circle corresponding to the angles $Y$ and $Z$; the $x$-coordinate of $R$ is equal to the $y$-coordinate of $S$ exactly when the angles $Y$ and $Z$ are complementary.\n\nTherefore, the following equations are equivalent:\n\n$$\n\\begin{aligned}\n\\cos \\left(\\frac{3}{2} \\cos x\\right) & =\\sin \\left(\\frac{3}{2} \\sin x\\right) \\\\\n\\frac{3}{2} \\cos x+\\frac{3}{2} \\sin x & =\\frac{\\pi}{2} \\\\\n\\cos x+\\sin x & =\\frac{\\pi}{3} \\\\\n(\\sin x+\\cos x)^{2} & =\\frac{\\pi^{2}}{9} \\\\\n\\sin ^{2} x+2 \\sin x \\cos x+\\sin ^{2} x & =\\frac{\\pi^{2}}{9} \\\\\n2 \\sin x \\cos x+\\left(\\sin ^{2} x+\\cos ^{2} x\\right) & =\\frac{\\pi^{2}}{9} \\\\\n\\sin 2 x+1 & =\\frac{\\pi^{2}}{9} \\\\\n\\sin 2 x & =\\frac{\\pi^{2}-9}{9}\n\\end{aligned}\n$$\n\nTherefore, the only possible value of $\\sin 2 x$ is $\\frac{\\pi^{2}-9}{9}$." ]

[ "$\\frac{\\pi^{2}-9}{9}$" ]

[ "By definition, $f(2,5)=\\frac{2}{5}+\\frac{5}{2}+\\frac{1}{2 \\cdot 5}=\\frac{2 \\cdot 2+5 \\cdot 5+1}{2 \\cdot 5}=\\frac{4+25+1}{10}=\\frac{30}{10}=3$." ]

[ "3" ]

[ "By definition, $f(a, a)=\\frac{a}{a}+\\frac{a}{a}+\\frac{1}{a^{2}}=2+\\frac{1}{a^{2}}$.\n\nFor $2+\\frac{1}{a^{2}}$ to be an integer, it must be the case that $\\frac{1}{a^{2}}$ is an integer.\n\nFor $\\frac{1}{a^{2}}$ to be an integer and since $a^{2}$ is an integer, $a^{2}$ needs to be a divisor of 1 .\n\nSince $a^{2}$ is positive, then $a^{2}=1$.\n\nSince $a$ is a positive integer, then $a=1$.\n\nThus, the only positive integer $a$ for which $f(a, a)$ is an integer is $a=1$." ]

[ "1" ]

[ "On her first two turns, Brigitte either chooses two cards of the same colour or two cards of different colours. If she chooses two cards of different colours, then on her third turn, she must choose a card that matches one of the cards that she already has.\n\nTherefore, the game ends on or before Brigitte's third turn.\n\nThus, if Amir wins, he wins on his second turn or on his third turn. (He cannot win on his first turn.)\n\nFor Amir to win on his second turn, the second card he chooses must match the first card that he chooses.\n\nOn this second turn, there will be 5 cards in his hand, of which 1 matches the colour of the first card that he chose.\n\nTherefore, the probability that Amir wins on his second turn is $\\frac{1}{5}$.\n\nNote that there is no restriction on the first card that he chooses or the first card that Brigitte chooses.\n\nFor Amir to win on his third turn, the following conditions must be true: (i) the colour of the second card that he chooses is different from the colour of the first card that he chooses, (ii) the colour of the second card that Brigitte chooses is different from the colour of the first card that she chooses, and (iii) the colour of the third card that Amir chooses matches the colour of one of the first two cards.\n\nThe probability of (i) is $\\frac{4}{5}$, since he must choose any card other than the one that matches the first one.\n\nThe probability of (ii) is $\\frac{2}{3}$, since Brigitte must choose either of the cards that does not match her first card.\n\nThe probability of (iii) is $\\frac{2}{4}$, since Amir can choose either of the 2 cards that matches one of the first two cards that he chose.\n\nAgain, the cards that Amir and Brigitte choose on their first turns do not matter.\n\nThus, the probability that Amir wins on his third turn is $\\frac{4}{5} \\cdot \\frac{2}{3} \\cdot \\frac{2}{4}$ which equals $\\frac{4}{15}$.\n\nFinally, the probabilty that Amir wins the game is thus $\\frac{1}{5}+\\frac{4}{15}$ which equals $\\frac{7}{15}$." ]

[ "$\\frac{7}{15}$" ]

[ "Calculating some terms, $t_{1}=1, t_{2}=-1, t_{3}=0, t_{4}=\\frac{-1}{3}, t_{5}=0, t_{6}=\\frac{-1}{5}$ etc.\n\nBy pattern recognition, $t_{1998}=\\frac{-1}{1997}$.", "$$\n\\begin{aligned}\nt_{1998} & =\\frac{1995}{1997} t_{1996}=\\frac{1995}{1997} \\times \\frac{1993}{1995} t_{1994} \\\\\n& =\\frac{1995}{1997} \\cdot \\frac{1993}{1995} \\cdot \\frac{1991}{1993} \\cdots \\frac{3}{5} \\cdot \\frac{1}{3} t_{2} \\\\\n& =\\frac{-1}{1997}\n\\end{aligned}\n$$" ]

[ "$\\frac{-1}{1997}$" ]

[ "This is an arithmetic sequence in which $a=548$ and $d=-7$.\n\nTherefore, $S_{n}=\\frac{n}{2}[2(548)+(n-1)(-7)]=\\frac{n}{2}[-7 n+1103]$.\n\nWe now want $\\frac{n}{2}(-7 n+1103)<0$.\n\nSince $n>0,-7 n+1103<0$\n\n$$\nn>157 \\frac{4}{7}\n$$\n\nTherefore the smallest value of $n$ is 158 .", "For this series we want, $\\sum_{k=1}^{n} t_{k}<0$, or $\\sum_{k=1}^{n}(555-7 k)<0$.\n\nRewriting, $555 n-7 \\frac{(n)(n+1)}{2}<0$\n\n$$\n\\begin{aligned}\n1110 n-7 n^{2}-7 n & <0 \\\\\n7 n^{2}-1103 n & >0 \\\\\n\\text { or, } n & >\\frac{1103}{7} .\n\\end{aligned}\n$$\n\nThe smallest value of $n$ is 158 .", "We generate the series as $548,541,534, \\ldots, 2,-5, \\ldots,-544,-551$.\n\nIf we pair the series from front to back the sum of each pair is -3 .\n\nIncluding all the pairs $548-551,541-544$ and so on there would be 79 pairs which give a sum of -237 .\n\nIf the last term, -551 , were omitted we would have a positive sum.\n\nTherefore we need all 79 pairs or 158 terms." ]

[ "158" ]

[ "Subtracting,\n\n$$\n\\begin{array}{r}\nx^{2}-x y+8=0 \\\\\nx^{2}-8 x+y=0 \\\\\n\\hline-x y+8 x+8-y=0 \\\\\n8(1+x)-y(1+x)=0 \\\\\n(8-y)(1+x)=0 \\\\\ny=8 \\text { or } x=-1\n\\end{array}\n$$\n\n\n\nIf $y=8$, both equations become $x^{2}-8 x+8=0, x=4 \\pm 2 \\sqrt{2}$.\n\nIf $x=-1$ both equations become $y+9=0, y=-9$.\n\nThe solutions are $(-1,-9),(4+2 \\sqrt{2}, 8)$ and $(4-2 \\sqrt{2}, 8)$.", "If $x^{2}-x y+8=0, y=\\frac{x^{2}+8}{x}$.\n\nAnd $x^{2}-8 x+y=0$ implies $y=8 x-x^{2}$.\n\nEquating, $\\frac{x^{2}+8}{x}=8 x-x^{2}$\n\n$$\n\\text { or, } x^{3}-7 x^{2}+8=0 \\text {. }\n$$\n\nBy inspection, $x=-1$ is a root.\n\nBy division, $x^{3}-7 x^{2}+8=(x+1)\\left(x^{2}-8 x+8\\right)$.\n\nAs before, the solutions are $(-1,-9),(4 \\pm 2 \\sqrt{2}, 8)$." ]

[ "$(-1,-9),(4+2 \\sqrt{2}, 8),(4-2 \\sqrt{2}, 8)$" ]