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[ "For each digit $a=1,2,\\ldots,9$ there are $10\\cdot10$ (ways of choosing $b$ and $c$ ) palindromes. So the $a$ s contribute $(1+2+\\cdots+9)(100)(10^4+1)$ to the sum.\nFor each digit $b=0,1,2,\\ldots,9$ there are $9\\cdot10$ (since $a \\neq 0$ ) palindromes. So the $b$ s contribute $(0+1+2+\\cdots+9)(90)(10^3+10)$ to the sum.\nSimilarly, for each $c=0,1,2,\\ldots,9$ there are $9\\cdot10$ palindromes, so the $c$ contributes $(0+1+2+\\cdots+9)(90)(10^2)$ to the sum.\nIt just so happens that \\[(1+2+\\cdots+9)(100)(10^4+1)+(1+2+\\cdots+9)(90)(10^3+10)+(1+2+\\cdots+9)(90)(10^2)=49500000\\] so the sum of the digits of the sum is $\\boxed{18}$", "Notice that $10001+ 99999 = 110000.$ In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is $110000.$ We have $9\\cdot 10\\cdot 10$ palindromes, or $450$ pairs of palindromes summing to $110000.$ Performing the multiplication gives $49500000$ , so the sum $\\boxed{18}$", "As shown above, there are a total of $900$ five-digit palindromes. We can calculate their sum by finding the expected value of a randomly selected palindrome satisfying the conditions given, then multiplying it by $900$ to get our sum. The expected value for the ten-thousands and the units digit is $\\frac{1+2+3+\\cdots+9}{9}=5$ , and the expected value for the thousands, hundreds, and tens digit is $\\frac{0+1+2+\\cdots+9}{10}=4.5$ . Therefore our expected value is $5\\times10^4+4.5\\times10^3+4.5\\times10^2+4.5\\times10^1+5\\times10^0=55,\\!000$ . Since the question asks for the sum of the digits of the resulting sum, we do not need to keep the trailing zeros of either $55,\\!000$ or $900$ . Thus we only need to calculate $55\\times9=495$ , and the desired sum is $\\boxed{18}$", "First, allow $a$ to be zero, and then subtract by how much we overcount. We'll also sum each palindrome with its $\\textit{complement}$ . If $\\overline{abcba}$ (the line means a, b, and c are digits and $abcba\\ne a\\cdot b\\cdot c\\cdot b\\cdot a$ ) is a palindrome, then its complement is $\\overline{defed}$ where $d=9-a$ $e=9-b$ $f=9-c$ . Notice how every palindrome has a unique compliment, and that the sum of a palindrome and its complement is $99999$ . Therefore, the sum of our palindromes is $99999\\times (10^3/2)$ . (There are $10^3/2$ pairs.)\nHowever, we have overcounted, as something like $05350$ $\\textit{isn't}$ a palindrome by the problem's definition, but we've still included it. So we must subtract the sum of numbers in the form $\\overline{0nmn0}$ . By the same argument as before, these sum to $9990\\times (10^2/2)$ . Therefore, the sum that the problem asks for is:\n\\[500\\times99999-50\\times 9990\\] \\[=500\\times99999-500\\times 999\\] \\[=500(99999-999)\\] \\[=500\\times 99000\\]\nSince all we care about is the sum of the digits, we can drop the $0$ 's.\n\\[5\\times99\\] \\[=5\\times(100-1)\\] \\[=495\\]\nAnd finally, $4+9+5=\\boxed{18}$" ]

[ "The Martian can pull out $12$ socks, $4$ of each color, without having $5$ of the same kind yet. However, the next one he pulls out must be the fifth of one of the colors so he must remove $\\boxed{13}$ socks." ]

[ "The broken flagpole forms a right triangle with legs $1$ and $x$ , and hypotenuse $5-x$ . The Pythagorean theorem now states that $1^2 + x^2 = (5-x)^2$ , hence $10x = 24$ , and $x=\\boxed{2.4}$", "Let $AB$ represent the flagpole in the diagram above. After the flagpole breaks at point $D$ , its tip lies at point $C$ . Since none of the flagpole is destroyed, we know that $DA=DC$ . Therefore, triangle $\\triangle ADC$ is isosceles.\nDraw the altitude $DE \\perp AC$ . Since $\\triangle ADC$ is isosceles, we know that $AE = EC$ . Also note that $\\triangle AED \\sim \\triangle ABC$ . Therefore, \\begin{align*} AD &= AE \\times \\frac{AD}{AE} \\\\ &= \\frac{AC}{2} \\times \\frac{AC}{AB} \\\\ &= \\frac{AC^2}{2 AB} \\\\ &= \\frac{AB^2 + BC^2}{2 AB} \\end{align*}\nSince $AB = 5$ and $BC = 1$ , we have that $AD = \\frac{5^2 + 1^2}{2 \\cdot 5} = 2.6$ , and thus $x = AB - AD = \\boxed{2.4}$" ]

[ " Set the common radius to $r$ . First, take the cross section of the sphere sitting in the hole of radius $1$ . If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse $r$ and base $1$ . Therefore, the height of this circle outside of the hole is $\\sqrt{r^2-1}$\nThe other circle follows similarly for a height (outside the hole) of $\\sqrt{r^2-4}$ . Now, if we take the cross section of the entire board, essentially making it two-dimensional, we can connect the centers of the two spheres, then form another right triangle with base $7$ , as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is $\\sqrt{r^2-1} - \\sqrt{r^2-4}$ . Now we can set up an equation in terms of $r$ with the Pythagorean theorem: \\[\\left(\\sqrt{r^2-1} - \\sqrt{r^2-4}\\right)^2 + 7^2 = (2r)^2.\\] Simplifying a few times, \\begin{align*} r^2 - 1 - 2\\left(\\sqrt{(r^2-1)(r^2-4)}\\right) + r^2 - 4 + 49 &= 4r^2 \\\\ 2r^2-44 &= -2\\left(\\sqrt{(r^2-1)(r^2-4)}\\right) \\\\ 22-r^2 &= \\left(\\sqrt{r^4 - 5r^2 + 4}\\right) \\\\ r^4 -44r^2 + 484 &= r^4 - 5r^2 + 4 \\\\ 39r^2&=480 \\\\ r^2&=\\frac{480}{39} = \\frac{160}{13}. \\end{align*} Therefore, our answer is $\\boxed{173}$" ]

[ "Let the total amount of flowers be $x$ . Thus, the number of pink flowers is $0.6x$ , and the number of red flowers is $0.4x$ . The number of pink carnations is $\\frac{2}{3}(0.6x) = 0.4x$ and the number of red carnations is $\\frac{3}{4}(0.4x) = 0.3x$ . Summing these, the total number of carnations is $0.4x+0.3x=0.7x$ . Dividing, we see that $\\frac{0.7x}{x} = 0.7 = \\boxed{70}$", "We have $\\dfrac15$ for pink roses, $1-\\dfrac6{10}=\\dfrac4{10}=\\dfrac25$ red flowers, $\\dfrac6{10}-\\dfrac15=\\dfrac35-\\dfrac15=\\dfrac25$ pink carnations, $\\dfrac25\\cdot \\dfrac34=\\dfrac6{20}=\\dfrac3{10}$ red carnations we add them up to get $\\dfrac25+\\dfrac3{10}=\\dfrac7{10}=70\\%$ so our final answer is 70% or $\\boxed{70}$" ]

[ "If the Cougars won by a margin of 14 points, then the Panthers' score would be half of (34-14). That's 10 $\\Rightarrow \\boxed{10}$", "Let the Panthers' score be $x$ . The Cougars then scored $x+14$ . Since the teams combined scored $34$ , we get $x+x+14=34 \\\\ \\rightarrow 2x+14=34 \\\\ \\rightarrow 2x=20 \\\\ \\rightarrow x = 10$\nand the answer is $\\boxed{10}$" ]

[ "Let $x$ be the number of points scored by the Cougars, and $y$ be the number of points scored by the Panthers. The problem is asking for the value of $y$ \\begin{align*} x+y &= 34 \\\\ x-y &= 14 \\\\ 2x &= 48 \\\\ x &= 24 \\\\ y &= \\boxed{10}" ]

[ "Let the arithmetic sequence be $a,a+d,a+2d,a+3d$ and the geometric sequence be $b,br,br^2,br^3.$\nWe are given that \\begin{align*} a+b&=57, \\\\ a+d+br&=60, \\\\ a+2d+br^2&=91, \\end{align*} and we wish to find $a+3d+br^3.$\nSubtracting the first equation from the second and the second equation from the third, we get \\begin{align*} d+b(r-1)&=3, \\\\ d+br(r-1)&=31. \\end{align*} Subtract these results, we get \\[b(r-1)^2=28.\\]\nNote that either $b=28$ or $b=7.$ We proceed with casework:\nTherefore, The answer is $a+3d+br^3=17+189=\\boxed{206}.$", "Start similarly to Solution 1 and deduce the three equations \\begin{align*} a+b&=57, \\\\ a+d+br&=60, \\\\ a+2d+br^2&=91. \\end{align*} Then, add the last two equations and take away the first equation to get $a+3d+br^2+br-b=94$ We can solve for this in terms of what we want: $a+3d=-br^2-br+b+94$\nWe're looking for $a+3d+br^3$ . We can substitute our value of $a+3d$ in here to get \\[br^3-br^2-br+b+94=b(r+1)(r-1)(r-1)+94.\\] Since our sequence only has positive integers we can now check by the answer choices. For each answer choice, we can subtract $94$ and factor it to see if it has a perfect square factor and at least one other factor and those should differ by $2$ \\begin{alignat*}{8} \\textbf{(A)} \\ 190-94&=96&&=2^5\\cdot3, \\\\ \\textbf{(B)} \\ 194-94&=100&&=2^2\\cdot5^2, \\\\ \\textbf{(C)} \\ 198-94&=104&&=2^3\\cdot13, \\\\ \\textbf{(D)} \\ 202-94&=108&&=2^2\\cdot3^3, \\\\ \\textbf{(E)} \\ 206-94&=112&&=2^4\\cdot7. \\end{alignat*} From this, the only possible answer choices are $\\textbf{(A)}$ and $\\textbf{(E)}$ , where $r=3$ . To solve for $b$ , we look back to the given equations above.\nWe are looking for $a+3d+27b$ . If $\\textbf{(A)}$ were the answer, then we know that $a$ would have to be divisible by $3$ and $b$ would equal $6$ . Looking at our second equation, if this were the case, then $d$ would also have to be divisible by $3$ . However, this contradicts the third equation, as all variables are divisible by $3$ , but their sum isn't. So, $\\boxed{206}$ is our answer." ]

[ "First of all, it is easy to see by induction that for any $P(x,y)$ in the frog's jump sequence, $x+y$ will be a multiple of $3$ and $x-y$ will be a multiple of $5.$ The base case $(x,y) = (0,0)$ obviously satisfies the constraints and if $x+y = 3n$ and $x-y = 5m,$ any of the four transformations will sustain this property:\n\\begin{align*} (x+7)+(y+2) = x+y+9 \\rightarrow 3(n+3) &\\text{ and } (x+7)-(y+2) = x-y+5 \\rightarrow 5(m+1)\\\\ (x+2)+(y+7) = x+y+9 \\rightarrow 3(n+3) &\\text{ and } (x+2)-(y+7) = x-y-5 \\rightarrow 5(m-1)\\\\ (x-5)+(y-10) = x+y-15 \\rightarrow 3(n-5) &\\text{ and } (x-5)-(y-10) = x-y+5 \\rightarrow 5(m+1)\\\\ (x-10)+(y-5) = x+y-15 \\rightarrow 3(n-5) &\\text{ and } (x-10)-(y-5) = x-y-5 \\rightarrow 5(m-1).\\\\ \\end{align*} So we know that any point the frog can reach will satisfy $x+y = 3n$ and $x-y = 5m.$\n$\\textbf{Lemma:}$ Any point $(x,y)$ such that there exists 2 integers $m$ and $n$ that satisfy $x+y = 3n$ and $x-y = 5m$ is reachable.\n$\\textbf{Proof:}$ Denote the total amounts of each specific transformation in the frog's jump sequence to be $a,$ $b,$ $c,$ and $d$ respectively. Then\n$x=7a+2b-5c-10d$\n$y=2a+7b-10c-5d$\n$x+y = 9(a+b)-15(c+d) = 3n$ , and\n$x-y = 5(a-b)+5(c-d) = 5m$\ntogether must have integral solutions. But\n$3(a+b)-5(c+d) = n$ implies\n$(c+d) \\equiv n \\mod 3$ and thus\n$(a+b) = \\lfloor{n/3}\\rfloor + 2(c+d).$\nSimilarly, $(a-b)+(c-d) = m$ implies that $(a-b)$ and $(c-d)$ have the same parity. Now in order for an integral solution to exist, there must always be a way to ensure that the pairs $(a+b)$ and $(a-b)$ and $(c+d)$ and $(c-d)$ have identical parities. The parity of $(a+b)$ is completely dependent on $n,$ so the parities of $(a-b)$ and $(c-d)$ must be chosen to match this value. But the parity of $(c+d)$ can then be adjusted by adding or subtracting $3$ until it is identical to the parity of $(c-d)$ as chosen before, so we conclude that it is always possible to find an integer solution for $(a,b,c,d)$ and thus any point that satisfies $x+y = 3n$ and $x-y = 5m$ can be reached by the frog.\nTo count the number of such points in the region $|x| + |y| \\le 100,$ we first note that any such point will lie on the intersection of one line of the form $y=x-5m$ and another line of the form $y=-x+3n.$ The intersection of two such lines will yield the point $\\left(\\frac{3n+5m}{2},\\frac{3n-5m}{2}\\right),$ which will be integral if and only if $m$ and $n$ have the same parity. Now since $|x| + |y| = |x \\pm y|,$ we find that\n\\begin{align*} |x + y| = |3n| \\le 100 &\\rightarrow -33 \\le n \\le 33\\\\ |x - y| = |5m| \\le 100 &\\rightarrow -20 \\le m \\le 20. \\end{align*}\nSo there are $34$ possible odd values and $33$ possible even values for $n,$ and $20$ possible odd values and $21$ possible even values for $m.$ Every pair of lines described above will yield a valid accessible point for all pairs of $m$ and $n$ with the same parity, and the number of points $M$ is thus $34 \\cdot 20 + 33 \\cdot 21 = 1373 \\rightarrow \\boxed{373}$" ]

[ "Another way would be to use a table representing the number of ways to reach a certain number\n$\\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\\\ \\hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\\\ \\end{tabular}$\nHow we came with each value is to just add in the number of ways that we can reach that number from previous numbers. For example, for $26$ , we can reach it from $13, 15, 18, 21, 24$ , so we add all those values to get the value for $26$ . For $27$ , it is only reachable from $24$ or $26$ , so we have $29 + 6 = 35$\nThe answer for $39$ can be computed in a similar way to get $35 * 4 + 29 = \\boxed{169}$", "Let $f(n)$ be the number of ways one can get to $39$ starting at position $n.$ We wish to compute $f(0).$ Now it's just a long simplifications until you get to $f(36) = 1.$ We have \\[f(0) = f(3) + f(13) = f(6) +2f(13) + f(9) + 3f(12) + f(12) + 4f(12) + f(15) + 5f(12).\\]\nMost of these steps are valid since at any $n$ that is a multiple of $3$ we can either go to the next multiple of $3$ or we can skip to the next multiple of $13$ which is simply $13.$\nFrom these equations we have deduced $f(0) = f(15) + 5f(12).$ Continuing we have \\[f(15) + 5f(12) = f(15) + 5(f(26) +f(15) = 5f(26) + 6f(15) = 5f(26)+ 6f(26) + 6f(18) = 5f(26) + 12f(26) + 6f(21) = 5f(26) + 18f(26) + 6f(24) = 5f(26) + 24f(26) + 6f(27) = 29f(26) + 6f(27).\\]\nFinally, note that $f(26) = 1 + f(27) = 2 + f(30) = 3+f(33) = 4+f(36) = 5$ since at any point we can either go to the next multiple of $3$ or go to the next multiple of $13$ which happens to be $39.$ Therefore $f(26) = 5.$ Similarly we find $f(27) = 1+f(30) = 2 + f(33) = 3+f(36) = 4$ so the end answer is $5 \\cdot 29 + 6 \\cdot 4 = \\boxed{169}.$", "Another way you can visualize the problem is by thinking of points $13$ $26$ , and $39$ as planets and all multiples of 3 as points at which your spaceship can jump to hyperspace. Given that you wish to visit planet $39$ , you can choose to visit planets $13$ or $26$ along the way.\nCase 1: Neither\nThere are $4$ ways to jump to $39$ from hyperspace.\nCase 2: Only planet $13$\nThere are $5$ ways to jump to planet $13$ and $4$ ways to jump to planet $39$ $20$ ways total.\nCase 3: Only planet $26$\nThere are $4$ ways to jump to planet $26$ and $5$ ways to jump to planet $39$ $20$ ways total.\nCase 4: Both\nThere are $5$ ways to jump to $13$ $5$ ways to jump to $26$ , and $5$ ways to jump to $39$ $125$ ways total.\nTherefore, there are $4+20+20+125=\\boxed{169}$ ways to jump to 39 through various journeys in hyperspace." ]

[ "We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$ . This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$ , recording down the number of ways to get to each point recursively.\n$(0,0): 1$\n$(1,0)=(0,1)=1$\n$(2,0)=(0, 2)=2$\n$(3,0)=(0, 3)=3$\n$(4,0)=(0, 4)=5$\n$(1,1)=2$ $(1,2)=(2,1)=5$ $(1,3)=(3,1)=10$ $(1,4)=(4,1)= 20$\n$(2,2)=14, (2,3)=(3,2)=32, (2,4)=(4,2)=71$\n$(3,3)=84, (3,4)=(4,3)=207$\n$(4,4)=2\\cdot \\left( (4,2)+(4,3)\\right) = 2\\cdot \\left( 207+71\\right)=2\\cdot 278=\\boxed{556}$", "We'll refer to the moves $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , and $(x, y + 2)$ as $R_1$ $R_2$ $U_1$ , and $U_2$ , respectively. Then the possible sequences of moves that will take the frog from $(0,0)$ to $(4,4)$ are all the permutations of $U_1U_1U_1U_1R_1R_1R_1R_1$ $U_2U_1U_1R_1R_1R_1R_1$ $U_1U_1U_1U_1R_2R_1R_1$ $U_2U_1U_1R_2R_1R_1$ $U_2U_2R_1R_1R_1R_1$ $U_1U_1U_1U_1R_2R_2$ $U_2U_2R_2R_1R_1$ $U_2U_1U_1R_2R_2$ , and $U_2U_2R_2R_2$ . We can reduce the number of cases using symmetry.\nCase 1: $U_1U_1U_1U_1R_1R_1R_1R_1$\nThere are $\\frac{8!}{4!4!} = 70$ possibilities for this case.\nCase 2: $U_2U_1U_1R_1R_1R_1R_1$ or $U_1U_1U_1U_1R_2R_1R_1$\nThere are $2 \\cdot \\frac{7!}{4!2!} = 210$ possibilities for this case.\nCase 3: $U_2U_1U_1R_2R_1R_1$\nThere are $\\frac{6!}{2!2!} = 180$ possibilities for this case.\nCase 4: $U_2U_2R_1R_1R_1R_1$ or $U_1U_1U_1U_1R_2R_2$\nThere are $2 \\cdot \\frac{6!}{2!4!} = 30$ possibilities for this case.\nCase 5: $U_2U_2R_2R_1R_1$ or $U_2U_1U_1R_2R_2$\nThere are $2 \\cdot \\frac{5!}{2!2!} = 60$ possibilities for this case.\nCase 6: $U_2U_2R_2R_2$\nThere are $\\frac{4!}{2!2!} = 6$ possibilities for this case.\nAdding up all these cases gives us $70+210+180+30+60+6=\\boxed{556}$ ways.", "Mark the total number of distinct sequences of jumps for the frog to reach the point $(x,y)$ as $\\varphi (x,y)$ . Consider for each point $(x,y)$ in the first quadrant, there are only $4$ possible points in the first quadrant for frog to reach point $(x,y)$ , and these $4$ points are \\[(x-1,y); (x-2,y); (x,y-1); (x,y-2)\\] . As a result, the way to count $\\varphi (x,y)$ is \\[\\varphi (x,y)=\\varphi (x-1,y)+\\varphi (x-2,y)+\\varphi (x,y-1)+\\varphi (x,y-2)\\]\nAlso, for special cases, \\[\\varphi (0,y)=\\varphi (0,y-1)+\\varphi (0,y-2)\\]\n\\[\\varphi (x,0)=\\varphi (x-1,0)+\\varphi (x-2,0)\\]\n\\[\\varphi (x,1)=\\varphi (x-1,1)+\\varphi (x-2,1)+\\varphi (x,0)\\]\n\\[\\varphi (1,y)=\\varphi (1,y-1)+\\varphi (1,y-2)+\\varphi (0,y)\\]\n\\[\\varphi (1,1)=\\varphi (1,0)+\\varphi (0,1)\\]\nStart with $\\varphi (0,0)=1$ , use this method and draw the figure below, we can finally get \\[\\varphi (4,4)=556\\] (In order to make the LaTeX thing more beautiful to look at, I put $0$ to make every number $3$ digits)\n\\[005-020-071-207-\\boxed{556}\\] \\[003-010-032-084-207\\] \\[002-005-014-032-071\\] \\[001-002-005-010-020\\] \\[001-001-002-003-005\\]", "Casework Solution:\nx-distribution: 1-1-1-1 (1 way to order)\ny-distribution: 1-1-1-1 (1 way to order) $\\dbinom{8}{4} = 70$ ways total\nx-distribution: 1-1-1-1 (1 way to order)\ny-distribution: 1-1-2 (3 ways to order) $\\dbinom{7}{3} \\times 3= 105$ ways total\nx-distribution: 1-1-1-1 (1 way to order)\ny-distribution: 2-2 (1 way to order) $\\dbinom{6}{4} = 15$ ways total\nx-distribution: 1-1-2 (3 ways to order)\ny-distribution: 1-1-1-1 (1 way to order) $\\dbinom{7}{3} \\times 3= 105$ ways total\nx-distribution: 1-1-2 (3 ways to order)\ny-distribution: 1-1-2 (3 ways to order) $\\dbinom{6}{3} \\times 9= 180$ ways total\nx-distribution: 1-1-2 (3 ways to order)\ny-distribution: 2-2 (1 way to order) $\\dbinom{5}{3} \\times 3 = 30$ ways total\nx-distribution: 2-2 (1 way to order)\ny-distribution: 1-1-1-1 (1 way to order) $\\dbinom{6}{4} = 15$ ways total\nx-distribution: 2-2 (1 way to order)\ny-distribution: 1-1-2 (3 ways to order) $\\dbinom{5}{3} \\times 3 = 30$ ways total\nx-distribution: 2-2 (1 way to order)\ny-distribution: 2-2 (1 way to order) $\\dbinom{4}{2} = 6$ ways total\n$6+30+15+105+180+70+30+15+105=\\boxed{556}$ -fidgetboss_4000" ]

[ "Since the frog always jumps in length $5$ and lands on a lattice point, the sum of its coordinates must change either by $5$ (by jumping parallel to the x- or y-axis), or by $3$ or $4$ (3-4-5 right triangle).\nBecause either $1$ $5$ , or $7$ is always the change of the sum of the coordinates, the sum of the coordinates will always change from odd to even or vice versa. Thus, it can't go from $(0,0)$ to $(1,0)$ in an even number of moves. Therefore, the frog cannot reach $(1,0)$ in two moves.\nHowever, a path is possible in 3 moves: from $(0,0)$ to $(3,4)$ to $(6,0)$ to $(1,0)$\nThus, the answer is $= \\boxed{3}$" ]

[ "Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is $\\frac{1}{4} \\cdot 1 = \\frac{1}{4}$ . If the frog goes to the right, it will be in the center of the square at $(2,2)$ , and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is $\\frac{1}{2}$ . The probability of this happening is $\\frac{1}{4} \\cdot \\frac{1}{2} = \\frac{1}{8}$\nIf the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is $\\frac{1}{2}$ . Because there's a $\\frac{1}{2}$ chance of the frog going up or down, the total probability for this case is $\\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4}$ and summing up all the cases, $\\frac{1}{4} + \\frac{1}{8} + \\frac{1}{4} = \\frac{5}{8} \\implies \\boxed{58}$", "If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes $\\frac{1}{2}$ . Since it starts on $(1,2)$ , there is a $\\frac{3}{4}$ chance (up, down, or right) it will reach a diagonal on the first jump and $\\frac{1}{4}$ chance (left) it will reach the vertical side. The probablity of landing on a vertical is $\\frac{1}{4}+\\frac{3}{4} \\cdot \\frac{1}{2}=\\boxed{58}$ .\n- Lingjun", "Let $P_{(x,y)}$ denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at $(x,y)$ . Note that $P_{(1,2)}=P_{(3,2)}$ by reflective symmetry over the line $x=2$ . Similarly, $P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}$ , and $P_{(2,1)}=P_{(2,3)}$ . \nNow we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: \\[P_{(1,2)}=\\frac{1}{4}+\\frac{1}{2}P_{(1,1)}+\\frac{1}{4}P_{(2,2)}\\] \\[P_{(2,2)}=\\frac{1}{2}P_{(1,2)}+\\frac{1}{2}P_{(2,1)}\\] \\[P_{(1,1)}=\\frac{1}{4}+\\frac{1}{4}P_{(1,2)}+\\frac{1}{4}P_{(2,1)}\\] \\[P_{(2,1)}=\\frac{1}{2}P_{(1,1)}+\\frac{1}{4}P_{(2,2)}\\] We have a system of $4$ equations in $4$ variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives \\[P_{(2,1)}=\\frac{1}{2}P_{(1,1)}+\\frac{1}{4}\\left(\\frac{1}{2}P_{(1,2)}+\\frac{1}{2}P_{(2,1)}\\right)\\] \\[P_{(2,1)}=\\frac{8}{7}\\left(\\frac{1}{2}P_{(1,1)}+\\frac{1}{8}P_{(1,2)}\\right)=\\frac{4}{7}P_{(1,1)}+\\frac{1}{7}P_{(1,2)}\\] Plugging in the third equation into this gives \\[P_{(2,1)}=\\frac{4}{7}\\left(\\frac{1}{4}+\\frac{1}{4}P_{(1,2)}+\\frac{1}{4}P_{(2,1)}\\right)+\\frac{1}{7}P_{(1,2)}\\] \\[P_{(2,1)}=\\frac{7}{6}\\left(\\frac{1}{7}+\\frac{2}{7}P_{(1,2)}\\right)=\\frac{1}{6}+\\frac{1}{3}P_{(1,2)}\\text{ (*)}\\] Next, plugging in the second and third equation into the first equation yields \\[P_{(1,2)}=\\frac{1}{4}+\\frac{1}{2}\\left(\\frac{1}{4}+\\frac{1}{4}P_{(1,2)}+\\frac{1}{4}P_{(2,1)}\\right)+\\frac{1}{4}\\left(\\frac{1}{2}P_{(1,2)}+\\frac{1}{2}P_{(2,1)}\\right)\\] \\[P_{(1,2)}=\\frac{3}{8}+\\frac{1}{4}P_{(1,2)}+\\frac{1}{4}P_{(2,1)}\\] Now plugging in (*) into this, we get \\[P_{(1,2)}=\\frac{3}{8}+\\frac{1}{4}P_{(1,2)}+\\frac{1}{4}\\left(\\frac{1}{6}+\\frac{1}{3}P_{(1,2)}\\right)\\] \\[P_{(1,2)}=\\frac{3}{2}\\cdot\\frac{5}{12}=\\boxed{58}\\] -mathisawesome2169", "this is basically another version of solution 4; shoutout to mathisawesome2169 :D\nFirst, we note the different places the frog can go at certain locations in the square:\nIf the frog is at a border vertical point ( $(1,2),(3,2)$ ), it moves with probability $\\frac{1}{4}$ to a vertical side of the square, probability $\\frac{1}{4}$ to the center of the square, and probability $\\frac{1}{2}$ to a corner square.\nIf the frog is at a border horizontal point ( $(2,1),(2,3)$ ), it moves with probability $\\frac{1}{4}$ to a horizontal side of the square, probability $\\frac{1}{4}$ to the center of the square, and probability $\\frac{1}{2}$ to a corner square.\nIf the frog is at a center square ( $(2,2)$ ), it moves with probability $\\frac{1}{2}$ to a border horizontal point and probability $\\frac{1}{2}$ to a border vertical point.\nIf the frog is at a corner ( $(1,1),(1,3),(3,3),(3,1)$ ), it moves with probability $\\frac{1}{4}$ to a vertical side of the square, probability $\\frac{1}{4}$ to a horizontal side, probability $\\frac{1}{4}$ to a border horizontal point, and probability $\\frac{1}{4}$ to a border vertical point.\nNow, let $x$ denote the probability of the frog reaching a vertical side when it is at a border vertical point. Similarly, let $y$ denote the probability of the frog reaching a vertical side when it is at a border horizontal point.\nNow, the probability of the frog reaching a vertical side of the square at any location inside the square can be expressed in terms of $x$ and $y$\nFirst, the two easier ones: $P_{center}=\\frac{1}{2}x+\\frac{1}{2}y$ , and $P_{corner}=\\frac{1}{4}+\\frac{1}{4}x+\\frac{1}{4}y$ .\nNow, we can write $x$ and $y$ in terms of $x$ and $y$ , allowing us to solve a system of two variables: \\[x=\\frac{1}{4}+\\frac{1}{4}P_{center}+\\frac{1}{2}P_{corner}=\\frac{1}{4}+\\frac{1}{4}\\left(\\frac{1}{2}x+\\frac{1}{2}y\\right)+\\frac{1}{2}\\left(\\frac{1}{4}+\\frac{1}{4}x+\\frac{1}{4}y\\right)\\] and \\[y=\\frac{1}{4}P_{center}+\\frac{1}{2}P_{corner}=\\frac{1}{4}\\left(\\frac{1}{2}x+\\frac{1}{2}y\\right)+\\frac{1}{2}\\left(\\frac{1}{4}+\\frac{1}{4}x+\\frac{1}{4}y\\right).\\] From these two equations, it is apparent that $y=x-\\frac{1}{4}$ . We can then substitute this value for $y$ back into any of the two equations above to get \\[x=\\frac{1}{4}+\\frac{1}{4}\\left(\\frac{1}{2}x+\\frac{1}{2}\\left(x-\\frac{1}{4}\\right)\\right)+\\frac{1}{2}\\left(\\frac{1}{4}+\\frac{1}{4}x+\\frac{1}{4}\\left(x-\\frac{1}{4}\\right)\\right).\\] Although this certainly looks intimidating, we can expand the parentheses and multiply both sides by 16 to eliminate the fractions, which upon simplification yields the equation \\[16x=5+8x,\\] giving us the desired probability $x=\\frac{5}{8}$ . The answer is then $\\boxed{58}$" ]

[ "Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is $\\frac{1}{4} \\cdot 1 = \\frac{1}{4}$ . If the frog goes to the right, it will be in the center of the square at $(2,2)$ , and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is $\\frac{1}{2}$ . The probability of this happening is $\\frac{1}{4} \\cdot \\frac{1}{2} = \\frac{1}{8}$\nIf the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is $\\frac{1}{2}$ . Because there's a $\\frac{1}{2}$ chance of the frog going up or down, the total probability for this case is $\\frac{1}{2} \\cdot \\frac{1}{2} = \\frac{1}{4}$ and summing up all the cases, $\\frac{1}{4} + \\frac{1}{8} + \\frac{1}{4} = \\frac{5}{8} \\implies \\boxed{58}$", "If the frog is on one of the 2 diagonals, the chance of landing on vertical or horizontal each becomes $\\frac{1}{2}$ . Since it starts on $(1,2)$ , there is a $\\frac{3}{4}$ chance (up, down, or right) it will reach a diagonal on the first jump and $\\frac{1}{4}$ chance (left) it will reach the vertical side. The probablity of landing on a vertical is $\\frac{1}{4}+\\frac{3}{4} \\cdot \\frac{1}{2}=\\boxed{58}$ .\n- Lingjun", "Let $P_{(x,y)}$ denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at $(x,y)$ . Note that $P_{(1,2)}=P_{(3,2)}$ by reflective symmetry over the line $x=2$ . Similarly, $P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}$ , and $P_{(2,1)}=P_{(2,3)}$ . \nNow we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point: \\[P_{(1,2)}=\\frac{1}{4}+\\frac{1}{2}P_{(1,1)}+\\frac{1}{4}P_{(2,2)}\\] \\[P_{(2,2)}=\\frac{1}{2}P_{(1,2)}+\\frac{1}{2}P_{(2,1)}\\] \\[P_{(1,1)}=\\frac{1}{4}+\\frac{1}{4}P_{(1,2)}+\\frac{1}{4}P_{(2,1)}\\] \\[P_{(2,1)}=\\frac{1}{2}P_{(1,1)}+\\frac{1}{4}P_{(2,2)}\\] We have a system of $4$ equations in $4$ variables, so we can solve for each of these probabilities. Plugging the second equation into the fourth equation gives \\[P_{(2,1)}=\\frac{1}{2}P_{(1,1)}+\\frac{1}{4}\\left(\\frac{1}{2}P_{(1,2)}+\\frac{1}{2}P_{(2,1)}\\right)\\] \\[P_{(2,1)}=\\frac{8}{7}\\left(\\frac{1}{2}P_{(1,1)}+\\frac{1}{8}P_{(1,2)}\\right)=\\frac{4}{7}P_{(1,1)}+\\frac{1}{7}P_{(1,2)}\\] Plugging in the third equation into this gives \\[P_{(2,1)}=\\frac{4}{7}\\left(\\frac{1}{4}+\\frac{1}{4}P_{(1,2)}+\\frac{1}{4}P_{(2,1)}\\right)+\\frac{1}{7}P_{(1,2)}\\] \\[P_{(2,1)}=\\frac{7}{6}\\left(\\frac{1}{7}+\\frac{2}{7}P_{(1,2)}\\right)=\\frac{1}{6}+\\frac{1}{3}P_{(1,2)}\\text{ (*)}\\] Next, plugging in the second and third equation into the first equation yields \\[P_{(1,2)}=\\frac{1}{4}+\\frac{1}{2}\\left(\\frac{1}{4}+\\frac{1}{4}P_{(1,2)}+\\frac{1}{4}P_{(2,1)}\\right)+\\frac{1}{4}\\left(\\frac{1}{2}P_{(1,2)}+\\frac{1}{2}P_{(2,1)}\\right)\\] \\[P_{(1,2)}=\\frac{3}{8}+\\frac{1}{4}P_{(1,2)}+\\frac{1}{4}P_{(2,1)}\\] Now plugging in (*) into this, we get \\[P_{(1,2)}=\\frac{3}{8}+\\frac{1}{4}P_{(1,2)}+\\frac{1}{4}\\left(\\frac{1}{6}+\\frac{1}{3}P_{(1,2)}\\right)\\] \\[P_{(1,2)}=\\frac{3}{2}\\cdot\\frac{5}{12}=\\boxed{58}\\] -mathisawesome2169", "this is basically another version of solution 4; shoutout to mathisawesome2169 :D\nFirst, we note the different places the frog can go at certain locations in the square:\nIf the frog is at a border vertical point ( $(1,2),(3,2)$ ), it moves with probability $\\frac{1}{4}$ to a vertical side of the square, probability $\\frac{1}{4}$ to the center of the square, and probability $\\frac{1}{2}$ to a corner square.\nIf the frog is at a border horizontal point ( $(2,1),(2,3)$ ), it moves with probability $\\frac{1}{4}$ to a horizontal side of the square, probability $\\frac{1}{4}$ to the center of the square, and probability $\\frac{1}{2}$ to a corner square.\nIf the frog is at a center square ( $(2,2)$ ), it moves with probability $\\frac{1}{2}$ to a border horizontal point and probability $\\frac{1}{2}$ to a border vertical point.\nIf the frog is at a corner ( $(1,1),(1,3),(3,3),(3,1)$ ), it moves with probability $\\frac{1}{4}$ to a vertical side of the square, probability $\\frac{1}{4}$ to a horizontal side, probability $\\frac{1}{4}$ to a border horizontal point, and probability $\\frac{1}{4}$ to a border vertical point.\nNow, let $x$ denote the probability of the frog reaching a vertical side when it is at a border vertical point. Similarly, let $y$ denote the probability of the frog reaching a vertical side when it is at a border horizontal point.\nNow, the probability of the frog reaching a vertical side of the square at any location inside the square can be expressed in terms of $x$ and $y$\nFirst, the two easier ones: $P_{center}=\\frac{1}{2}x+\\frac{1}{2}y$ , and $P_{corner}=\\frac{1}{4}+\\frac{1}{4}x+\\frac{1}{4}y$ .\nNow, we can write $x$ and $y$ in terms of $x$ and $y$ , allowing us to solve a system of two variables: \\[x=\\frac{1}{4}+\\frac{1}{4}P_{center}+\\frac{1}{2}P_{corner}=\\frac{1}{4}+\\frac{1}{4}\\left(\\frac{1}{2}x+\\frac{1}{2}y\\right)+\\frac{1}{2}\\left(\\frac{1}{4}+\\frac{1}{4}x+\\frac{1}{4}y\\right)\\] and \\[y=\\frac{1}{4}P_{center}+\\frac{1}{2}P_{corner}=\\frac{1}{4}\\left(\\frac{1}{2}x+\\frac{1}{2}y\\right)+\\frac{1}{2}\\left(\\frac{1}{4}+\\frac{1}{4}x+\\frac{1}{4}y\\right).\\] From these two equations, it is apparent that $y=x-\\frac{1}{4}$ . We can then substitute this value for $y$ back into any of the two equations above to get \\[x=\\frac{1}{4}+\\frac{1}{4}\\left(\\frac{1}{2}x+\\frac{1}{2}\\left(x-\\frac{1}{4}\\right)\\right)+\\frac{1}{2}\\left(\\frac{1}{4}+\\frac{1}{4}x+\\frac{1}{4}\\left(x-\\frac{1}{4}\\right)\\right).\\] Although this certainly looks intimidating, we can expand the parentheses and multiply both sides by 16 to eliminate the fractions, which upon simplification yields the equation \\[16x=5+8x,\\] giving us the desired probability $x=\\frac{5}{8}$ . The answer is then $\\boxed{58}$" ]

[ "So let the number of blueberries be $b,$ the number of raspberries be $r,$ the number of grapes be $g,$ and finally the number of cherries be $c.$\nObserve that since there are $280$ pieces of fruit, \\[b+r+g+c=280.\\]\nSince there are twice as many raspberries as blueberries, \\[2b=r.\\]\nThe fact that there are three times as many grapes as cherries implies, \\[3c=g.\\]\nBecause there are four times as many cherries as raspberries, we deduce the following: \\[4r=c.\\]\nNote that we are looking for $c.$ So, we try to rewrite all of the other variables in terms of $c.$ The third equation gives us the value of $g$ in terms of $c$ already. We divide the fourth equation by $4$ to get that $r=\\frac{c}{4}.$ Finally, substituting this value of $r$ into the first equation provides us with the equation $b=\\frac{c}{8}$ and substituting yields: \\[\\frac{c}{4}+\\frac{c}{8}+3c+c=280\\] Multiply this equation by $8$ to get: \\[2c+c+24c+8c=8\\cdot 280,\\] \\[35c=8\\cdot 280,\\] \\[c=64.\\] \\[\\boxed{64}\\]" ]

[ "Going out the final step, if you have $f(f(f(k))) = 27$ , you would have to have $f(f(k))) = 51$ or $f(f(k)) = 108$\nIf you doubled either of these, $k$ would not be odd. So you must subtract $3$\nIf you subtract $3$ from $51$ , you would compute $f(48)$ , which would halve it, and not add the $3$ back.\nIf you subtract $3$ from $108$ , you would compute $f(105)$ , which would add the $3$ back.\nThus, $f(f(f(105))) = f(f(108)) = f(54) = 27$ , and $105$ is odd. The desired sum of the digits is $6$ , and the answer is $\\boxed{6}$" ]

[ "Suppose we pick an arbitrary point on the complex plane , say $(1,1)$ . According to the definition of $f(z)$ \\[f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i,\\] this image must be equidistant to $(1,1)$ and $(0,0)$ . Thus the image must lie on the line with slope $-1$ and which passes through $\\left(\\frac 12, \\frac12\\right)$ , so its graph is $x + y = 1$ . Substituting $x = (a-b)$ and $y = (a+b)$ , we get $2a = 1 \\Rightarrow a = \\frac 12$\nBy the Pythagorean Theorem , we have $\\left(\\frac{1}{2}\\right)^2 + b^2 = 8^2 \\Longrightarrow b^2 = \\frac{255}{4}$ , and the answer is $\\boxed{259}$", "Plugging in $z=1$ yields $f(1) = a+bi$ . This implies that $a+bi$ must fall on the line $Re(z)=a=\\frac{1}{2}$ , given the equidistant rule. By $|a+bi|=8$ , we get $a^2 + b^2 = 64$ , and plugging in $a=\\frac{1}{2}$ yields $b^2=\\frac{255}{4}$ . The answer is thus $\\boxed{259}$", "We are given that $(a + bi)z$ is equidistant from the origin and $z.$ This translates to \\begin{eqnarray*} |(a + bi)z - z| & = & |(a + bi)z| \\\\ |z(a - 1 + bi)| & = & |z(a + bi)| \\\\ |z|\\cdot|(a - 1) + bi| & = & |z|\\cdot|a + bi| \\\\ |(a - 1) + bi| & = & |a + bi| \\\\ (a - 1)^2 + b^2 & = & a^2 + b^2 \\\\ & \\Rightarrow & a = \\frac 12 \\end{eqnarray*} Since $|a + bi| = 8,$ $a^2 + b^2 = 64.$ Because $a = \\frac 12,$ thus $b^2 = \\frac {255}4.$ So the answer is $\\boxed{259}$", "Similarly to in Solution 3, we see that $|(a + bi)z - z| = |(a + bi)z|$ . Letting the point $z = c + di$ , we have $\\sqrt{(ab+bc-d)^2+(ac-bd-c)^2} = \\sqrt{(ac-bd)^2+(ad+bc)^2}$ . Expanding both sides of this equation (after squaring, of course) and canceling terms, we get $(d^2+c^2)(-2a+1) = 0$ . Of course, $(d^2+c^2)$ can't be zero because this property of the function holds for all complex $z$ . Therefore, $a = \\frac{1}{2}$ and we proceed as above to get $\\boxed{259}$", "This is a solution that minimizes the use of complex numbers, turning this into an introductory algebra analytic geometry problem.\nConsider any complex number $z=c+di$ . Let $z$ denote point $P$ on the complex plane. Then $P=(c,d)$ on the complex plane. The equation for the line $OP$ is $y=\\frac{d}{c}x$\nLet the image of point $P$ be $Q$ , after the point undergoes the function. Since each image is equidistant from the preimage and the origin, $Q$ must be on the perpendicular bisector of $OP$ .Given $z=c+di$ $f(z)=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$ . Then $Q=(ac-bd,ad+bc)$ . The midpoint of $OP$ is $(0.5c, 0.5d)$ . Since the slopes of two respectively nonvertical and nonhorizontal lines have a product of $-1$ , using the point-slope form, the equation of the perpendicular line to $OP$ is $y-0.5d=-\\frac{c}{d}(x-0.5c)$ . Rearranging, we have $y=-\\frac{cx}{d}+\\frac{c^2}{2d}+\\frac{d}{2}$\nSince we know that $Q=(ac-bd,ad+bc)$ , thus we plug in $Q$ into the line: $ad+bc=-\\frac{ac^2-bcd}{d}+\\frac{c^2}{2d}+\\frac{d}{2}$\nLet's start canceling. $2ad^2+2bcd=-2ac^2+2bcd+c^2+d^2$ . Subtracting, $c^2+d^2-2ac^2=2ad^2$ . Thus $c^2+d^2=2ac^2+2ad^2$ . Since this is an identity for any $(c,d)$ , thus $2a=1$ $a=\\frac{1}{2}$ . Since $|a+bi|=8$ , thus $a^2+b^2=64$ (or simply think of $a+bi$ as the point $(a,b)$ , and $|a+bi|$ being the distance of $(a,b)$ to the origin). Thus plug in $a=\\frac{1}{2}, b^2=\\frac{255}{4}$ . Since $255$ and $4$ are relatively prime, the final result is $255+4=\\boxed{259}$" ]

[ "For all integers $n \\geq 7,$ note that \\begin{align*} f(n)&=f(n-1)-f(n-2)+n \\\\ &=[f(n-2)-f(n-3)+n-1]-f(n-2)+n \\\\ &=-f(n-3)+2n-1 \\\\ &=-[f(n-4)-f(n-5)+n-3]+2n-1 \\\\ &=-f(n-4)+f(n-5)+n+2 \\\\ &=-[f(n-5)-f(n-6)+n-4]+f(n-5)+n+2 \\\\ &=f(n-6)+6. \\end{align*} It follows that \\begin{align*} f(2018)&=f(2012)+6 \\\\ &=f(2006)+12 \\\\ &=f(2000)+18 \\\\ & \\ \\vdots \\\\ &=f(2)+2016 \\\\ &=\\boxed{2017} ~MRENTHUSIASM", "For all integers $n\\geq3,$ we rearrange the given equation: \\[f(n)-f(n-1)+f(n-2)=n. \\hspace{28.25mm}(1)\\] For all integers $n\\geq4,$ it follows that \\[f(n-1)-f(n-2)+f(n-3)=n-1. \\hspace{15mm}(2)\\] For all integers $n\\geq4,$ we add $(1)$ and $(2):$ \\[f(n)+f(n-3)=2n-1. \\hspace{38.625mm}(3)\\] For all integers $n\\geq7,$ it follows that \\[f(n-3)+f(n-6)=2n-7. \\hspace{32mm}(4)\\] For all integers $n\\geq7,$ we subtract $(4)$ from $(3):$ \\[f(n)-f(n-6)=6. \\hspace{47.5mm}(5)\\] From $(5),$ we have the following system of $336$ equations: \\begin{align*} f(2018)-f(2012)&=6, \\\\ f(2012)-f(2006)&=6, \\\\ f(2006)-f(2000)&=6, \\\\ & \\ \\vdots \\\\ f(8)-f(2)&=6. \\end{align*} We add these equations up to get \\[f(2018)-f(2)=6\\cdot336=2016,\\] from which $f(2018)=f(2)+2016=\\boxed{2017}.$", "Preamble: In this solution, we define the sequence $A$ to satisfy $a_n = f(n),$ where $a_n$ represents the $n$ th term of the sequence $A.$ This solution will show a few different perspectives. Even though it may not be as quick as some of the solutions above, I feel like it is an interesting concept, and may be more motivated.\nTo begin, we consider the sequence $B$ formed when we take the difference of consecutive terms between $A.$ Define $b_n = a_{n+1} - a_n.$ Notice that for $n \\ge 4,$ we have\nNotice that subtracting the second equation from the first, we see that $b_{n} = b_{n-1} - b_{n-2} + 1.$\nIf you didn’t notice that $B$ repeated directly in the solution above, you could also, possibly more naturally, take the finite differences of the sequence $b_n$ so that you could define $c_n = b_{n+1} - b_n.$ Using a similar method as above through reindexing and then subtracting, you could find that $c_n = c_{n-1} - c_{n-2}.$ The sum of any six consecutive terms of a sequence which satisfies such a recursion is $0,$ in which you have that $b_{n} = b_{n+6}.$ In the case in which finite differences didn’t reduce to such a special recursion, you could still find the first few terms of $C$ to see if there are any patterns, now that you have a much simpler sequence. Doing so in this case, it can also be seen by seeing that the sequence $C$ looks like \\[\\underbrace{2, 1, -1, -2, -1, 1,}_{\\text{cycle period}} 2, 1, -1, -2, -1, 1, \\ldots\\] in which the same result follows.\nUsing the fact that $B$ repeats every six terms, this motivates us to look at the sequence $B$ more carefully. Doing so, we see that $B$ looks like \\[\\underbrace{2, 3, 2, 0, -1, 0,}_{\\text{cycle period}} 2, 3, 2, 0, -1, 0, \\ldots\\] (If you tried pattern finding on sequence $B$ directly, you could also arrive at this result, although I figured defining a second sequence based on finite differences was more motivated.)\nNow, there are two ways to finish.\nFinish Method #1: Notice that any six consecutive terms of $B$ sum to $6,$ after which we see that $a_n = a_{n-6} + 6.$ Therefore, $a_{2018} = a_{2012} + 6 = \\cdots = a_{2} + 2016 = \\boxed{2017}.$", "Start out by listing some terms of the sequence. \\begin{align*} f(1)&=1 \\\\ f(2)&=1 \\\\ f(3)&=3 \\\\ f(4)&=6 \\\\ f(5)&=8 \\\\ f(6)&=8 \\\\ f(7)&=7 \\\\ f(8)&=7 \\\\ f(9)&=9 \\\\ f(10)&=12 \\\\ f(11)&=14 \\\\ f(12)&=14 \\\\ f(13)&=13 \\\\ f(14)&=13 \\\\ f(15)&=15 \\\\ & \\ \\vdots \\end{align*} Notice that $f(n)=n$ whenever $n$ is an odd multiple of $3$ , and the pattern of numbers that follow will always be $+2$ $+3$ $+2$ $+0$ $-1$ $+0$ .\nThe largest odd multiple of $3$ smaller than $2018$ is $2013$ , so we have \\begin{align*} f(2013)&=2013 \\\\ f(2014)&=2016 \\\\ f(2015)&=2018 \\\\ f(2016)&=2018 \\\\ f(2017)&=2017 \\\\ f(2018)&=\\boxed{2017}", "Writing out the first few values, we get \\[1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19,\\ldots.\\] We see that every number $x$ where $x \\equiv 1\\pmod 6$ has $f(x)=x,f(x+1)=f(x)=x,$ and $f(x-1)=f(x-2)=x+1.$ The greatest number that's $1\\pmod{6}$ and less than $2018$ is $2017,$ so we have $f(2017)=f(2018)=\\boxed{2017}.$", "\\begin{align*} f(n)&=f(n-1)-f(n-2)+n \\\\ f(n-1)&=f(n-2)-f(n-3)+n-1 \\end{align*} Subtracting those two and rearranging gives \\begin{align*} f(n)-2f(n-1)+2f(n-2)-f(n-3)&=1 \\\\ f(n-1)-2f(n-2)+2f(n-3)-f(n-4)&=1 \\end{align*} Subtracting those two gives $f(n)-3f(n-1)+4f(n-2)-3f(n-3)+f(n-4)=0.$\nThe characteristic polynomial is $x^4-3x^3+4x^2-3x+1=0.$\n$x=1$ is a root, so using synthetic division results in $(x-1)(x^3-2x^2+2x-1)=0.$\n$x=1$ is a root, so using synthetic division results in $(x-1)^2(x^2-x+1)=0.$\n$x^2-x+1=0$ has roots $x=\\frac{1}{2}\\pm\\frac{i\\sqrt{3}}{2}.$\nAnd \\[f(n)=(An+D)\\cdot1^n+B\\cdot\\left(\\frac{1}{2}-\\frac{i\\sqrt{3}}{2}\\right)^n+C\\cdot\\left(\\frac{1}{2}+\\frac{i\\sqrt{3}}{2}\\right)^n.\\] Plugging in $n=1$ $n=2$ $n=3$ , and $n=4$ results in a system of $4$ linear equations $\\newline$ Solving them gives $A=1, \\ B=\\frac{1}{2}-\\frac{i\\sqrt{3}}{2}, \\ C=\\frac{1}{2}+\\frac{i\\sqrt{3}}{2}, \\ D=1.$ Note that you can guess $A=1$ by answer choices.\nSo plugging in $n=2018$ results in \\begin{align*} 2018+1+\\left(\\frac{1}{2}-\\frac{i\\sqrt{3}}{2}\\right)^{2019}+\\left(\\frac{1}{2}+\\frac{i\\sqrt{3}}{2}\\right)^{2019}&=2019+(\\cos(-60^{\\circ})+\\sin(-60^{\\circ}))^{2019})+(\\cos(60^{\\circ})+\\sin(60^{\\circ}))^{2019}) \\\\ &=2019+(\\cos(-60^{\\circ}\\cdot2019)+\\sin(-60^{\\circ}\\cdot2019))+(\\cos(60^{\\circ}\\cdot2019)+sin(60^{\\circ}\\cdot2019)) \\\\ &=\\boxed{2017} ~ryanbear", "We utilize patterns to solve this equation: \\begin{align*} f(3)&=3, \\\\ f(4)&=6, \\\\ f(5)&=8, \\\\ f(6)&=8, \\\\ f(7)&=7, \\\\ f(8)&=8. \\end{align*} We realize that the pattern repeats itself. For every six terms, there will be four terms that we repeat, and two terms that we don't repeat. We will exclude the first two for now, because they don't follow this pattern.\nFirst, we need to know whether or not $2016$ is part of the skip or repeat. We notice that $f(6),f(12), \\ldots,f(6n)$ all satisfy $6+6(n-1)=n,$ and we know that $2016$ satisfies this, leaving $n=336.$ Therefore, we know that $2016$ is part of the repeat section. But what number does it repeat?\nWe know that the repeat period is $2,$ and it follows that pattern of $1,1,8,8,7,7.$ Again, since $f(6) = f(5)$ and so on for the repeat section, $f(2016)=f(2015),$ so we don't need to worry about which one, since it repeats with period $2.$ We see that the repeat pattern of $f(6),f(12),\\ldots,f(6n)$ follows $8,14,20,$ it is an arithmetic sequence with common difference $6.$ Therefore, $2016$ is the $335$ th term of this, but including $1,$ it is $336\\cdot6+1=\\boxed{2017}.$" ]

[ "For all integers $n \\geq 7,$ note that \\begin{align*} f(n)&=f(n-1)-f(n-2)+n \\\\ &=[f(n-2)-f(n-3)+n-1]-f(n-2)+n \\\\ &=-f(n-3)+2n-1 \\\\ &=-[f(n-4)-f(n-5)+n-3]+2n-1 \\\\ &=-f(n-4)+f(n-5)+n+2 \\\\ &=-[f(n-5)-f(n-6)+n-4]+f(n-5)+n+2 \\\\ &=f(n-6)+6. \\end{align*} It follows that \\begin{align*} f(2018)&=f(2012)+6 \\\\ &=f(2006)+12 \\\\ &=f(2000)+18 \\\\ & \\ \\vdots \\\\ &=f(2)+2016 \\\\ &=\\boxed{2017} ~MRENTHUSIASM", "For all integers $n\\geq3,$ we rearrange the given equation: \\[f(n)-f(n-1)+f(n-2)=n. \\hspace{28.25mm}(1)\\] For all integers $n\\geq4,$ it follows that \\[f(n-1)-f(n-2)+f(n-3)=n-1. \\hspace{15mm}(2)\\] For all integers $n\\geq4,$ we add $(1)$ and $(2):$ \\[f(n)+f(n-3)=2n-1. \\hspace{38.625mm}(3)\\] For all integers $n\\geq7,$ it follows that \\[f(n-3)+f(n-6)=2n-7. \\hspace{32mm}(4)\\] For all integers $n\\geq7,$ we subtract $(4)$ from $(3):$ \\[f(n)-f(n-6)=6. \\hspace{47.5mm}(5)\\] From $(5),$ we have the following system of $336$ equations: \\begin{align*} f(2018)-f(2012)&=6, \\\\ f(2012)-f(2006)&=6, \\\\ f(2006)-f(2000)&=6, \\\\ & \\ \\vdots \\\\ f(8)-f(2)&=6. \\end{align*} We add these equations up to get \\[f(2018)-f(2)=6\\cdot336=2016,\\] from which $f(2018)=f(2)+2016=\\boxed{2017}.$", "Preamble: In this solution, we define the sequence $A$ to satisfy $a_n = f(n),$ where $a_n$ represents the $n$ th term of the sequence $A.$ This solution will show a few different perspectives. Even though it may not be as quick as some of the solutions above, I feel like it is an interesting concept, and may be more motivated.\nTo begin, we consider the sequence $B$ formed when we take the difference of consecutive terms between $A.$ Define $b_n = a_{n+1} - a_n.$ Notice that for $n \\ge 4,$ we have\nNotice that subtracting the second equation from the first, we see that $b_{n} = b_{n-1} - b_{n-2} + 1.$\nIf you didn’t notice that $B$ repeated directly in the solution above, you could also, possibly more naturally, take the finite differences of the sequence $b_n$ so that you could define $c_n = b_{n+1} - b_n.$ Using a similar method as above through reindexing and then subtracting, you could find that $c_n = c_{n-1} - c_{n-2}.$ The sum of any six consecutive terms of a sequence which satisfies such a recursion is $0,$ in which you have that $b_{n} = b_{n+6}.$ In the case in which finite differences didn’t reduce to such a special recursion, you could still find the first few terms of $C$ to see if there are any patterns, now that you have a much simpler sequence. Doing so in this case, it can also be seen by seeing that the sequence $C$ looks like \\[\\underbrace{2, 1, -1, -2, -1, 1,}_{\\text{cycle period}} 2, 1, -1, -2, -1, 1, \\ldots\\] in which the same result follows.\nUsing the fact that $B$ repeats every six terms, this motivates us to look at the sequence $B$ more carefully. Doing so, we see that $B$ looks like \\[\\underbrace{2, 3, 2, 0, -1, 0,}_{\\text{cycle period}} 2, 3, 2, 0, -1, 0, \\ldots\\] (If you tried pattern finding on sequence $B$ directly, you could also arrive at this result, although I figured defining a second sequence based on finite differences was more motivated.)\nNow, there are two ways to finish.\nFinish Method #1: Notice that any six consecutive terms of $B$ sum to $6,$ after which we see that $a_n = a_{n-6} + 6.$ Therefore, $a_{2018} = a_{2012} + 6 = \\cdots = a_{2} + 2016 = \\boxed{2017}.$", "Start out by listing some terms of the sequence. \\begin{align*} f(1)&=1 \\\\ f(2)&=1 \\\\ f(3)&=3 \\\\ f(4)&=6 \\\\ f(5)&=8 \\\\ f(6)&=8 \\\\ f(7)&=7 \\\\ f(8)&=7 \\\\ f(9)&=9 \\\\ f(10)&=12 \\\\ f(11)&=14 \\\\ f(12)&=14 \\\\ f(13)&=13 \\\\ f(14)&=13 \\\\ f(15)&=15 \\\\ & \\ \\vdots \\end{align*} Notice that $f(n)=n$ whenever $n$ is an odd multiple of $3$ , and the pattern of numbers that follow will always be $+2$ $+3$ $+2$ $+0$ $-1$ $+0$ .\nThe largest odd multiple of $3$ smaller than $2018$ is $2013$ , so we have \\begin{align*} f(2013)&=2013 \\\\ f(2014)&=2016 \\\\ f(2015)&=2018 \\\\ f(2016)&=2018 \\\\ f(2017)&=2017 \\\\ f(2018)&=\\boxed{2017}", "Writing out the first few values, we get \\[1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19,\\ldots.\\] We see that every number $x$ where $x \\equiv 1\\pmod 6$ has $f(x)=x,f(x+1)=f(x)=x,$ and $f(x-1)=f(x-2)=x+1.$ The greatest number that's $1\\pmod{6}$ and less than $2018$ is $2017,$ so we have $f(2017)=f(2018)=\\boxed{2017}.$", "\\begin{align*} f(n)&=f(n-1)-f(n-2)+n \\\\ f(n-1)&=f(n-2)-f(n-3)+n-1 \\end{align*} Subtracting those two and rearranging gives \\begin{align*} f(n)-2f(n-1)+2f(n-2)-f(n-3)&=1 \\\\ f(n-1)-2f(n-2)+2f(n-3)-f(n-4)&=1 \\end{align*} Subtracting those two gives $f(n)-3f(n-1)+4f(n-2)-3f(n-3)+f(n-4)=0.$\nThe characteristic polynomial is $x^4-3x^3+4x^2-3x+1=0.$\n$x=1$ is a root, so using synthetic division results in $(x-1)(x^3-2x^2+2x-1)=0.$\n$x=1$ is a root, so using synthetic division results in $(x-1)^2(x^2-x+1)=0.$\n$x^2-x+1=0$ has roots $x=\\frac{1}{2}\\pm\\frac{i\\sqrt{3}}{2}.$\nAnd \\[f(n)=(An+D)\\cdot1^n+B\\cdot\\left(\\frac{1}{2}-\\frac{i\\sqrt{3}}{2}\\right)^n+C\\cdot\\left(\\frac{1}{2}+\\frac{i\\sqrt{3}}{2}\\right)^n.\\] Plugging in $n=1$ $n=2$ $n=3$ , and $n=4$ results in a system of $4$ linear equations $\\newline$ Solving them gives $A=1, \\ B=\\frac{1}{2}-\\frac{i\\sqrt{3}}{2}, \\ C=\\frac{1}{2}+\\frac{i\\sqrt{3}}{2}, \\ D=1.$ Note that you can guess $A=1$ by answer choices.\nSo plugging in $n=2018$ results in \\begin{align*} 2018+1+\\left(\\frac{1}{2}-\\frac{i\\sqrt{3}}{2}\\right)^{2019}+\\left(\\frac{1}{2}+\\frac{i\\sqrt{3}}{2}\\right)^{2019}&=2019+(\\cos(-60^{\\circ})+\\sin(-60^{\\circ}))^{2019})+(\\cos(60^{\\circ})+\\sin(60^{\\circ}))^{2019}) \\\\ &=2019+(\\cos(-60^{\\circ}\\cdot2019)+\\sin(-60^{\\circ}\\cdot2019))+(\\cos(60^{\\circ}\\cdot2019)+sin(60^{\\circ}\\cdot2019)) \\\\ &=\\boxed{2017} ~ryanbear", "We utilize patterns to solve this equation: \\begin{align*} f(3)&=3, \\\\ f(4)&=6, \\\\ f(5)&=8, \\\\ f(6)&=8, \\\\ f(7)&=7, \\\\ f(8)&=8. \\end{align*} We realize that the pattern repeats itself. For every six terms, there will be four terms that we repeat, and two terms that we don't repeat. We will exclude the first two for now, because they don't follow this pattern.\nFirst, we need to know whether or not $2016$ is part of the skip or repeat. We notice that $f(6),f(12), \\ldots,f(6n)$ all satisfy $6+6(n-1)=n,$ and we know that $2016$ satisfies this, leaving $n=336.$ Therefore, we know that $2016$ is part of the repeat section. But what number does it repeat?\nWe know that the repeat period is $2,$ and it follows that pattern of $1,1,8,8,7,7.$ Again, since $f(6) = f(5)$ and so on for the repeat section, $f(2016)=f(2015),$ so we don't need to worry about which one, since it repeats with period $2.$ We see that the repeat pattern of $f(6),f(12),\\ldots,f(6n)$ follows $8,14,20,$ it is an arithmetic sequence with common difference $6.$ Therefore, $2016$ is the $335$ th term of this, but including $1,$ it is $336\\cdot6+1=\\boxed{2017}.$" ]

[ "Notice that, in order to step onto any particular white square, the marker must have come from one of the $1$ or $2$ white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move from $P$ to each of the white squares immediately beneath it(also called the Water Fall Method). To solve the problem, we can accordingly construct the following diagram, where each number in a square is calculated as the sum of the numbers on the white squares immediately beneath that square (and thus will represent the number of ways to remove from $P$ to that square, as already stated).\n\nThe answer is therefore $\\boxed{28}$", "Suppose we \"extend\" the chessboard infinitely with $2$ additional columns to the right, as shown below. The red line shows the right-hand edge of the original board.\n\nThe total number of paths from $P$ to $Q$ , including invalid paths which cross over the red line, is then the number of paths which make $4$ steps up-and-right and $3$ steps up-and-left, which is $\\binom{4+3}{3} = \\binom{7}{3} = 35$ . We need to subtract the number of invalid paths, i.e. the number of paths that pass through $X$ or $Y$ . To get to $X$ , the marker has to make $3$ up-and-right steps, after which it can proceed to $Q$ with $3$ steps up-and-left and $1$ step up-and-right. Thus, the number of paths from $P$ to $Q$ that pass through $X$ is $1 \\cdot \\binom{3+1}{3} = 4$ . Similarly, the number of paths that pass through $Y$ is $\\binom{4+1}{1}\\cdot 1 = 5$ . However, we have now double-counted the invalid paths which pass through both $X$ and $Y$ ; from the diagram, it is clear that there are only $2$ of these (as the marker can get from $X$ to $Y$ by a step up-and-left and a step-up-and-right in either order). Hence the number of invalid paths is $4+5-2=7$ , and the number of valid paths from $P$ to $Q$ is $35-7 = \\boxed{28}$", "On any white square, we may choose to go left or right, as long as we do not cross over the border of the board. Call the moves $L$ and $R$ respectively. Every single legal path consists of $4$ $R's$ and $3$ $L's$ , so now all we have to find is the number of ways to order $4 R's$ and $3 L's$ in any way, which is ${7 \\choose 3}=35$ . However, we originally promised that we will not go over the border, and now we have to subtract the paths that do go over the border. The paths that go over the border are any paths that start with RRR(1 path), RR(5 paths) and LRRR(1 path) so our final number of paths is $35-7=\\boxed{28}.$ ~PEKKA", "We label the rows starting from the bottom. At row 1, there is $1$ way: at P. We draw all the possible ways to get to Q. There are two ways to choose for row 2, and another two ways to choose for row 3. However, you can go to the \"edge\" or the farthest possible square westward of Q, so you can't multiply by 2 again. Notice how, at the first step, we figured that the answer was even, so choice D and E are eliminated, and after the second row, we realized it must be a multiple of 4, so choice B is eliminated. When we get to the fourth row, we do not multiply by 2 again, since we have limited possibilities rather than multiplying by 2 again. Choice C implies that there are two possibilities per row; however, we know that if you go to the farthest possible, you only have one possibility, so it is not $2^5 = 32$ so we know that the answer is choice $\\boxed{28}$ . \n~hh99754539" ]

[ "We look at a set of three rounds, where the players begin with $x+1$ $x$ , and $x-1$ tokens.\nAfter three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$ $2$ , and $1$ tokens, respectively. After $1$ more round, player $A$ will give away $3$ tokens, leaving them empty-handed, and thus the game will end. We then have there are $36+1=\\boxed{37}$ rounds until the game ends.", "Let's bash a few rounds. The amounts are for players $1,2,$ and $3$ , respectively.\nFirst round: $15,14,13$ (given)\nSecond round: $12,15,14$ Third round: $13,12,15$ Fourth round: $14,13,12$\nWe see that after $3$ rounds are played, we have the exact same scenario as the first round but with one token less per player. So, the sequence $1,4,7,10...$ where each of the next members are $3$ greater than the previous one corresponds with the sequence $15,14,13,12...$ where the first sequence represents the round and the second sequence represents the number of tokens player $1$ has. But we note that once player $1$ reaches $3$ coins, the game will end on his next turn as he must give away all his coins. Therefore, we want the $15-3+1=13$ th number in the sequence $1,4,7,10...$ which is $\\boxed{37}$", "Looking at a set of five rounds, you'll see $A$ has $4$ fewer tokens than in the beginning. Looking at four more rounds, you'll notice $A$ has the same amount of tokens, namely $11$ , compared to round five. If you keep doing this process, you'll see a pattern: Every four rounds, the amount of tokens $A$ has either decreased by $4$ or stayed the same compared to the previous four rounds. For example, in round nine, $A$ had $11$ tokens, in round $13$ $A$ had $11$ tokens, and in round $17$ $A$ had $7$ tokens, etc. Using this weird pattern, you can find out that in round $37$ $A$ should have $3$ tokens, but since they would have given them away in that round, the game would end on $\\boxed{37}$" ]

[ "We look at a set of three rounds, where the players begin with $x+1$ $x$ , and $x-1$ tokens.\nAfter three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$ $2$ , and $1$ tokens, respectively. After $1$ more round, player $A$ will give away $3$ tokens, leaving them empty-handed, and thus the game will end. We then have there are $36+1=\\boxed{37}$ rounds until the game ends.", "Let's bash a few rounds. The amounts are for players $1,2,$ and $3$ , respectively.\nFirst round: $15,14,13$ (given)\nSecond round: $12,15,14$ Third round: $13,12,15$ Fourth round: $14,13,12$\nWe see that after $3$ rounds are played, we have the exact same scenario as the first round but with one token less per player. So, the sequence $1,4,7,10...$ where each of the next members are $3$ greater than the previous one corresponds with the sequence $15,14,13,12...$ where the first sequence represents the round and the second sequence represents the number of tokens player $1$ has. But we note that once player $1$ reaches $3$ coins, the game will end on his next turn as he must give away all his coins. Therefore, we want the $15-3+1=13$ th number in the sequence $1,4,7,10...$ which is $\\boxed{37}$", "Looking at a set of five rounds, you'll see $A$ has $4$ fewer tokens than in the beginning. Looking at four more rounds, you'll notice $A$ has the same amount of tokens, namely $11$ , compared to round five. If you keep doing this process, you'll see a pattern: Every four rounds, the amount of tokens $A$ has either decreased by $4$ or stayed the same compared to the previous four rounds. For example, in round nine, $A$ had $11$ tokens, in round $13$ $A$ had $11$ tokens, and in round $17$ $A$ had $7$ tokens, etc. Using this weird pattern, you can find out that in round $37$ $A$ should have $3$ tokens, but since they would have given them away in that round, the game would end on $\\boxed{37}$" ]

[ "[Clarification: You are supposed to find the number of all possible tuples of prices, $(A, B, C)$ , that could have been on that day.]\nSince we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$ , then the string is\n\\[1131331.\\]\nSince the strings have seven digits and three threes, there are $\\binom{7}{3}=35$ arrangements of all such strings.\nIn order to obtain all combination of A,B,C, we partition all the possible strings into 3 groups.\nLet's look at the example. We have to partition it into 3 groups with each group having at least 1 digit. In other words, we need to find the solution to\n\\[x+y+z=7, x,y,z>0.\\]\nThis gives us\n\\[\\binom{6}{2}=15\\]\nways by balls and urns. But we have counted the one with 5 digit numbers; that is, $(5,1,1),(1,1,5),(1,5,1)$\nThus, each arrangement has \\[\\binom{6}{2}-3=12\\] ways per arrangement, and there are $12\\times35=\\boxed{420}$ ways." ]

[ "First notice that there is no difference between the maple trees and the oak trees; we have only two types, birch trees and \"non-birch\" trees. (If you don't believe this reasoning, think about it. You could also differentiate the tall oak trees from the short oak trees, and the maple trees with many branches as opposed to those with few branches. Indeed, you could keep dividing until you have them each in their own category, but in the end it will not change the probability of the birch trees being near each other. That is, in the end, you multiply the numerator by the number of ways to arrange the oak and maple trees and you also multiply the denominator by the number of ways to arrange the oak and maple trees, making them cancel out.)\nThe five birch trees must be placed amongst the seven previous trees. We can think of these trees as 5 dividers of 8 slots that the birch trees can go in, making ${8\\choose5} = 56$ different ways to arrange this.\nThere are ${12 \\choose 5} = 792$ total ways to arrange the twelve trees, so the probability is $\\frac{56}{792} = \\frac{7}{99}$\nThe answer is $7 + 99 = \\boxed{106}$", "Let $b$ $n$ denote birch tree and not-birch tree, respectively. Notice that we only need $4$ $n$ s to separate the $5$ $b$ s. Specifically, \\[b,n,b,n,b,n,b,n,b\\] Since we have $7$ $n$ s, we are placing the extra $3$ $n$ s into the $6$ intervals beside the $b$ s.\nNow doing simple casework.\nIf all $3$ $n$ s are in the same interval, there are $6$ ways.\nIf $2$ of the $3$ $n$ s are in the same interval, there are $6\\cdot5=30$ ways.\nIf the $n$ s are in $3$ different intervals, there are ${6 \\choose 3} =20$ ways.\nIn total there are $6+30+20=56$ ways.\nThere are ${12\\choose5}=792$ ways to distribute the birch trees among all $12$ trees.\nThus the probability equals $\\frac{56}{792}=\\frac{7}{99}\\Longrightarrow m+n=7+99=\\boxed{106}$", "Note that the requested probability is computed by dividing the number of configurations with no adjacent Birch trees by the total number of configurations. We can compute the number of configurations with no adjacent Birch trees using complementary counting and then the Principle of Inclusion-Exclusion.\nThe number of configurations with no adjacent Birch trees is equal to the total number of configurations minus the number of configurations with at least one pair of adjacent Birch trees.\nThe total number of configurations is given by $\\frac{12!}{3! \\cdot 4! \\cdot 5!}$ . To compute the number of configurations with at least one pair of adjacent Birch trees, we use PIE.\n$\\#$ (configurations with at least one pair of adjacent Birch trees) $=$ $\\#$ (configurations with one pair) $-$ $\\#$ (configurations with two pairs) $+$ $\\#$ (configurations with three pairs) $-$ $\\#$ (configurations with four pairs).\nTo compute the first term, note that we can treat the adjacent pair of Birch trees as one separate tree. This then gives $\\frac{11!}{3! \\cdot 3! \\cdot 4!}$ configurations.\nFor the second term, we have two cases. The two pairs could either happen consecutively (BBB) or separately (BB BB). They both give $\\frac{10!}{2! \\cdot 3! \\cdot 4!}$ cases. So our second term is $\\frac{2 \\cdot 10!}{2! \\cdot 3! \\cdot 4!}$\nThe third term can also happen in two ways. The three pairs could be arranged like BBBB or BBB BB. Both cases together give $\\frac{2 \\cdot 9!}{3! \\cdot 4!}$ arrangements.\nThe final term can happen in one way (BBBBB). This gives $\\frac{8!}{3! \\cdot 4!}$ arrangements.\nSubstituting these into our PIE expression, we find that there are $25760$ configurations with at least one pair of adjacent Birch trees. Therefore, there are a total of $\\frac{12!}{3! \\cdot 4! \\cdot 5!} - 25760 = 1960$ configurations with no adjacent Birch trees.\nThus, the probability of a given configuration having no two adjacent Birch trees is given by $\\frac{1960}{\\frac{12!}{3! \\cdot 4! \\cdot 5!}} = \\frac{7}{99}$\nTherefore, the desired result is given by $7+99 = \\boxed{106}$", "Here is a solution leaving out nothing. This solution is dedicated to those that are in self study and wish to learn the most they can. I will make it as elementary as possible and intuition based.\nArrange first the $3$ maple and $4$ oaks as $MMMOOOO$ . We then notice that for none of the $5$ birch trees to be adjacent, they must be put in between these $M$ 's and $O$ 's. We then see that there are $8$ spots to put these $5$ birch trees in. So we can select $5$ spots for these birch trees in $\\binom{8}{5}$ . But then, we can rearrange the $M$ 's and $O$ 's in $7!/(3!4!)=\\binom{7}{3}$ ways. So then there are $\\binom{8}{5}\\binom{7}{3}$ valid arrangements with no given consecutive birch trees.\nThere are then a total of $\\frac{12!}{3!4!5!}$ different total arrangements. Therefore the probability is given as $\\frac{\\binom{8}{5}\\binom{7}{3}}{\\frac{12!}{3!4!5!}}=\\frac{7}{99}$ , so the answer is $7+99=\\boxed{106}$" ]

[ "By the defintion of a geometric sequence, we have $\\cos^2x=\\sin x \\tan x$ . Since $\\tan x=\\frac{\\sin x}{\\cos x}$ , we can rewrite this as $\\cos^3x=\\sin^2x$\nThe common ratio of the sequence is $\\frac{\\cos x}{\\sin x}$ , so we can write\n\\[a_1= \\sin x\\] \\[a_2= \\cos x\\] \\[a_3= \\frac{\\cos^2x}{\\sin x}\\] \\[a_4=\\frac{\\cos^3x}{\\sin^2x}=1\\] \\[a_5=\\frac{\\cos x}{\\sin x}\\] \\[a_6=\\frac{\\cos^2x}{\\sin^2x}\\] \\[a_7=\\frac{\\cos^3x}{\\sin^3x}=\\frac{1}{\\sin x}\\] \\[a_8=\\frac{\\cos x}{\\sin^2 x}=\\frac{1}{\\cos^2 x}\\]\nSince $\\cos^3x=\\sin^2x=1-\\cos^2x$ , we have $\\cos^3x+\\cos^2x=1 \\implies \\cos^2x(\\cos x+1)=1 \\implies \\cos x+1=\\frac{1}{\\cos^2 x}$ , which is $a_8$ , making our answer $8 \\Rightarrow \\boxed{8}$", "Notice that the common ratio is $r=\\frac{\\cos(x)}{\\sin(x)}$ ; multiplying it to $\\tan(x)=\\frac{\\sin(x)}{\\cos(x)}$ gives $a_4=1$ . Then, working backwards we have $a_3=\\frac{1}{r}$ $a_2=\\frac{1}{r^2}$ and $a_1=\\frac{1}{r^3}$ . Now notice that since $a_1=\\sin(x)$ and $a_2=\\cos(x)$ , we need $a_1^2+a_2^2=1$ , so $\\frac{1}{r^6}+\\frac{1}{r^4}=\\frac{r^2+1}{r^6}=1\\implies r^2+1=r^6$ . Dividing both sides by $r^2$ gives $1+\\frac{1}{r^2}=r^4$ , which the left side is equal to $1+\\cos(x)$ ; we see as well that the right hand side is equal to $a_8$ given $a_4=1$ , so the answer is $\\boxed{8}$ . - mathleticguyyy" ]

[ "If any of $r_1, \\ldots, r_{19}$ is larger than $r_{20}$ , one of these numbers will be compared with $r_{20}$ on the 19th step of the first bubble pass and $r_{20}$ will be moved back to the 19th position. Thus, $r_{20}$ must be the largest of the first 20 terms. In addition, $r_{20}$ must be larger than $r_{21}, r_{22}, \\ldots, r_{30}$ but smaller than $r_{31}$ in order that it move right to the 30th position but then not continue moving right to the 31st.\nThus, our problem can be restated: What is the probability that in a sequence of 31 distinct real numbers, the largest is in position 31 and the second-largest is in position 20 (the other 29 numbers are irrelevant)?\nThis is much easier to solve: there are $31!$ ways to order the first thirty-one numbers and $29!$ ways to arrange them so that the largest number is in the 31st position and the second-largest is in the 20th. This gives us a desired probability of $\\frac{29!}{31!} = \\frac{1}{31\\cdot 30} = \\frac{1}{930}$ , so the answer is $\\boxed{931}$", "Note that you can solve the restated problem differently:\nWe see that the numerator is $1$ because there is only $1$ way to make the $31^{\\text{st}}$ term the largest and the $20^{\\text{th}}$ the second-largest. To calculate the denominator, we note that there are $31\\times30$ ways to pick the largest term and the second-largest term. Therefore, our answer is $\\boxed{931}$ .\n~Yiyj1" ]

[ "Let $a$ be number of cents $A$ originally had, $b$ be number of cents $B$ originally had, and $c$ be number of cents $C$ originally had.\nAfter $A$ gave his money away, $A$ has $a-b-c$ cents, $B$ has $2b$ cents, and $C$ has $2c$ cents.\nAfter $B$ gave his money away, $A$ has $2a-2b-2c$ cents, $B$ has $-a+3b-c$ cents, and $C$ has $4c$ cents.\nAfter $C$ gave his money away, $A$ has $4a-4b-4c$ cents, $B$ has $-2a+6b-2c$ cents, and $C$ has $-a-b+7c$ cents.\nSince all of them have $16$ cents in the end, we can write a system of equations \\[4a-4b-4c=16\\] \\[-2a+6b-2c=16\\] \\[-a-b+7c=16\\] Note that adding the three equation yields $a+b+c=48$ , so $4a+4b+4c=192$ . Therefore, $8a=208$ , so $a = 26$ . Solving for $a$ can also be done traditionally.\nThus, $A$ started out with $26$ cents, which is answer choice $\\boxed{26}$", "We know that people not giving away money on the previous turn now have twice what they previously had. Using the fact that the sum of their money is $48$ cents, we can work backward \\[16,16,16\\] \\[8,8,32\\] \\[4,28,16\\] \\[26,14,8\\] Thus at the beginning $A$ has $26\\boxed{26}$ cents." ]

[ "The sum of the first $n$ odd numbers is $n^2$ . As in our case $n^2=100$ , we have $n=\\boxed{10}$" ]

[ "The sum of the first $n$ odd numbers is $n^2$ . As in our case $n^2=100$ , we have $n=\\boxed{10}$" ]

[ "There are $5\\times8=40$ oranges on the $1^{\\text{st}}$ layer of the stack. The $2^{\\text{nd}}$ layer that is added on top of the first will be a layer of $4\\times7=28$ oranges. When the third layer is added on top of the $2^{\\text{nd}}$ , it will be a layer of $3\\times6=18$ oranges, etc.\nTherefore, there are $5\\times8+4\\times7+3\\times6+2\\times5+1\\times4=40+28+18+10+4=100$ oranges in the stack. $\\boxed{100}$" ]

[ "Let $x$ be the number of coins. After the $k^{\\text{th}}$ pirate takes his share, $\\frac{12-k}{12}$ of the original amount is left. Thus, we know that\n$x \\cdot \\frac{11}{12} \\cdot \\frac{10}{12} \\cdot \\frac{9}{12} \\cdot \\frac{8}{12} \\cdot \\frac{7}{12} \\cdot \\frac{6}{12} \\cdot \\frac{5}{12} \\cdot \\frac{4}{12} \\cdot \\frac{3}{12} \\cdot \\frac{2}{12} \\cdot \\frac{1}{12}$ must be an integer. Simplifying, we get\n$x \\cdot \\frac{11}{12} \\cdot \\frac{5}{6} \\cdot \\frac{1}{2} \\cdot \\frac{7}{12} \\cdot \\frac{1}{2} \\cdot \\frac{5}{12} \\cdot \\frac{1}{3} \\cdot \\frac{1}{4} \\cdot \\frac{1}{6} \\cdot \\frac{1}{12}$ . Now, the minimal $x$ is the denominator of this fraction multiplied out, obviously. We mentioned before that this product must be an integer. Specifically, it is an integer and it is the amount that the $12^{\\text{th}}$ pirate receives, as he receives $\\frac{12}{12} = 1 =$ all of what is remaining.\nThus, we know the denominator is canceled out, so the number of gold coins received is going to be the product of the numerators, $11 \\cdot 5 \\cdot 7 \\cdot 5 = \\boxed{1925}$", "Solution $1$ mentioned the expression $x \\cdot \\frac{11}{12} \\cdot \\frac{10}{12} \\cdot ... \\cdot \\frac{1}{12}$ . Note that this is equivalent to $\\frac{x \\cdot 11!}{12^{11}}$\nWe can compute the amount of factors of $2$ $3$ $5$ , etc. but this is not necessary. To minimize this expression, we must take out factors of $2$ and $3$ , since $12^{11}=2^{22} \\cdot 3^{11}$ $11!$ has neither $22$ factors of $2$ , nor $11$ factors of $3$ . This means that if $11!$ contains $a$ factors of $2$ , then $x$ will contain $22-a$ factors of $2$ . This also holds for factors of $3$\nThus, once simplified, the expression will have no factors of $2$ . It will also have no factors of $3$\nLooking at the answer choices, there is only one answer which is not even, which is $\\boxed{1925}$", "We know that the 11th pirate takes $\\frac{11}{12}$ of what is left from the 10th pirate, so we have the 12th pirate taking \\[\\frac{12}{12}\\cdot(1-\\frac{11}{12})=1\\cdot\\frac{1}{12}\\] of what is left from the 10th pirate. Similarly, since the 10th pirate takes $\\frac{10}{12}$ of what is left from the 9th pirate, we have the 11th pirate taking \\[\\frac{11}{12}\\cdot(1-\\frac{10}{12})=\\frac{11}{12}\\cdot\\frac{2}{12}\\] of what is left from the 9th pirate. Thus, the 12th pirate takes \\[1\\cdot\\frac{1}{12}\\cdot\\frac{2}{12}\\] of what is left from the 9th pirate. Repeating the method, we can find that the 12th pirate takes \\[1\\cdot\\frac{1}{12}\\cdot\\frac{2}{12}\\cdot\\frac{3}{12}\\cdot...\\cdot\\frac{10}{12}\\cdot\\frac{11}{12}\\] of what is left from the 1st pirate, or \\[1\\cdot\\frac{1}{12}\\cdot\\frac{2}{12}\\cdot\\frac{3}{12}\\cdot...\\cdot\\frac{10}{12}\\cdot\\frac{11}{12}\\cdot\\frac{12}{12}\\] of the total amount of coins.\nNow canceling out the denominator with the numerator as possible, we are left with $\\frac{1\\cdot5\\cdot7\\cdot5\\cdot11}{...}=\\frac{1925}{...}$ with some factors of 12 in the denominator. For this fraction to be an integer, the smallest possible number of coins is the same as the denominator, so the numerator is the number of coins taken by the 12th pirate, or $1925 \\,\\boxed{1925}$" ]

[ "The first pirate takes $\\frac{1}{12}$ of the $x$ coins, leaving $\\frac{11}{12} x$\nThe second pirate takes $\\frac{2}{12}$ of the remaining coins, leaving $\\frac{10}{12}\\cdot \\frac{11}{12}*x$\nNote that\n$12^{11} = (2^2 \\cdot 3)^{11} = 2^{22} \\cdot 3^{11}$\n$11! = 11 \\cdot 10 \\cdot 9 \\cdot 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2$\nAll the $2$ s and $3$ s cancel out of $11!$ , leaving\n$11 \\cdot 5 \\cdot 7 \\cdot 5 = 1925$\nin the numerator.\nWe know there were just enough coins to cancel out the denominator in the fraction. So, at minimum, $x$ is the denominator, leaving $\\boxed{1925}$ coins for the twelfth pirate.", "The answer cannot be an even number. Here is why:\nConsider the highest power of 2 that divides the starting number of coins, and consider how this value changes as each pirate takes their share. At each step, the size of the pile is multiplied by some $\\frac{n}{12}$ . This means the highest power of 2 that divides the number of coins is continually decreasing or staying the same (except once, briefly, when we multiply by $\\frac{2}{3}$ for pirate 4, but it immediately drops again in the next step).\nTherefore, if the 12th pirate's coin total were even, then it can't be the smallest possible value, because we can safely cut the initial pot (and all the intermediate totals) in half. We could continue halving the result until the 12th pirate's total is finally odd.\nOnly one of the choices given is odd, $\\boxed{1925}$", "Let $x$ be the number of coins the $12$ th pirate takes. Then the number of coins the $k<12$ th pirate takes is $\\frac{12}{1} \\cdot \\frac{12}{2} \\cdots \\frac{12}{12-k} x$ . For all these to be an integer, we need the denominators to divide into the numerators. Looking at prime factors, obviously there are sufficient $2$ s and $3$ s, so we just need $5^2 \\cdot 7 \\cdot 11 = \\boxed{1925}$ to divide into $x$ . -Frestho" ]

[ "The total number of grapes eaten can be computed as the sum of the arithmetic progression with initial term $n$ (the number of grapes eaten by the child in $1$ -st place), difference $d=-2$ , and number of terms $c$ . We can easily compute that this sum is equal to $c(n-c+1)$\nHence we have the equation $2009=c(n-c+1)$ , and we are looking for a solution $(n,c)$ , where both $n$ and $c$ are positive integers, $n\\geq 2(c-1)$ , and $n$ is minimized. (The condition $n\\geq 2(c-1)$ states that even the last child had to eat a non-negative number of grapes.)\nThe prime factorization of $2009$ is $2009=7^2 \\cdot 41$ . Hence there are $6$ ways how to factor $2009$ into two positive terms $c$ and $n-c+1$\nThe smallest valid solution is therefore $c=41$ $n=\\boxed{089}$", "If the first child ate $n=2m$ grapes, then the maximum number of grapes eaten by all the children together is $2m + (2m-2) + (2m-4) + \\cdots + 4 + 2 = m(m+1)$ . Similarly, if the first child ate $2m-1$ grapes, the maximum total number of grapes eaten is $(2m-1)+(2m-3)+\\cdots+3+1 = m^2$\nFor $m=44$ the value $m(m+1)=44\\cdot 45 =1980$ is less than $2009$ . Hence $n$ must be at least $2\\cdot 44+1=89$ . For $n=89$ , the maximum possible sum is $45^2=2025$ . And we can easily see that $2009 = 2025 - 16 = 2025 - (1+3+5+7)$ , hence $2009$ grapes can indeed be achieved for $n=89$ by dropping the last four children.\nHence we found a solution with $n=89$ and $45-4=41$ kids, and we also showed that no smaller solution exists. Therefore the answer is $\\boxed{089}$", "If the winner ate n grapes, then 2nd place ate $n+2-4=n-2$ grapes, 3rd place ate $n+2-6=n-4$ grapes, 4th place ate $n-6$ grapes, and so on. Our sum can be written as $n+(n-2)+(n-4)+(n-6)\\dots$ . If there are x places, we can express this sum as $(x+1)n-x(x+1)$ , as there are $(x+1)$ occurrences of n, and $(2+4+6+\\dots)$ is equal to $x(x+1)$ . This can be factored as $(x+1)(n-x)=2009$ . Our factor pairs are (1,2009), (7,287), and (41,49). To minimize n we take (41,49). If $x+1=41$ , then $x=40$ and $n=40+49=\\boxed{089}$ . (Note we would have come upon the same result had we used $x+1=49$ .)\n~MC413551" ]

[ "If all the children were riding bicycles, there would be $2 \\times 7=14$ wheels. Each tricycle adds an extra wheel and $19-14=5$ extra wheels are needed, so there are $\\boxed{5}$ tricycles.", "Setting up an equation, we have $a+b=7$ children and $3a+2b=19$ . Solving for the variables, we get, $a=\\boxed{5}$ tricycles." ]

[ "There are $x$ clerks at the beginning, and $t$ clerks are reassigned to another task at the end of each hour. So, $30x+30(x-t)+30(x-2t)+30\\cdot\\frac{10}{60} \\cdot (x-3t)=1775$ , and simplify that we get $19x-21t=355$ .\nNow the problem is to find a reasonable integer solution. Now we know $x= \\frac{355+21t}{19}$ , so $19$ divides $355+21t$ , AND as long as $t$ is a integer, $19$ must divide $2t+355$ . Now, we suppose that $19m=2t+355$ , similarly we get $t=\\frac{19m-355}{2}$ , and so in order to get a minimum integer solution for $t$ , it is obvious that $m=19$ works. So we get $t=3$ and $x=22$ . One and a half hour's work should be $30x+15(x-t)$ , so the answer is $\\boxed{945}$", "We start with the same approach as solution 1 to get $19x-21t=355$ . Then notice that $21t + 355 \\equiv 0 \\pmod{19}$ , or $2t-6 \\equiv 0 \\pmod{19}$ , giving the smallest solution at $t=3$ . We find that $x=22$ . Then the number of files they sorted will be $30x+15(x-t)=660+285=\\boxed{945}.$" ]

[ "Since the original ratio is $3:1$ and the new ratio is $4:1$ , the number of frogs must be a multiple of $12$ , the only solutions left are $(B)$ and $(E)$\nLet's start with $12$ frogs:\nWe must have $9$ frogs in the shade and $3$ frogs in the sun. After the change, there would be $11$ frogs in the shade and $1$ frog in the sun, which is not a $4:1$ ratio.\nTherefore the answer is: $\\boxed{24}$" ]

[ "If a person gets three gumballs of each of the three colors, that is, $9$ gumballs, then the $10^{\\text{th}}$ gumball must be the fourth one for one of the colors. Therefore, the person must buy $\\boxed{10}$ gumballs." ]

[ "Georgie can enter the haunted house through any of the six windows. Then, he can leave through any of the remaining five windows.\nSo, Georgie has a total of $6 \\cdot 5$ ways he can enter the house by one window and leave\nby a different window.\nTherefore, we have $\\boxed{30}$ ways." ]

[ "All angle measures are in degrees.\nLet the first trapezoid be $ABCD$ , where $AB=BC=CD=3$ . Then the second trapezoid is $AFED$ , where $AF=FE=ED=5$ . We look for $AD$\nSince $ABCD$ is an isosceles trapezoid, we know that $\\angle BAD=\\angle CDA$ and, since $AB=BC$ , if we drew $AC$ , we would see $\\angle BCA=\\angle BAC$ . Anyway, $\\widehat{AB}=\\widehat{BC}=\\widehat{CD}$ $\\widehat{AB}$ means arc AB). Using similar reasoning, $\\widehat{AF}=\\widehat{FE}=\\widehat{ED}$\nLet $\\widehat{AB}=2\\phi$ and $\\widehat{AF}=2\\theta$ . Since $6\\theta+6\\phi=360$ (add up the angles), $2\\theta+2\\phi=120$ and thus $\\widehat{AB}+\\widehat{AF}=\\widehat{BF}=120$ . Therefore, $\\angle FAB=\\frac{1}{2}\\widehat{BDF}=\\frac{1}{2}(240)=120$ $\\angle CDE=120$ as well.\nNow I focus on triangle $FAB$ . By the Law of Cosines, $BF^2=3^2+5^2-30\\cos{120}=9+25+15=49$ , so $BF=7$ . Seeing $\\angle ABF=\\theta$ and $\\angle AFB=\\phi$ , we can now use the Law of Sines to get: \\[\\sin{\\phi}=\\frac{3\\sqrt{3}}{14}\\;\\text{and}\\;\\sin{\\theta}=\\frac{5\\sqrt{3}}{14}.\\]\nNow I focus on triangle $AFD$ $\\angle AFD=3\\phi$ and $\\angle ADF=\\theta$ , and we are given that $AF=5$ , so \\[\\frac{\\sin{\\theta}}{5}=\\frac{\\sin{3\\phi}}{AD}.\\] We know $\\sin{\\theta}=\\frac{5\\sqrt{3}}{14}$ , but we need to find $\\sin{3\\phi}$ . Using various identities, we see \\begin{align*}\\sin{3\\phi}&=\\sin{(\\phi+2\\phi)}=\\sin{\\phi}\\cos{2\\phi}+\\cos{\\phi}\\sin{2\\phi}\\\\ &=\\sin{\\phi}(1-2\\sin^2{\\phi})+2\\sin{\\phi}\\cos^2{\\phi}\\\\ &=\\sin{\\phi}\\left(1-2\\sin^2{\\phi}+2(1-\\sin^2{\\phi})\\right)\\\\ &=\\sin{\\phi}(3-4\\sin^2{\\phi})\\\\ &=\\frac{3\\sqrt{3}}{14}\\left(3-\\frac{27}{49}\\right)=\\frac{3\\sqrt{3}}{14}\\left(\\frac{120}{49}\\right)=\\frac{180\\sqrt{3}}{343} \\end{align*} Returning to finding $AD$ , we remember \\[\\frac{\\sin{\\theta}}{5}=\\frac{\\sin{3\\phi}}{AD}\\;\\text{so}\\;AD=\\frac{5\\sin{3\\phi}}{\\sin{\\theta}}.\\] Plugging in and solving, we see $AD=\\frac{360}{49}$ . Thus, the answer is $360 + 49 = 409$ , which is answer choice $\\boxed{409}$", "Note that minor arc $\\overarc{AB}$ is a third of the circumference, therefore, $\\angle AOB = 120^{\\circ}$ . Major arc $\\overarc{AB}$ $=240^{\\circ}$ $\\angle ACB = 120^{\\circ}$\nBy the Law of Cosine, $AB = \\sqrt{ 3^2 + 5^2 - 2 \\cdot 3 \\cdot 5 \\cdot \\cos 120^{\\circ}} = \\sqrt{ 9 + 25 + 15 } = 7$\n$\\frac{\\angle AOB}{2} = 60^{\\circ}$ , therefore, $r = \\frac{\\frac{AB}{2}}{\\sin 60^{\\circ}} = \\frac{\\frac{7}{2}}{ \\frac{\\sqrt{3}}{2} } = \\frac{7\\sqrt{3}}{3}$\n$\\sin \\frac{\\theta}{2} = \\frac{\\frac32}{r} = \\frac{3}{2r} = \\frac{3}{2 \\cdot \\frac{7\\sqrt{3}}{3}} = \\frac{3 \\sqrt{3}}{14}$\nLet $x$ be the length of the chord, $\\sin \\frac{3 \\theta}{2} = \\frac{\\frac{x}{2}}{r}$\nBy the triple angle formula, $\\sin \\frac{3 \\theta}{2} = 3 \\cdot \\sin \\frac{\\theta}{2} - 4 \\cdot \\sin(\\frac{ \\theta}{2})^3 = 3 \\cdot \\frac{3 \\sqrt{3}}{14} - 4 \\cdot (\\frac{3 \\sqrt{3}}{14})^3$\n$x = 2 \\cdot \\frac{7\\sqrt{3}}{3} \\cdot [3 \\cdot \\frac{3 \\sqrt{3}}{14} - 4 \\cdot (\\frac{3 \\sqrt{3}}{14})^3] = 2 \\cdot \\frac{7\\sqrt{3}}{3} \\cdot (\\frac{9\\sqrt{3}}{14} - \\frac{82\\sqrt{3}}{2 \\cdot 7^3}) = 9 - \\frac{81}{49} = \\frac{360}{49}$\nTherefore, the answer is $\\boxed{409}$", "Note that minor arc $\\overarc{AB}$ is a third of the circumference, therefore, $\\angle AOB = 120^{\\circ}$\n$\\sin \\frac{\\alpha}{2} = \\frac{\\frac32}{r}$ $\\sin \\frac{\\alpha}{2} = \\frac{3}{2r}$\n$\\sin \\frac{120^{\\circ}-\\alpha}{2} = \\frac{\\frac52}{r}$ $\\sin (60^{\\circ} - \\frac{\\alpha}{2}) = \\frac{5}{2r}$\n$\\frac{\\sin \\frac{\\alpha}{2}}{\\sin (60^{\\circ} - \\frac{\\alpha}{2}) } = \\frac{\\frac{3}{2r}}{\\frac{5}{2r}} = \\frac35$ $5 \\cdot \\sin \\frac{\\alpha}{2} = 3 \\cdot \\sin (60^{\\circ} - \\frac{\\alpha}{2})$\n$5 \\cdot \\sin \\frac{\\alpha}{2} = 3 ( \\sin 60^{\\circ} \\cos \\frac{\\alpha}{2} - \\sin \\frac{\\alpha}{2} \\cos 60^{\\circ}) = 3 ( \\frac{\\sqrt{3}}{2} \\cdot \\cos \\frac{\\alpha}{2} - \\frac12 \\cdot \\sin \\frac{\\alpha}{2})$\n$13 \\cdot \\sin \\frac{\\alpha}{2} = \\frac{3\\sqrt{3}}{2} \\cdot \\cos \\frac{\\alpha}{2}$\nLet $\\sin \\frac{\\alpha}{2} = a$ $\\cos \\frac{\\alpha}{2} = \\sqrt{1-a^2}$ $13a = \\frac{3\\sqrt{3}}{2} \\cdot \\sqrt{1-a^2}$\n$169a^2 = 27-27a^2$ $196a^2=27$ $\\sin \\frac{\\alpha}{2} = a = \\sqrt{\\frac{27}{196}} = \\frac{3 \\sqrt{3}}{14}$\nLet $x$ be the length of the chord, $\\sin \\frac{3 \\alpha}{2} = \\frac{\\frac{x}{2}}{r}$\nBy the triple angle formula, $\\sin \\frac{3 \\alpha}{2} = 3 \\cdot \\sin \\frac{\\alpha}{2} - 4 \\cdot \\sin(\\frac{ \\alpha}{2})^3 = 3 \\cdot \\frac{3 \\sqrt{3}}{14} - 4 \\cdot (\\frac{3 \\sqrt{3}}{14})^3$\n$x = 2 \\cdot \\frac{7\\sqrt{3}}{3} \\cdot [3 \\cdot \\frac{3 \\sqrt{3}}{14} - 4 \\cdot (\\frac{3 \\sqrt{3}}{14})^3] = 2 \\cdot \\frac{7\\sqrt{3}}{3} \\cdot (\\frac{9\\sqrt{3}}{14} - \\frac{82\\sqrt{3}}{2 \\cdot 7^3}) = 9 - \\frac{81}{49} = \\frac{360}{49}$\nTherefore, the answer is $\\boxed{409}$", "Note that major arc $\\overarc{AE}$ is two thirds of the circumference, therefore, $\\angle AFE = 120^{\\circ}$\nBy the Law of Cosine, $AE= \\sqrt{ 3^2 + 5^2 - 2 \\cdot 3 \\cdot 5 \\cdot \\cos 120^{\\circ}} = \\sqrt{ 9 + 25 + 15 } = 7$\nBy the Ptolemy's theorem of quadrilateral $ABDE$ $AD \\cdot BE = AB \\cdot DE + BD \\cdot AE$ $AD = BE$ $AD^2= 3 \\cdot 5 + 7^2 = 64$ $AD = 8$\nBy the Ptolemy's theorem of quadrilateral $ABCD$ $AC \\cdot BD = BC \\cdot AD + AB \\cdot CD$ $7AC = 3 \\cdot 8 + 3 \\cdot 5 = 39$ $AC = \\frac{39}{7}$\nBy the Ptolemy's theorem of quadrilateral $ABCF$ $AC \\cdot BF = AB \\cdot CF + BC \\cdot AF$ $AC = BF$ $(\\frac{39}{7})^2 = 3 \\cdot CF + 3 \\cdot 3$ $CF = \\frac{360}{49}$\nTherefore, the answer is $\\boxed{409}$" ]

[ "In hexagon $ABCDEF$ , let $AB=BC=CD=3$ and let $DE=EF=FA=5$ . Since arc $BAF$ is one third of the circumference of the circle, it follows that $\\angle BCF = \\angle BEF=60^{\\circ}$ . Similarly, $\\angle CBE =\\angle CFE=60^{\\circ}$ . Let $P$ be the intersection of $\\overline{BE}$ and $\\overline{CF}$ $Q$ that of $\\overline{BE}$ and $\\overline{AD}$ , and $R$ that of $\\overline{CF}$ and $\\overline{AD}$ . Triangles $EFP$ and $BCP$ are equilateral, and by symmetry, triangle $PQR$ is isosceles and thus also equilateral. \nFurthermore, $\\angle BAD$ and $\\angle BED$ subtend the same arc, as do $\\angle ABE$ and $\\angle ADE$ . Hence triangles $ABQ$ and $EDQ$ are similar. Therefore, \\[\\frac{AQ}{EQ}=\\frac{BQ}{DQ}=\\frac{AB}{ED}=\\frac{3}{5}.\\] It follows that \\[\\frac{\\frac{AD-PQ}{2}}{PQ+5} =\\frac{3}{5}\\quad \\mbox {and}\\quad \\frac{3-PQ}{\\frac{AD+PQ}{2}}=\\frac{3}{5}.\\] Solving the two equations simultaneously yields $AD=360/49,$ so $m+n=\\boxed{409}. \\blacksquare$" ]

[ "Let us call the hexagon $ABCDEF$ , where $AB=CD=DE=AF=22$ , and $BC=EF=20$ . \nWe can just consider one half of the hexagon, $ABCD$ , to make matters simpler. \nDraw a line from the center of the circle, $O$ , to the midpoint of $BC$ $X$ . Now, draw a line from $O$ to the midpoint of $AB$ $Y$ . Clearly, $\\angle BXO=90^{\\circ}$ , because $BO=CO$ , and $\\angle BYO=90^{\\circ}$ , for similar reasons. Also notice that $\\angle AOX=90^{\\circ}$ . \nLet us call $\\angle BOY=\\theta$ . Therefore, $\\angle AOB=2\\theta$ , and so $\\angle BOX=90-2\\theta$ . Let us label the radius of the circle $r$ . This means \\[\\sin{\\theta}=\\frac{BY}{r}=\\frac{11}{r}\\] \\[\\sin{(90-2\\theta)}=\\frac{BX}{r}=\\frac{10}{r}\\] Now we can use simple trigonometry to solve for $r$ .\nRecall that $\\sin{(90-\\alpha)}=\\cos(\\alpha)$ : That means $\\sin{(90-2\\theta)}=\\cos{2\\theta}=\\frac{10}{r}$ .\nRecall that $\\cos{2\\alpha}=1-2\\sin^2{\\alpha}$ : That means $\\cos{2\\theta}=1-2\\sin^2{\\theta}=\\frac{10}{r}$ .\nLet $\\sin{\\theta}=x$ .\nSubstitute to get $x=\\frac{11}{r}$ and $1-2x^2=\\frac{10}{r}$ Now substitute the first equation into the second equation: $1-2\\left(\\frac{11}{r}\\right)^2=\\frac{10}{r}$ Multiplying both sides by $r^2$ and reordering gives us the quadratic \\[r^2-10r-242=0\\] Using the quadratic equation to solve, we get that $r=5+\\sqrt{267}$ (because $5-\\sqrt{267}$ gives a negative value), so the answer is $5+267=\\boxed{272}$", "Using the trapezoid $ABCD$ mentioned above, draw an altitude of the trapezoid passing through point $B$ onto $AD$ at point $J$ . Now, we can use the pythagorean theorem: $(22^2-(r-10)^2)+10^2=r^2$ . Expanding and combining like terms gives us the quadratic \\[r^2-10r-242=0\\] and solving for $r$ gives $r=5+\\sqrt{267}$ . So the solution is $5+267=\\boxed{272}$", "Join the diameter of the circle $AD$ and let the length be $d$ . By Ptolemy's Theorem on trapezoid $ADEF$ $(AD)(EF) + (AF)(DE) = (AE)(DF)$ . Since it is an isosceles trapezoid, both diagonals are equal. Let them be equal to $x$ each. Then\n\\[20d + 22^2 = x^2\\]\nSince $\\angle AED$ is subtended by the diameter, it is right. Hence by the Pythagorean Theorem with right $\\triangle AED$\n\\[(AE)^2 + (ED)^2 = (AD)^2\\] \\[x^2 + 22^2 = d^2\\]\nFrom the above equations, we have: \\[x^2 = d^2 - 22^2 = 20d + 22^2\\] \\[d^2 - 20d = 2\\times22^2\\] \\[d^2 - 20d + 100 = 968+100 = 1068\\] \\[(d-10) = \\sqrt{1068}\\] \\[d = \\sqrt{1068} + 10 = 2\\times(\\sqrt{267}+5)\\]\nSince the radius is half the diameter, it is $\\sqrt{267}+5$ , so the answer is $5+267 \\Rightarrow \\boxed{272}$", "As we can see this image, it is symmetrical hence the diameter divides the hexagon into two congruent quadrilateral. Now we can apply the Ptolemy's theorem. Denote the radius is r, we can get \\[22*2x+440=\\sqrt{4x^2-400}\\sqrt{4x^2-484}\\] , after simple factorization, we can get \\[x^4-342x^2-2420x=0\\] , it is easy to see that $x=-10, x=0$ are two solutions for the equation, so we can factorize that into \\[x(x+10)(x^2-10x-242)\\] so we only need to find the solution for \\[x^2-10x-242=0\\] and we can get $x=(\\sqrt{267}+5)$ is the desired answer for the problem, and our answer is $5+267 \\Rightarrow \\boxed{272}$ .～bluesoul", "Using solution 1's diagram, extend line segments $AB$ and $CD$ upwards until they meet at point $G$ . Let point $O$ be the center of the hexagon. By the $AA$ postulate, $\\Delta ADG \\sim \\Delta BCG \\sim \\Delta CDO$ . This means $\\frac{DC}{AD} = \\frac{22}{2r} = \\frac{CO}{AG}$ , so $AG = r \\times \\frac{2r}{22} = \\frac{r^2}{11}$ . We then solve for $AB$ \\[\\frac{AD-BC}{AD} = \\frac{2r-20}{2r} = \\frac{AB}{AG}\\] \\[AB = \\frac{r^2}{11} \\times \\frac{2r-20}{2r} = \\frac{r^2-10r}{11}\\] Remember that $AB=22$ as well, so $22 = \\frac{r^2-10r}{11} \\Rightarrow r^2-10r-242=0$ . Solving for $r$ gives $r=5+\\sqrt{267}$ . So the solution is $5+267=\\boxed{272}$", "Let $\\angle{AOB} = \\theta$ . So, we have $\\sin \\dfrac{\\theta}{2} = \\dfrac{11}{r}$ and $\\cos \\dfrac{\\theta}{2} = \\dfrac{\\sqrt{r^{2} - 121}}{r}$ . So, $\\sin \\theta = 2 \\sin \\dfrac{\\theta}{2} \\cos \\dfrac{\\theta}{2} = \\dfrac{22 \\sqrt{r^{2} - 121}}{r^{2}}$ . Let $H$ be the foot of the perpendicular from $B$ to $\\overline{AD}$ . We have $BF = 2 BH = 2 r \\sin \\theta = \\dfrac{44 \\sqrt{r^{2} - 121}}{r}$ . Using Pythagorean theorem on $\\triangle BCF$ , to get $(\\dfrac{44 \\sqrt{r^{2} - 121}}{r})^{2} + 20^{2} = (2r)^{2}$ , or $\\dfrac{44^{2}r^{2} - 44^{2} \\cdot 121}{r^{2}} + 20^{2} = 4r^{4}$ . Multiplying by $r^{2}$ , we get $44^{2} r^{2} - 44^{2} \\cdot 121 + 20^{2} r^{2} = 4r^{4}$ . Rearranging and simplifying, we get a quadratic in $r^{2}$ \\[r^{4} - 584r^{2} + 242^{2} = 0 \\text\\] which gives us $r^{2} = 292 \\pm 10\\sqrt{267}$ . Because $r$ is in the form $p + \\sqrt{q}$ , we know to choose the larger option, meaning $r^2 = 292 + 10\\sqrt{267}$ , so $p\\sqrt{q} = 5\\sqrt{267}$ and $p^2 + q = 292$ . By inspection, we get $(p, q) = (5, 267)$ , so our answer is $5 + 267 = \\boxed{272}$", "\nWe know that $AD=x$ is a diameter, hence $ABD$ and $ACD$ are right triangles. Let $AB=BC=22$ , and $CD=20.$ Hence, $ABD$ is a right triangle with legs $22,\\sqrt{x^2-484},$ and hypotenuse, $x,$ and $ACD$ is a right triangle with legs $20, \\sqrt{x^2-400},$ with hypotenuse $x$ . By Ptolemy's we have \\[22(x+20)=\\sqrt{x^2-400}\\sqrt{x^2-484}\\] .\nWe square both sides to get \\[484(x+20)^2=(x^2-400)(x^2-484) \\implies 484(x+20)=(x-20)(x^2-484) \\implies 484x=x^3-20x^2-484x \\implies x(x^2-20x-968)=0\\]\nWe solve for $x$ via the Quadratic Formula and receive $x=10+2\\sqrt{267}$ , but we must divide by $2$ since we want the radius, and hence $267+5=\\boxed{272}.$ ~SirAppel" ]

[ "Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know $r > 1$ because the sequence is said to be increasing. We also know that each of $a, ar, ar^2, ar^3$ is an integer. We start by showing that $r$ must also be an integer.\nSuppose not, and say $r = m/n$ where $m>n>1$ , and $\\gcd(m,n)=1$ . Then $n, n^2, n^3$ must all divide $a$ so $a=n^3k$ for some integer $k$ . Then $S_R = n^3k + n^2mk + nm^2k + m^3k < 100$ and we see that even if $k=1$ and $n=2$ , we get $m < 4$ , which means that the only option for $r$ is $r=3/2$ . A quick check shows that even this doesn't work. Thus $r$ must be an integer.\nLet $a, a+d, a+2d, a+3d$ be the quarterly scores for the Wildcats. Let $S_W = a+(a+d) + (a+2d)+(a+3d) = 4a+6d$ . Let $S_R = a+ar+ar^2+ar^3 = a(1+r)(1+r^2)$ . Then $S_R<100$ implies that $r<5$ , so $r\\in \\{2, 3, 4\\}$ . The Raiders win by one point, so \\[a(1+r)(1+r^2) = 4a+6d+1.\\]\nThen the quarterly scores for the Raiders are $5, 10, 20, 40$ , and those for the Wildcats are $5, 14, 23, 32$ . Also $S_R = 75 = S_W + 1$ . The total number of points scored by the two teams in the first half is $5+10+5+14=\\boxed{34}$", "Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let $a,a+d,a+2d,a+3d$ be the quarterly scores for the Wildcats. The sum of the Raiders scores is $a(1+r+r^{2}+r^{3})$ and the sum of the Wildcats scores is $4a+6d$ . Now we can narrow our search for the values of $a,d$ , and $r$ . Because points are always measured in positive integers, we can conclude that $a$ and $d$ are positive integers. We can also conclude that $r$ is a positive integer by writing down the equation:\n\\[a(1+r+r^{2}+r^{3})=4a+6d+1\\]\nNow we can start trying out some values of $r$ . We try $r=2$ , which gives\n\\[15a=4a+6d+1\\]\n\\[11a=6d+1\\]\nWe need the smallest multiple of $11$ (to satisfy the <100 condition) that is $\\equiv 1 \\pmod{6}$ . We see that this is $55$ , and therefore $a=5$ and $d=9$\nSo the Raiders' first two scores were $5$ and $10$ and the Wildcats' first two scores were $5$ and $14$\n\\[5+10+5+14=34 \\longrightarrow \\boxed{34}\\]" ]

[ "Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know $r > 1$ because the sequence is said to be increasing. We also know that each of $a, ar, ar^2, ar^3$ is an integer. We start by showing that $r$ must also be an integer.\nSuppose not, and say $r = m/n$ where $m>n>1$ , and $\\gcd(m,n)=1$ . Then $n, n^2, n^3$ must all divide $a$ so $a=n^3k$ for some integer $k$ . Then $S_R = n^3k + n^2mk + nm^2k + m^3k < 100$ and we see that even if $k=1$ and $n=2$ , we get $m < 4$ , which means that the only option for $r$ is $r=3/2$ . A quick check shows that even this doesn't work. Thus $r$ must be an integer.\nLet $a, a+d, a+2d, a+3d$ be the quarterly scores for the Wildcats. Let $S_W = a+(a+d) + (a+2d)+(a+3d) = 4a+6d$ . Let $S_R = a+ar+ar^2+ar^3 = a(1+r)(1+r^2)$ . Then $S_R<100$ implies that $r<5$ , so $r\\in \\{2, 3, 4\\}$ . The Raiders win by one point, so \\[a(1+r)(1+r^2) = 4a+6d+1.\\]\nThen the quarterly scores for the Raiders are $5, 10, 20, 40$ , and those for the Wildcats are $5, 14, 23, 32$ . Also $S_R = 75 = S_W + 1$ . The total number of points scored by the two teams in the first half is $5+10+5+14=\\boxed{34}$", "Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let $a,a+d,a+2d,a+3d$ be the quarterly scores for the Wildcats. The sum of the Raiders scores is $a(1+r+r^{2}+r^{3})$ and the sum of the Wildcats scores is $4a+6d$ . Now we can narrow our search for the values of $a,d$ , and $r$ . Because points are always measured in positive integers, we can conclude that $a$ and $d$ are positive integers. We can also conclude that $r$ is a positive integer by writing down the equation:\n\\[a(1+r+r^{2}+r^{3})=4a+6d+1\\]\nNow we can start trying out some values of $r$ . We try $r=2$ , which gives\n\\[15a=4a+6d+1\\]\n\\[11a=6d+1\\]\nWe need the smallest multiple of $11$ (to satisfy the <100 condition) that is $\\equiv 1 \\pmod{6}$ . We see that this is $55$ , and therefore $a=5$ and $d=9$\nSo the Raiders' first two scores were $5$ and $10$ and the Wildcats' first two scores were $5$ and $14$\n\\[5+10+5+14=34 \\longrightarrow \\boxed{34}\\]" ]

[ "Use construction . We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined.\nOur answer is thus $\\frac{9}{28} \\cdot \\frac{2}{5} = \\frac{9}{70}$ , and $m + n = \\boxed{79}$", "Call the three different types of rolls as A, B, and C. We need to arrange 3As, 3Bs, and 3Cs in a string such that A, B, and C appear in the first three, second three, and the third three like ABCABCABC or BCABACCAB. This can occur in $\\left(\\frac{3!}{1!1!1!}\\right)^3 = 6^3 = 216$ different manners. The total number of possible strings is $\\frac{9!}{3!3!3!} = 1680$ . The solution is therefore $\\frac{216}{1680} = \\frac{9}{70}$ , and $m + n = \\boxed{79}$", "The denominator of m/n is equal to the total amount of possible roll configurations given to the three people. This is equal to ${9 \\choose 3}{6 \\choose3}$ as the amount of ways to select three rolls out of 9 to give to the first person is ${9 \\choose 3}$ , and three rolls out of 6 is ${6 \\choose3}$ . After that, the three remaining rolls have no more configurations.\nThe numerator is the amount of ways to give one roll of each type to each of the three people, which can be done by defining the three types of rolls as x flavored, y flavored, and z flavored.\nxxx, yyy, zzz\nSo you have to choose one x, one y, and one z to give to the first person. There are 3 xs, 3 ys, and 3 zs to select from, giving $3^3$ combinations. Multiply that by the combinations of xs, ys, and zs for the second person, which is evidently $2^3$ since there are two of each letter left.\n$(27*8)/{9 \\choose 3}{6 \\choose3}$ simplifies down to our fraction m/n, which is $9/70$ . Adding them up gives $9 + 70 = \\boxed{79}$" ]

[ "Denote the original price of the house and the store as $h$ and $s$ , respectively. It is given that $\\frac{4h}{5}=\\textdollar 12,000$ , and that $\\frac{6s}{5}=\\textdollar 12,000$ . Thus, $h=\\textdollar 15,000$ $s=\\textdollar10,000$ , and $h+s=\\textdollar25,000$ . This value is $\\textdollar1000$ higher than the current price of the property, $2\\cdot \\textdollar12,000$ . Hence, the transaction resulted in a $\\boxed{1000}$" ]

[ "We do $100-30-20-15-10$ to find the percent of gumdrops that are green. We find that $25\\%$ of the gumdrops are green. That means there are $120$ gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then $15\\%$ of the jar's gumdrops are brown. $\\dfrac{35}{100} \\cdot 120=42 \\Rightarrow \\boxed{42}$" ]

[ "The probability that Terry picks two red candies is $\\frac{10 \\cdot 9}{20 \\cdot 19} = \\frac{9}{38}$ , and the probability that Mary picks two red candies after Terry chooses two red candies is $\\frac{7\\cdot8}{18\\cdot17} = \\frac{28}{153}$ . So the probability that they both pick two red candies is $\\frac{9}{38} \\cdot \\frac{28}{153} = \\frac{14}{323}$ . The same calculation works for the blue candies.\nThe probability that Terry picks two different candies is $\\frac{20\\cdot10}{20\\cdot19} = \\frac{10}{19}$ , and the probability that Mary picks two different candies after Terry picks two different candies is $\\frac{18\\cdot 9}{18\\cdot 17} = \\frac{9}{17}$ . Thus, the probability that they both choose two different candies is $\\frac{10}{19}\\cdot\\frac{9}{17} = \\frac{90}{323}$ . Then the total probability is\n\\[2 \\cdot \\frac{14}{323} + \\frac{90}{323} = \\frac{118}{323}\\]\nand so the answer is $118 + 323 = \\boxed{441}$" ]

[ "Note that \\[\\frac{\\text{number of trout}}{\\text{total number of fish}} = \\frac{30}{180} = \\frac16.\\] So, the total number of fish is $6$ times the number of trout. Since the lake contains $250$ trout, there are $250\\cdot6=\\boxed{1500}$ fish in the lake." ]

[ "If there are $x$ pennies in the bag, then there are $2x$ dimes and $3(2x) = 6x$ quarters. Since pennies are $$0.01$ , dimes are $$0.10$ , and quarters are $$0.25$ , the total amount of money in the bag is \\[$ \\left(0.01x+(0.10)(2x)+(0.25)(6x)\\right) = $1.71x.\\] Therefore, the possible amounts of money are precisely the integer multiples of $$1.71$\nSince the answer choices are all integer numbers of dollars, we multiply by $100$ to deduce that the answer must be an integer multiple of $$171$ . The only such multiple among the answer choices is $$(2 \\cdot 171) = \\boxed{342}$" ]

[ "We find that $T=10(1+2+\\cdots +119)$ . From Gauss's formula, we find that the value of $T$ is $10(7140)=71400$ . The value of $\\frac{T}{2}$ is therefore $35700$ . We find that $35700$ is $10(3570)=10\\cdot \\frac{k(k+1)}{2}$ , so $3570=\\frac{k(k+1)}{2}$ . As a result, $7140=k(k+1)$ , which leads to $0=k^2+k-7140$ . We notice that $k=84$ , so the answer is $10(119-84)=\\boxed{350}$" ]

[ "Place the cube so that its space diagonal is perpendicular to the ground. The space diagonal has length of $3\\sqrt{3}$ , the altitude of the top vertex of the newly placed cube is $3\\sqrt{3}$ . The plane perpendicular and bisecting the space diagonal is now parallel to the ground and also bisecting the space diagonal into $\\frac{3\\sqrt{3}}{2}$ , so that the height of the plane is $\\frac{3\\sqrt{3}}{2}$\nBy symmetry, the space diagonal is trisected by the pyramid at the top of the cube and the pyramid at the bottom of the cube.\nWe can prove that the space diagonal is trisected. Let the altitude of the pyramid at the top of the cube be $h$ . The base of the pyramid is an equilateral triangle with side length of $\\sqrt{3^2+3^2}=3\\sqrt{2}$ . The height of the triangle is $\\frac{ \\sqrt{3} }{2} \\cdot 3\\sqrt{2}$ . The distance of the center of the triangle to the vertex is $\\frac23 \\cdot \\frac{ \\sqrt{3} }{2} \\cdot 3\\sqrt{2} = \\sqrt {6}$ . Therefore, $h = \\sqrt{3^2-(\\sqrt {6})^2} = \\sqrt{9-6}=\\sqrt{3}$ . The altitude of the pyramid at the bottom of the cube is also $h$ . The altitude in the middle is $3\\sqrt{3}-\\sqrt{3}-\\sqrt{3}=\\sqrt{3}$\nThe altitude of the vertex at the top is $3\\sqrt{3}$ . The altitude of the second highest $3$ vertices are all $2\\sqrt{3}$ . The altitude of the third highest $3$ vertices are all $\\sqrt{3}$ . The altitude of the bottom-most vertex is $0$\nBy scale, for the unit cube, place the cube so that its space diagonal is perpendicular to the ground. The altitude of the vertex at the top is $\\sqrt{3}$ . The altitude of the second highest $3$ vertices are all $\\frac{2\\sqrt{3}}{3}$ . The altitude of the third highest $3$ vertices are all $\\frac{\\sqrt{3}}{3}$ . The altitude of the bottom-most vertex is $0$\nThe length of the space diagonal of a unit cube is $\\sqrt{3}$ . The highest vertex of the bottom-most unit cube has an altitude of $\\sqrt{3}$ . As $\\sqrt{3} < \\frac{3\\sqrt{3}}{2}$ , therefore, the plane will not pass through the unit cube at the bottom.\nFor the next $3$ cubes from the bottom, the altitude of their highest vertex is $\\frac{\\sqrt{3}}{3} + \\sqrt{3} = \\frac{4\\sqrt{3}}{3}$ . As $\\frac{4\\sqrt{3}}{3} < \\frac{3\\sqrt{3}}{2}$ , therefore, the plane will not pass through the next $3$ unit cubes.\nFor the next $3$ cubes from the bottom, the altitude of their highest vertex is $\\frac{2\\sqrt{3}}{3} + \\frac{\\sqrt{3}}{3} = \\frac{5\\sqrt{3}}{3}$ . As $\\frac{5\\sqrt{3}}{3} > \\frac{3\\sqrt{3}}{2}$ , therefore, the plane will pass through the next $3$ unit cubes.\nSo at the bottom half of the cube, there are only $1+3 = 4$ unit cubes that the plane does not passes through. By symmetry, the plane will not pass through $4$ unit cubes at the top half of the cube.\nThus, the plane does not pass through $4+4 = 8$ unit cubes, it passes through $27-8=\\boxed{19}$ unit cubes." ]

[ "Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.\nNow consider the diagonal from $(0,0,0)$ to $(3,3,3)$ . The midpoint of this diagonal is at $\\left(\\frac 32,\\frac 32,\\frac 32\\right)$ . The plane that passes through this point and is orthogonal to the diagonal has the equation $x+y+z=\\frac 92$\nThe unit cube with opposite corners at $(x,y,z)$ and $(x+1,y+1,z+1)$ is intersected by this plane if and only if $x+y+z < \\frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3$ . Therefore the cube is intersected by this plane if and only if $x+y+z\\in\\{2,3,4\\}$\nThere are six cubes such that $x+y+z=2$ : permutations of $(1,1,0)$ and $(2,0,0)$ Symmetrically, there are six cubes such that $x+y+z=4$ Finally, there are seven cubes such that $x+y+z=3$ : permutations of $(2,1,0)$ and the central cube $(1,1,1)$\nThat gives a total of $\\boxed{19}$ intersected cubes." ]

[ "There are $1+3+5+...+2003=1002^{2}=1004004$ small equilateral triangles.\nEach small equilateral triangle needs $3$ toothpicks to make it.\nBut, each toothpick that isn't one of the $1002\\cdot3=3006$ toothpicks on the outside of the large equilateral triangle is a side for $2$ small equilateral triangles.\nSo, the number of toothpicks on the inside of the large equilateral triangle is $\\frac{10040004\\cdot3-3006}{2}=1504503$\nTherefore the total number of toothpicks is $1504503+3006=\\boxed{1,507,509}$ ~dolphin7", "We just need to count upward facing triangles because if we exclude the downward-facing triangles, we won't be overcounting any toothpicks. The first row of triangles has $1$ upward-facing triangle, the second row has $2$ upward-facing triangles, the third row has $3$ upward-facing triangles, and so on having $n$ upward-facing triangles in the $n^\\text{th}$ row. The last row with $2003$ small triangles has $1002$ upward-facing triangles. By Gauss's formula, the number of the upward-facing triangles in the entire triangle are now $\\frac{1002\\times1003}{2}$ , meaning that the number of toothpicks are $\\frac{1002\\times1003}{2}\\times3$ , or $\\boxed{1,507,509}$", "You don't have to calculate the value of $\\frac{1002\\times1003}{2}\\times3$ , and you can use units digits to find the answer easily. The units digit of $1002\\times1003$ is $6$ , and has a unit digit of $3$ after being divided by $2$ . Then this is multiplied by $3$ , now the final number ending with a $9$ . This leaves only one answer choice possible, which is $\\boxed{1,507,509}$" ]

[ "Test out some fewer cases first.\nWhen there is just 1 equilateral triangle in the base, you need $3$ toothpicks.\nWhen there are 3 equilateral triangles in the base, you need $9$ toothpicks in all.\nWhen there are 5 equilateral triangles in the base, you need $18$ toothpicks in all.\nWhen there are 7 equilateral triangles in the base, you need $30$ toothpicks in all.\nTaking the finite differences, we get $6, 9, 12.$ It forms a linear equations. This means the original numbers $(3, 9, 18, 30)$ form a quadratic.\nLet the quadratic be $y = ax^2 + bx + c$ where $y = 2* \\text{equilateral triangles in base} - 1.$\nThen, we have the following points: $(1, 3), (2, 9), (3, 18), (4, 30).$\nWe can plug these values into $y = ax^2 + bx + c$ , giving:\n\\[a + b + c = 3, 4a + b + c = 9, 9a + 3b + c = 18.\\]\nSolving gives $a = b = 1.5, c = 0.$ So, \\[y = 1.5x^2 + 1.5x.\\]\nFor our problem, we need it when there are $2003$ equilateral triangles in the base. For the quadratic, the corresponding $x$ -value would be $\\frac{2003 + 1}{2} = 1002.$ . So, our answer is simply: \\[1.5 * 1002^2 + 1.5*1002 = \\boxed{1507509}.\\]" ]

[ "\nWe can easily see the dimensions of each small rectangle. So the area of the last rectangle is $3\\times 5=\\boxed{15}$" ]

[ "Let the length of the longer side be $x$ , and the length of the shorter side be $y$ . We are given that $2x+2y=14\\implies x+y=7$ . However, note that $x+y$ is also the length of a side of the larger square. Thus the area of the larger square is $(x+y)^2=7^2=\\boxed{49}$", "Expand the small square so it basically equals the area of the large square. Two of the sides of the rectangles shrink to zero. The other two sides expand to equal the length of the large outer square, and have a length of $\\frac{14}{2} = 7$ . Thus, the area of the larger square is $7^2=\\boxed{49}$" ]

[ "There are $36$ red balls; for these red balls to comprise $72 \\%$ of the urn, there must be only $14$ blue balls. Since there are currently $64$ blue balls, this means we must remove $\\boxed{50}$", "There are $36$ red balls and $64$ blue balls. For the percentage of the red balls to double from $36 \\%$ to $72 \\%$ of the urn, half of the total number of balls must be removed. Therefore, the number of blue balls that need to be removed is $\\boxed{50}$", "There are $36$ red balls out of the total $100$ balls. \nWe want to continuously remove blue balls until the percentage of red balls in the urn is 72%.\nTherefore, we want \\[\\frac{36}{100-x}=\\frac{72}{100}.\\] Solving for $x$ gives that we must remove $\\boxed{50}$ blue balls." ]

[ "The difference in the $y$ -coordinates is $281 - 17 = 264$ , and the difference in the $x$ -coordinates is $48 - 3 = 45$ .\nThe gcd of 264 and 45 is 3, so the line segment joining $(3,17)$ and $(48,281)$ has slope \\[\\frac{88}{15}.\\] The points on the line have coordinates \\[\\left(3+t,17+\\frac{88}{15}t\\right).\\] If $t$ is an integer, the $y$ -coordinate of this point is an integer if and only if $t$ is a multiple of 15. The points where $t$ is a multiple of 15 on the segment $3\\leq x\\leq 48$ are $3$ $3+15$ $3+30$ , and $3+45$ . There are 4 lattice points on this line. Hence the answer is $\\boxed{4}$" ]

[ "The shortest segments would be perpendicular to the square. The lemming went $x$ meters horizontally and $y$ meters vertically. No matter how much it went, the lemming would have been $x$ and $y$ meters from the sides and $10-x$ and $10-y$ meters from the remaining two. To find the average, add the lengths of the four segments and divide by four: $\\frac {\\cancel{x}+10-\\cancel{x}+\\cancel{y}+10-\\cancel{y}}{4} =$ $\\boxed{5}$", "For any point in the square, the sum of its distance to the left edge and right edge is equal to $10$ , and the sum of its distance to the up edge and down edge is also equal to $10$ . Thus, the answer is $\\boxed{5}$ , and the moving progress is misguide at all." ]

[ "Since the line $x=k$ is vertical, we are only concerned with vertical distance.\nIn other words, we want to find the value of $k$ for which the distance $|\\log_5 x - \\log_5 (x+4)| = \\frac{1}{2}$\nSince $\\log_5 x$ is a strictly increasing function, we have:\n$\\log_5 (x + 4) - \\log_5 x = \\frac{1}{2}$\n$\\log_5 (\\frac{x+4}{x}) = \\frac{1}{2}$\n$\\frac{x+4}{x} = 5^\\frac{1}{2}$\n$x + 4 = x\\sqrt{5}$\n$x\\sqrt{5} - x = 4$\n$x = \\frac{4}{\\sqrt{5} - 1}$\n$x = \\frac{4(\\sqrt{5} + 1)}{5 - 1^2}$\n$x = 1 + \\sqrt{5}$\nThe desired quantity is $1 + 5 = 6$ , and the answer is $\\boxed{6}$" ]

[ "Let $w$ be the length of the shorter segment and $l$ be the length of the longer segment. We're given that $\\frac{w}{l}=\\frac{l}{w+l}$ . Cross-multiplying, we find that $w^2+wl=l^2\\implies w^2+wl-l^2=0$ . Now we divide both sides by $l^2$ to get $\\left(\\frac{w}{l}\\right)^2+\\left(\\frac{w}{l}\\right)-1=0$ . Therefore, $R^2+R-1=0$\nFrom this, we have $R^2=-R+1$ . Dividing both sides by $R$ , we get $R=-1+\\frac{1}{R}\\implies R^{-1}=R+1$ . Therefore, $R^2+R^{-1}=-R+1+R+1=2$ . Finally, we have \\[R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}=R^{[R^2+R^{-1}]}+R^{-1}=R^2+R^{-1}= \\boxed{2}.\\]" ]

[ "Using point slope form, we get the equations $y-30 = 6(x-40)$ and $y-30 = 2(x-40)$ . Simplifying, we get $6x-y=210$ and $2x-y=50$ . Letting $y=0$ in both equations and solving for $x$ gives the $x$ -intercepts: $x=35$ and $x=25$ , respectively. Thus the distance between them is $35-25=\\boxed{10}$", "In order for the line with slope $2$ to travel \"up\" $30$ units (from $y=0$ ), it must have traveled $30/2=15$ units to the right. Thus, the $x$ -intercept is at $x=40-15=25$ . As for the line with slope $6$ , in order for it to travel \"up\" $30$ units it must have traveled $30/6=5$ units to the right. Thus its $x$ -intercept is at $x=40-5=35$ . Then the distance between them is $35-25=\\boxed{10}$" ]

[ "Using the point-slope form, the equation of each line is\n\\[y-15=3(x-10) \\longrightarrow y=3x-15\\] \\[y-15=5(x-10) \\longrightarrow y=5x-35\\]\nSubstitute in $y=0$ to find the $x$ -intercepts.\n\\[0=3x-15\\longrightarrow x=5\\] \\[0=5x-35\\longrightarrow x=7\\] The difference between them is $7-5=\\boxed{2}$", "The $x$ -intercepts of a line is where the line's $y$ -coordinate is $=0$ . The slope is defined as $\\frac{\\text{change in } y\\text{-coordinate}}{\\text{change in } x\\text{-coordinate}}$ . Therefore, the line with slope $3$ can be expressed as\n\\[\\frac{\\text{change in }y\\text{-coordinate}}{\\text{change in }x\\text{-coordinate}}=3\\]\nFrom $(10,15)$ to the $x$ -intercept, the line must move $-15$ units to the $x$ -axis. Therefore, \\[\\frac{\\text{change in }y\\text{-coordinate}}{\\text{change in }x\\text{-coordinate}}=\\frac{-15}{x_1}=3\\implies x_1=-5\\]\nTherefore, the $x$ -intercept of the line with slope $3$ is $(10-5,0)=(5,0)$ .Note that some sources may state \"the $x$ -intercept is $5$ \" instead of \"the $x$ -intercept is $(5,0)$\nIf an idea functions great for one part of the problem but does not find the solution to the problem, we want to use the idea again. Using the idea for the line with slope $5$ , we find \\[\\frac{\\text{change in }y\\text{-coordinate}}{\\text{change in }x\\text{-coordinate}}=\\frac{-15}{x_2}=5\\implies x_2=-3\\] .\nTherefore, the $x$ -intercept of the line with slope $5$ is $(10-3,0)=(7,0)$ . The distance between $\\boxed{2}$" ]

[ "We start off with the fact that the median is $9$ , so we must have $a, b, c, d, e, 9, f, g, h, i, j$ , listed in ascending order. Note that the integers do not have to be distinct.\nSince the mode is $8$ , we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we will have $a, b, c, 8, 8, 9, f, g, h, i, j$ . In this case, since $8$ is the unique mode, the rest of the integers have to be distinct. So we minimize $a,b,c,f,g,h,i$ in order to maximize $j$ . If we let the list be $1,2,3,8,8,9,10,11,12,13,j$ , then $j = 11 \\times 10 - (1+2+3+8+8+9+10+11+12+13) = 33$\nNext, consider the case where there are $3$ occurrences of $8$ in the list. Now, we can have two occurrences of another integer in the list. We try $1,1,8,8,8,9,9,10,10,11,j$ . Following the same process as above, we get $j = 11 \\times 10 - (1+1+8+8+8+9+9+10+10+11) = 35$ . As this is the highest choice in the list, we know this is our answer. Therefore, the answer is $\\boxed{35}$", "Note that $x_1 + \\ldots + x_{11} = 110$ let $x_6 = 9$ so $x_1 + \\ldots + x_5 + x_7 + \\ldots + x_{11} = 101$ . To maximize the value of $x_i$ where $i$ ranges from $1$ to $11$ , we let any $7$ elements be $1,2,\\ldots,7$ so $x_1 + x_2 + x_3 = 57$ . Now we have to let one of above $3$ values = $8$ hence $x_1 + x_2 = 49$ now let $x_1 = 35$ $x_2 = 14$ hence $\\boxed{35}$ is the answer." ]

[ "We start off with the fact that the median is $9$ , so we must have $a, b, c, d, e, 9, f, g, h, i, j$ , listed in ascending order. Note that the integers do not have to be distinct.\nSince the mode is $8$ , we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we will have $a, b, c, 8, 8, 9, f, g, h, i, j$ . In this case, since $8$ is the unique mode, the rest of the integers have to be distinct. So we minimize $a,b,c,f,g,h,i$ in order to maximize $j$ . If we let the list be $1,2,3,8,8,9,10,11,12,13,j$ , then $j = 11 \\times 10 - (1+2+3+8+8+9+10+11+12+13) = 33$\nNext, consider the case where there are $3$ occurrences of $8$ in the list. Now, we can have two occurrences of another integer in the list. We try $1,1,8,8,8,9,9,10,10,11,j$ . Following the same process as above, we get $j = 11 \\times 10 - (1+1+8+8+8+9+9+10+10+11) = 35$ . As this is the highest choice in the list, we know this is our answer. Therefore, the answer is $\\boxed{35}$" ]

[ "To minimize the number of distinct values, we want to maximize the number of times a number appears. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223 + 1 + 1 = \\boxed{225}.$", "As in Solution 1, we want to maximize the number of time each number appears to do so. We can set up an equation $10 + 9( x - 1 )\\geq2018,$ where $x$ is the number of values. Notice how we can then rearrange the equation into $1 + 9 ( 1 )+9 ( x - 1 )\\geq2018,$ which becomes $9 x\\geq2017,$ or $x\\geq224\\frac19.$ We cannot have a fraction of a value so we must round up to $\\boxed{225}.$" ]

[ "To minimize the number of distinct values, we want to maximize the number of times a number appears. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223 + 1 + 1 = \\boxed{225}.$", "As in Solution 1, we want to maximize the number of time each number appears to do so. We can set up an equation $10 + 9( x - 1 )\\geq2018,$ where $x$ is the number of values. Notice how we can then rearrange the equation into $1 + 9 ( 1 )+9 ( x - 1 )\\geq2018,$ which becomes $9 x\\geq2017,$ or $x\\geq224\\frac19.$ We cannot have a fraction of a value so we must round up to $\\boxed{225}.$" ]

[ "Let $a$ be the smallest element, so $a+18$ is the largest element. Since the mode is $8$ , at least two of the five numbers must be $8$ . The last number we denote as $b$\nThen their average is $\\frac{a + (8) + (8) + b + (a+18)}5 = 12 \\Longrightarrow 2a + b = 26$ . Clearly $a \\le 8$ . Also we have $b \\le a + 18 \\Longrightarrow 26-2a \\le a + 18 \\Longrightarrow 8/3 < 3 \\le a$ . Thus there are a maximum of $6$ values of $a$ which corresponds to $6$ values of $b$ ; listing shows that all such values work. The answer is $\\boxed{6}$" ]

[ "Let there be $n$ integers on the list. The list of $n$ integers has mean $22$ , so the sum of the integers is $22n$\nReplacing $m$ with $m+10$ will increase the sum of the list from $22n$ to $22n + 10$\nThe new mean of the list is $24$ , so the new sum of the list is also $24n$\nThus, we get $22n + 10 = 24n$ , leading to $n=5$ numbers on the list.\nIf there are $5$ numbers on the list with mode $32$ and smallest number $10$ , then the list is $\\{10, x, m, 32, 32\\}$\nSince replacing $m$ with $m-8$ gives a new median of $m-4$ , and $m-4$ must be on the list of $5$ integers since $5$ is odd, $x = m-4$ , and the list is now $\\{10, m-4, m, 32, 32\\}$\nThe sum of the numbers on this list is $22n = 22\\cdot 5 = 110$ , so we get:\n$10 + m - 4 + m + 32 + 32 = 110$\n$70 + 2m = 110$\n$m = 20$ , giving answer $\\boxed{20}$" ]

[ "The third condition implies that the list's size must be an even number, as if it were an odd number, the median of hte list would surely appear in the list itself.\nTherefore, we can casework on what even numbers work.\nSay the size is 2. Clearly, this doesn't work as the only list would be $\\{9, 9\\}$ , which doesn't satisfy condition 1.\nIf the size is 4, then we can have two $9$ s, and a remaining sum of $12$ . Since the other two values in the list must be distinct, and their sum must equal $30-18=12$ , we have that the two numbers are in the form $a$ and $12-a$ . Note that we cannot have both values greater than $9$ , and we cannot have only one value greater than $9$ , because this would make the median $9$ , which violates condition 3. Since the median of the list is a positive integer, this means that the greater of $a$ and $12-a$ must be an odd number. The only valid solution to this is $a=5$ . Thus, our answer is $5^2+7^2+9^2+9^2 = \\boxed{236}$ . ~akliu", "If there were an odd number of elements, the median would be in the set. Thus, we start with 4 elements. For 9 to be the mode, there must be 2 9s. For 9 to not be the median, either both numbers are greater than 9, or both numbers are less than 9. Clearly, both numbers must be less. From here, the numbers are clearly $(5,7,9,9)$ , and we add their squares to get $\\boxed{236}$ -westwoodmonster", "Since the median is not in the list, there must not be an odd number of elements. Suppose the list has two elements. To meet the mode condition, both must equal $9$ , but this does not satisfy the other conditions.\nNext, suppose the list has six elements. If there were at least three $9$ s, then the other elements would sum to at most $30-27=3$ . Since the elements are positive integers, this can only be achieved with the set $\\{1,1,1,9,9,9\\}$ , which violates the unique mode condition. Therefore, there must be exactly two $9$ s, and the other four elements must be distinct to satisfy the unique mode condition. Two sets of four unique positive integers add to $12$ $\\{1,2,3,6\\}$ and $\\{1,2,4,5\\}$ . Neither can act as the remaining four elements since both possibilities violate the constraint that the median is an integer.\nNext, suppose the list had at least eight elements. For the sake of contradiction, suppose the third-largest element was at least $9$ . Then, since every element is a positive integer, the minimum sum would be $1+1+1+1+1+9+9+9>30$ . So, to satisfy the unique mode condition, there must be exactly two $9$ s, and the other elements must be distinct. But then the minimum sum is $1+2+3+4+5+6+9+9>30$ , so the sum constraint can never be satisfied. From these deductions, we conclude that the list has exactly four elements.\nNote that no element can appear three times in the list, or else the middle-two-largest elements would be equal, violating the condition that the median is not in the list. Therefore, to satisfy the unique mode condition, the list contains two $9$ s and two other distinct integers that add to $30-18=12$ . Five sets of two unique positive integers add to $12$ $\\{1,11\\}$ $\\{2,10\\}$ $\\{3,9\\}$ $\\{4,8\\}$ , and $\\{5,7\\}$ . The first four options violate the median condition (either they make the median one of the list elements, or they make the median a non-integer). Thus, the set must be $\\{5,7,9,9\\}$ , and the sum of the squares of these elements is $25+49+81+81=\\boxed{236}$" ]

[ "Notice (by imagining unfolding the roll), that the length of the paper is equal to the sum of the circumferences of the concentric circles, which is $\\pi$ times the sum of the diameters. Now the, the diameters form an arithmetic series with first term $2$ , last term $10$ , and $600$ terms in total, so using the formula $\\frac{1}{2}n(a+l)$ , the sum is $300 \\times 12 = 3600$ , so the length is $3600\\pi$ centimetres, or $36\\pi$ metres, which is answer $\\boxed{36}$" ]

[ "Number the squares $0, 1, 2, 3, ... 2^{k} - 1$ . In this case $k = 10$ , but we will consider more generally to find an inductive solution. Call $s_{n, k}$ the number of squares below the $n$ square after the final fold in a strip of length $2^{k}$\nNow, consider the strip of length $1024$ . The problem asks for $s_{941, 10}$ . We can derive some useful recurrences for $s_{n, k}$ as follows: Consider the first fold. Each square $s$ is now paired with the square $2^{k} - s - 1$ . Now, imagine that we relabel these pairs with the indices $0, 1, 2, 3... 2^{k - 1} - 1$ - then the $s_{n, k}$ value of the pairs correspond with the $s_{n, k - 1}$ values - specifically, double, and maybe $+ 1$ (if the member of the pair that you're looking for is the top one at the final step).\nSo, after the first fold on the strip of length $1024$ , the $941$ square is on top of the $82$ square. We can then write\n\\[s_{941, 10} = 2s_{82, 9} + 1\\]\n(We add one because $941$ is the odd member of the pair, and it will be on top. This is more easily visually demonstrated than proven.) We can repeat this recurrence, adding one every time we pair an odd to an even (but ignoring the pairing if our current square is the smaller of the two):\n\\[s_{82, 9} = 2s_{82, 8} = 4s_{82, 7} = 8s_{127 - 82, 6} = 8s_{45, 6}\\]\n\\[s_{45, 6} = 2s_{63 - 45, 5} + 1 = 2s_{18, 5} + 1 = 4s_{31 - 18, 4} + 1 = 4s_{13, 4} + 1\\]\n\\[s_{13, 4} = 2s_{15 - 13, 3} + 1 = 2s_{2, 3}+1\\]\nWe can easily calculate $s_{2, 3} = 4$ from a diagram. Plugging back in,\n\\begin{align*} s_{13, 4} &= 9 \\\\ s_{45, 6} &= 37 \\\\ s_{82, 9} &= 296 \\\\ s_{941, 10} &= \\boxed{593}", "We can keep track of the position of the square labeled 942 in each step. We use an $(x,y)$ coordinate system, so originally the 942 square is in the position $(942,1)$ . In general, suppose that we've folded the strip into an array $r=2^k$ squares wide and $c=1024/r=2^{10-k}$ squares tall (so we've made $10-k$ folds). Then if a square occupies the location $(x,y)$ , we find that after the next fold, it will be in the location described by the procedure \\[(x,y)\\to\\begin{cases}(x,y)&\\text{if }x\\le 2^{k-1}\\\\ (r+1-x,2c+1-y)&\\text{otherwise}.\\end{cases}\\] Therefore, we can keep track of the square's location in the following table. \\[\\begin{array}{c|c|c} (x,y)&\\text{rows}&\\text{columns}\\\\\\hline (942,1)&1024&1\\\\ (83,2)&512&2\\\\ (83,2)&256&4\\\\ (83,2)&128&8\\\\ (46,15)&64&16\\\\ (19,18)&32&32\\\\ (14,47)&16&64\\\\ (3,82)&8&128\\\\ (3,82)&4&256\\\\ (2,431)&2&512\\\\ (1,594)&1&1024.\\\\ \\end{array}\\] Therefore, at the end of the process, the square labeled 942 will be in the position $(1,594)$ , i.e., it will be above $\\boxed{593}$ squares." ]

[ "Because we are given a right angle, we look for ways to apply the Pythagorean Theorem . Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$ . Let $OE=x$ and $OD=y$ . We're trying to find $x^2+y^2$\nApplying the Pythagorean Theorem, $OA^2 = OD^2 + AD^2$ and $OC^2 = EC^2 + EO^2$\nThus, $\\left(\\sqrt{50}\\right)^2 = y^2 + (6-x)^2$ , and $\\left(\\sqrt{50}\\right)^2 = x^2 + (y+2)^2$ . We solve this system to get $x = 1$ and $y = 5$ , such that the answer is $1^2 + 5^2 = \\boxed{026}$", "Drop perpendiculars from $O$ to $AB$ (with foot $T_1$ ), $M$ to $OT_1$ (with foot $T_2$ ), and $M$ to $AB$ (with foot $T_3$ ).\nAlso, mark the midpoint $M$ of $AC$\nFirst notice that by computation, $OAC$ is a $\\sqrt {50} - \\sqrt {40} - \\sqrt {50}$ isosceles triangle, so $AC = MO$ .\nThen, notice that $\\angle MOT_2 = \\angle T_3MO = \\angle BAC$ . Therefore, the two blue triangles are congruent, from which we deduce $MT_2 = 2$ and $OT_2 = 6$ . As $T_3B = 3$ and $MT_3 = 1$ , we subtract and get $OT_1 = 5,T_1B = 1$ . Then the Pythagorean Theorem tells us that $OB^2 = \\boxed{026}$", "Draw segment $OB$ with length $x$ , and draw radius $OQ$ such that $OQ$ bisects chord $AC$ at point $M$ . This also means that $OQ$ is perpendicular to $AC$ . By the Pythagorean Theorem, we get that $AC=\\sqrt{(BC)^2+(AB)^2}=2\\sqrt{10}$ , and therefore $AM=\\sqrt{10}$ . Also by the Pythagorean theorem, we can find that $OM=\\sqrt{50-10}=2\\sqrt{10}$\nNext, find $\\angle BAC=\\arctan{\\left(\\frac{2}{6}\\right)}$ and $\\angle OAM=\\arctan{\\left(\\frac{2\\sqrt{10}}{\\sqrt{10}}\\right)}$ . Since $\\angle OAB=\\angle OAM-\\angle BAC$ , we get \\[\\angle OAB=\\arctan{2}-\\arctan{\\frac{1}{3}}\\] \\[\\tan{(\\angle OAB)}=\\tan{(\\arctan{2}-\\arctan{\\frac{1}{3}})}\\] By the subtraction formula for $\\tan$ , we get \\[\\tan{(\\angle OAB)}=\\frac{2-\\frac{1}{3}}{1+2\\cdot \\frac{1}{3}}\\] \\[\\tan{(\\angle OAB)}=1\\] \\[\\cos{(\\angle OAB)}=\\frac{1}{\\sqrt{2}}\\] Finally, by the Law of Cosines on $\\triangle OAB$ , we get \\[x^2=50+36-2(6)\\sqrt{50}\\frac{1}{\\sqrt{2}}\\] \\[x^2=\\boxed{026}.\\]", "We use coordinates. Let the circle have center $(0,0)$ and radius $\\sqrt{50}$ ; this circle has equation $x^2 + y^2 = 50$ . Let the coordinates of $B$ be $(a,b)$ . We want to find $a^2 + b^2$ $A$ and $C$ with coordinates $(a,b+6)$ and $(a+2,b)$ , respectively, both lie on the circle. From this we obtain the system of equations\n$a^2 + (b+6)^2 = 50$\n$(a+2)^2 + b^2 = 50$\nAfter expanding these terms, we notice by subtracting the first and second equations, we can cancel out $a^2$ and $b^2$ . after substituting $a=3b+8$ and plugging back in, we realize that $(a,b)=(-7,-5)$ or $(5,-1)$ . Since the first point is out of the circle, we find that $(5,-1)$ is the only relevant answer. This paragraph is written by ~hastapasta.\nSolving, we get $a=5$ and $b=-1$ , so the distance is $a^2 + b^2 = \\boxed{026}$", "I will use the law of cosines in triangle $\\triangle OAC$ and $\\triangle OBC$\n$AC = \\sqrt{AB^2 + BC^2} = \\sqrt{6^2 + 2^2} = 2 \\sqrt{10}$\n$\\cos \\angle ACB = \\frac{2}{2\\sqrt{10}} = \\frac{1}{\\sqrt{10}}$\n$\\cos \\angle ACO = \\frac{AC^2+OC^2-OA^2}{2 \\cdot AC \\cdot OC} = \\frac{(2\\sqrt{10})^2+(\\sqrt{50})^2-(\\sqrt{50})^2}{2 \\cdot 2\\sqrt{10} \\cdot \\sqrt{50}} = \\frac{1}{\\sqrt{5}}$\n$\\sin \\angle ACB = \\sqrt{1-\\cos^2 \\angle ACB} = \\sqrt{1-(\\frac{1}{\\sqrt{10}})^2} = \\frac{3}{\\sqrt{10}}$\n$\\sin \\angle ACO = \\sqrt{1-\\cos^2 \\angle ACO} = \\sqrt{1-(\\frac{1}{\\sqrt{5}})^2} = \\frac{2}{\\sqrt{5}}$\n$\\cos \\angle OCB = \\cos (\\angle ACB - \\angle ACO) = \\cos \\angle ACB \\cdot \\cos \\angle ACO + \\sin \\angle ACB \\cdot \\sin \\angle ACO = \\frac{1}{\\sqrt{10}} \\cdot \\frac{1}{\\sqrt{5}} + \\frac{3}{\\sqrt{10}} \\cdot \\frac{2}{\\sqrt{5}} = \\frac{7}{5\\sqrt{2}}$\n$OB^2 = OC^2 + BC^2 - 2 \\cdot OC \\cdot BC \\cdot \\cos \\angle OCB = (\\sqrt{50})^2 + 2^2 - 2 \\cdot \\sqrt{50} \\cdot 2 \\cdot \\frac{7}{5\\sqrt{2}} = 50 + 4 - 28 = \\boxed{026}$", "Notice that $50=5^2+5^2=7^2+1^2$ , and by the size of the diagram, it seems reasonable that $OA$ represents $5^2+5^2$ , and $OC$ means the $7^1+1^2$ , and indeed, the values work ( $7-5=2$ and $5+1=6$ ), so $OB^2=5^2+1^2=\\boxed{026}$" ]

[ "Let $0$ represent a house that does not receive mail and $1$ represent a house that does receive mail. This problem is now asking for the number of $19$ -digit strings of $0$ 's and $1$ 's such that there are no two consecutive $1$ 's and no three consecutive $0$ 's.\nThe last two digits of any $n$ -digit string can't be $11$ , so the only possibilities are $00$ $01$ , and $10$\nLet $a_n$ be the number of $n$ -digit strings ending in $00$ $b_n$ be the number of $n$ -digit strings ending in $01$ , and $c_n$ be the number of $n$ -digit strings ending in $10$\nIf an $n$ -digit string ends in $00$ , then the previous digit must be a $1$ , and the last two digits of the $n-1$ digits substring will be $10$ . So \\[a_{n} = c_{n-1}.\\]\nIf an $n$ -digit string ends in $01$ , then the previous digit can be either a $0$ or a $1$ , and the last two digits of the $n-1$ digits substring can be either $00$ or $10$ . So \\[b_{n} = a_{n-1} + c_{n-1}.\\]\nIf an $n$ -digit string ends in $10$ , then the previous digit must be a $0$ , and the last two digits of the $n-1$ digits substring will be $01$ . So \\[c_{n} = b_{n-1}.\\]\nClearly, $a_2=b_2=c_2=1$ . Using the recursive equations and initial values: \\[\\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \\multicolumn{19}{c}{}\\\\\\hline n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\\\\\hline a_n&1&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86\\\\\\hline b_n&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114&151\\\\\\hline c_n&1&1&2&2&3&4&5&7&9&12&16&21&28&37&49&65&86&114\\\\\\hline \\end{array}\\]\nAs a result $a_{19}+b_{19}+c_{19}=\\boxed{351}$", "We split the problem into cases using the number of houses that get mail. Let \"|\" represent a house that gets mail, and \"o\" represent a house that doesn't. With a fixed number of |, an o can be inserted between 2 |'s or on the very left or right. There cannot be more than one o that is free to arrange to be placed between two |'s because no three o's can be adjacent, but there can be a maximum of two o's placed on the very left or right. Note that according to the Pigeonhole Principle , no more than 10 houses can get mail on the same day.\nCase 1: 10 houses get mail. No 2 adjacent houses can get mail on the same day, so there must be an o between every two |. $10-1=9$ o's are fixed so we count the number of ways to insert $19 - 10 - 9 = 0$ o's to $10+1 = 11$ spots, or $\\binom{11}{0} = 1$\nCase 2: 9 houses get mail. In this case, $9-1 = 8$ o's are fixed so we count the number of ways to insert $19 - 9 - 8 = 2$ o's to $9+1=10$ spots. However, there is also the case where two o's are both on the very left / right. When both o's that are free to arrange are put on a side, there are $10-1=9$ spots left to insert $2-2=0$ o's. Hence the total number of ways in this case is $\\binom{10}{2} + 2\\binom{9}{0} = 47$\nCase 3: 8 houses get mail. In this case, $8-1=7$ o's are fixed so we count the number of ways to insert $19-8-7=4$ o's to $8+1=9$ spots. When two o's are put to the very left / right, there are $9-1=8$ spots left to insert $4-2=2$ o's. We also need to take care of the case where two o's are on the very left and two o's are on the very right: we have $9-1-1=7$ spots to insert $4-2-2=0$ o's. Hence the total number of ways in this case is $\\binom{9}{4} + 2\\binom{8}{2} + \\binom{7}{0} = 183$\nCase 4: 7 houses get mail. In this case, $7-1=6$ o's are fixed so we count the number of ways to insert $19-7-6=6$ o's to $7+1=8$ spots. When two o's are put to the very left / right, there are $8-1=7$ spots left to insert $6-2=4$ o's. When two o's are on the very left and two o's are on the very right, we have $8-1-1=6$ spots to insert $6-2-2=2$ o's. Hence the total number of ways in this case is $\\binom{8}{6} + 2\\binom{7}{4} + \\binom{6}{2} = 113$\nCase 5: 6 houses get mail. We have to be careful in this case: $6-1=5$ o's are fixed so we are inserting $19-6-5=8$ o's to $6+1=7$ spots, which means that at least 1 of the 2 sides must have two o's. When 1 of the 2 sides have two o's, there are $7-1=6$ spots to insert $8-2=6$ o's. When both sides have two o's, there are $7-1-1=5$ spots to insert $8-2-2=4$ o's. Hence the total number of ways in this case is $2\\binom{6}{6} + \\binom{5}{4} = 7$\nWhen less than 6 houses get(s) mail, it's again not possible since at least three o's must be together (again, according to the Pigeonhole Principle). Therefore, the desired answer is $1+47+183+113+7=\\boxed{351}$", "There doesn't seem to be anything especially noticeable about the number nineteen in this problem, meaning that we can replace the number nineteen with any number without a big effect on the logic that we use to solve the problem. This pits the problem as a likely candidate for recursion.\nAt first, it's not immediately clear how to relate the state of $n$ houses in general to that of $n - 1, n - 2,$ or $n - 3.$ We thus break it up into cases, based on whether the first house gets mail or not.\nLet $p_n$ be the number of ways to distribute the mail to $n$ houses. Assume that the first house gets mail. Therefore, since no two adjacent houses get mail on the same day, the second house must not get mail. Starting from the third house, however, things start to look messy, and it looks like we have to break our recurrence down into even smaller cases, which is something that we don't like -- we want to keep our relations as simple as possible. Therefore, seeing that we can't work forwards anymore, we try to work backwards.\nOnce the mail carrier delivers the mail to the first and (lack of mail) to the second houses, have him deliver mail to the remaining $n - 3$ houses at the end of the row, skipping the third house. There are $p_{n - 3}$ ways to do this. Now, we see that the availability of mail at the third house is fixed -- if the fourth house doesn't receive mail, the third one must, and if the fourth house receives mail, the third one can't. Therefore, there are simply $p_{n-3}$ ways to deliver the mail if the first house gets mail.\nIf the first house doesn't get mail, then we use the same logic -- have the mail carrier skip the second house and deliver the remaining mail to the $n - 2$ houses in $p_{n-2}$ ways. Then, the availability of mail for the second house is fixed, so there are $p_{n - 2}$ ways to deliver the mail in this case.\nWe thus have established a recurrence relation -- since the first house either gets mail or it doesn't, and cannot achieve both at the same time, we are confident about the validity of our relation: \\[p_n = p_{n-2} + p_{n-3}.\\] Now, we simply calculate $p_1, p_2,$ and $p_3.$ Then, it's off to the races for computation!\n$p_1 = 2,$ because the first house can either gets mail or it doesn't -- there are no restrictions.\n$p_2 = 3,$ because all of the possible deliveries are valid (of which there are $2 \\cdot 2 = 4$ ) except the one where both houses receive mail.\n$p_3 = 4,$ as there are $4$ possible ways (here, M represents that that house gets mail and N represents no mail): MNM, MNN, NNM, NMN.\nUsing our recurrence relation, we eventually get that $p_{19} = \\boxed{351},$ and we're done.", "Let $w_n$ be the number of possible ways if the last house has mail, and $b_n$ be the number of possible ways if the last house does not have mail.\nIf the last house has mail, then, the next house can't have mail, meaning that $b_n = w_{n - 1}$\nIf the last house doesn't have mail, then the next house can either have mail or not have mail. If the next house has mail, then we simply count the number of ways that the row ends in a house with mail, so that means so far, our recursive rule is $w_n = b_{n - 1} + \\text{something}$ . If the next house does not have mail, then the next house after that must have mail, meaning that $w_n = b_{n - 1} + b_{n - 2}$\nRecursing all the way up to $b_{19}$ and $w_{19}$ , we get $100 + 251 = \\boxed{351}$", "Let $a_n$ be the number of ways if the first house has mail, and let $b_n$ be the number of ways if the first house does not get mail.\n$a_n=a_{n-2}+a_{n-3}$ because if the first house gets mail, the next house that gets mail must either be the third or fourth house.\n$b_n=a_{n-1}+a_{n-2}$ because if the first house does not get mail, the next house that gets mail must either be the second or third house.\nNote that we only need list out values of $a_n$ as $b$ depends on $a$ $a_1=1, a_2=1, a_3=2, a_4=2, \\ldots$\n$a_{19}+b_{19}=a_{17}+a_{16}+a_{18}+a_{17}=a_{16}+2\\cdot a_{17}+a_{18}=65+2\\cdot 86+114=\\boxed{351}$" ]

[ "The total cost of the pencils can be found by $(\\text{students}\\cdot\\text{pencils purchased by each}\\cdot\\text{price of each pencil})$\nSince $1771$ is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: $7, 11, 23$ . Since neither $(C)$ nor $(E)$ are any of these factors, they can be eliminated immediately, leaving $(A)$ $(B)$ , and $(D)$\nBeginning with $(A) 7$ , we see that the number of pencils purchased by each student must be either $11$ or $23$ . However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.\nContinuing with $(B) 11$ , we can conclude that the only case that fulfils the restrictions are that there are $23$ students who each purchased $7$ such pencils, so the answer is $\\boxed{11}$ . We can apply the same logic to $(E)$ as we applied to $(A)$ if one wants to make doubly sure.", "We know the total cost of the pencils can be found by $(\\text{students}\\cdot\\text{pencils purchased by each}\\cdot\\text{price of each pencil})$\nUsing prime factorization like in the solution above, we see that there is only one combination of three whole numbers whose product is equal to $1771$ $7, 11, 23$ (without using $1$ ). So we know that $7, 11$ , and $23$ must be the number of students, the number of pencils purchased by each student and the price of each pencil in cents.\nWe know that $23$ must be the number of students, as it is the only number that makes up the majority of 30.\nWe pick the greater of the remaining numbers for the price of each pencil in cents, which is $11$\nTherefore, our answer is $\\boxed{11}$", "We let $s$ be the number of students that bought pencils at the bookstore, $c$ be the cost of each pencil in cents, and $n$ the number of pencils each student bought. Thus, we are looking for $c.$ Since a majority of the students in the class bought pencils at the bookstore, $s>\\dfrac{30}2=15.$ (s>15) We also know that $n>1$ and $c>n>1.$ Finally, $s\\cdot c\\cdot n=1771.$ We can factor $1771$ as $7\\cdot11\\cdot23.$ Since $n>15$ and $c>n>1,$ none of them can be $1,$ and therefore $c,n,$ and $s$ are $7,11,$ and $23$ in some order. We know that $n>15,$ so $n$ must be $23.$ $c>n,$ so $c=11$ and $n=7.$ Thus, $c=\\boxed{11}$ cents.\n~Technodoggo" ]

[ "The total cost of the pencils can be found by $(\\text{students}\\cdot\\text{pencils purchased by each}\\cdot\\text{price of each pencil})$\nSince $1771$ is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: $7, 11, 23$ . Since neither $(C)$ nor $(E)$ are any of these factors, they can be eliminated immediately, leaving $(A)$ $(B)$ , and $(D)$\nBeginning with $(A) 7$ , we see that the number of pencils purchased by each student must be either $11$ or $23$ . However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.\nContinuing with $(B) 11$ , we can conclude that the only case that fulfils the restrictions are that there are $23$ students who each purchased $7$ such pencils, so the answer is $\\boxed{11}$ . We can apply the same logic to $(E)$ as we applied to $(A)$ if one wants to make doubly sure.", "We know the total cost of the pencils can be found by $(\\text{students}\\cdot\\text{pencils purchased by each}\\cdot\\text{price of each pencil})$\nUsing prime factorization like in the solution above, we see that there is only one combination of three whole numbers whose product is equal to $1771$ $7, 11, 23$ (without using $1$ ). So we know that $7, 11$ , and $23$ must be the number of students, the number of pencils purchased by each student and the price of each pencil in cents.\nWe know that $23$ must be the number of students, as it is the only number that makes up the majority of 30.\nWe pick the greater of the remaining numbers for the price of each pencil in cents, which is $11$\nTherefore, our answer is $\\boxed{11}$", "We let $s$ be the number of students that bought pencils at the bookstore, $c$ be the cost of each pencil in cents, and $n$ the number of pencils each student bought. Thus, we are looking for $c.$ Since a majority of the students in the class bought pencils at the bookstore, $s>\\dfrac{30}2=15.$ (s>15) We also know that $n>1$ and $c>n>1.$ Finally, $s\\cdot c\\cdot n=1771.$ We can factor $1771$ as $7\\cdot11\\cdot23.$ Since $n>15$ and $c>n>1,$ none of them can be $1,$ and therefore $c,n,$ and $s$ are $7,11,$ and $23$ in some order. We know that $n>15,$ so $n$ must be $23.$ $c>n,$ so $c=11$ and $n=7.$ Thus, $c=\\boxed{11}$ cents.\n~Technodoggo" ]

[ "If a man born in the 19th century was $x$ years of in the year $x^2$ , it implies that the year the man was born was $x^2-x$ . So, if the man was born in the first half of the 19th century, it means that $x^2-x < 1850$ . Noticing that $40^2 - 40 = 1560$ and $50^2-50 = 2450$ , we see that $40 < x < 50$ . We can guess values until we hit a solution. $43^2-43 = 1806$ , so we see that the man had to have been $43$ years old in the year $1849=43^2$ , so the answer is $\\boxed{1806}$" ]

[ "$12\\frac{1}{2}\\%$ is the same as $\\frac{1}{8}$ , so the man sets one eighth of each month's rent aside, so he only gains $\\frac{7}{8}$ of his rent. He also pays $325 each year, and he realizes $5.5\\%$ , or $550, on his investment. Therefore he must have collected a total of $325 +$550 = $875 in rent. This was for the whole year, so he collected $\\frac{875}{12}$ dollars each month as rent. This is only $\\frac{7}{8}$ of the monthly rent, so the monthly rent in dollars is $\\frac{875}{12}\\cdot \\frac{8}{7}=\\boxed{83.33}$" ]

[ "The man has three possible combinations of transportation: \\[\\text{Morning train, Afternoon bus (m.t., a.b.)}\\] \\[\\text{Morning bus, Afternoon train (m.b., a.t.)}\\] \\[\\text{Morning bus, Afternoon bus (m.b, a.b.)}\\]\nLet $y$ be the number of times the man takes the $\\text{a.t.}$ . Then, $9-y$ is the number of times he takes the $\\text{m.t.}$ . Keep in mind that $\\text{m.b.}=y$ and $\\text{a.b.}=9-y$\nLet $z$ be the number of times the man takes the $\\text{m.b.}$ and $\\text{a.b.}$ . Now, we get the two equations \\[y+z=8\\] and \\[9-y+z=15.\\]\nSolving the system of equations, we get $y=1$ and $z=7$\nSo during the $x$ working days, the man took the $\\text{(m.t., a.b.)}$ on $9-1=8$ days, the $\\text{(m.b., a.t.)}$ on $1$ day, and the $\\text{(m.b., a.b.)}$ on $7$ days.\nTherefore, $x=8+1+7= \\boxed{16}$ . ~ jiang147369" ]

[ "The man currently earns $4000 \\cdot \\frac{5}{1000} + 3500 \\cdot \\frac{4}{1000} = 340$ dollars. So, we need to find the value of $x$ such that \\[2500 \\cdot \\frac{x}{1000} = 160.\\] Solving, we get $x = \\boxed{6.4}$" ]

[ "You are trying to find $\\frac{2(0.06x)}{4500}$ , where $x$ is the principle for one investment. To find $x$ , solve $0.04(4500-x) = 0.06x$ $X$ will come out to be $1800$ . Then, plug in x into the first equation, $\\frac{2(0.06)(1800)}{4500}$ , to get $0.048$ . Finally, convert that to a percentage and you get $\\boxed{4.8}$" ]

[ "Let $n$ be the number of minutes after 6:00. Let $h(n)=180+\\frac{n}2$ be the angle, in degrees, of the hour hand (with $0^{\\circ}$ at the top and increasing in the clockwise direction); similarly, let $m(n)=6n$ be the angle of the minute hand. We want $|h(n)-m(n)|=110$ . This is equivalent to \\[180-\\frac{11n}2=\\pm110\\] \\[-\\frac{11n}2\\in\\{-70,-290\\}\\] \\[\\frac{11n}2\\in\\{70,290\\}\\] \\[11n\\in\\{140,580\\}\\] \\[n\\in\\{\\frac{140}{11},\\frac{580}{11}\\}\\] The difference between the two values of $n$ is $\\frac{440}{11}=\\boxed{40}$" ]

[ "There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.\nThus, $N = 2520 + 1200 = 3720$ , and $\\frac{N}{10} = \\boxed{372}$" ]

[ "Let $C$ represent the cost of the goods, and let $L$ $S$ , and $M$ represent the list, selling, and marked prices of the goods, respectively. Hence, we have three equations, which we need to manipulate in order to relate $M$ and $L$\n$C=\\frac{4}{5}L$\n$S=C+\\frac{1}{5}S$\n$S=\\frac{4}{5}M$\nWe find that $M=\\frac{5}{4}L$ . Hence, the percent of the list price which should be marked is $\\boxed{125}$" ]

[ "Without loss of generality, we can set the list price equal to $100$ . The merchant buys the goods for $100*.75=75$ . Let $x$ be the marked price.\nWe then use the equation $0.8x-75=25$ to solve for $x$ and get a marked price of $\\boxed{125}$" ]

[ "Let's assume that each item is $100$ dollars. First we take off $30\\%$ off of $100$ dollars. $100\\cdot0.7=70$\nNext, we take off the extra $20\\%$ as asked by the problem. $70\\cdot0.80=56$\nSo the final price of an item is $56. We have to do $100-56$ because $56$ was the final price and we wanted the discount. So the final answer is $44\\%$ , which is answer choice $\\boxed{44}$" ]

[ "Since $30\\%$ of the original $30$ liters of paint was yellow, and 5 liters of yellow paint were added to make the new mixture, there are $9+5=14$ liters of yellow tint in the new mixture. Since only 5 liters of paint were added to the original 30, there are a total of 35 liters of paint in the new mixture. This gives $40\\%$ of yellow tint in the new mixture, which is $\\boxed{40}$" ]

[ "$31 \\equiv 3 \\pmod {7}$ so the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays. Therefore, Monday, Thursday, and Friday are valid so the answer is $\\boxed{3}$" ]

[ "$31 \\equiv 3 \\pmod {7}$ so the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays. Therefore, Monday, Thursday, and Friday are valid so the answer is $\\boxed{3}$" ]

[ "One could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as \\[P(x,y) = \\frac{1}{3} P(x-1,y) + \\frac{1}{3} P(x,y-1) + \\frac{1}{3} P(x-1,y-1)\\] for $x,y \\geq 1,$ and the base cases are $P(0,0) = 1, P(x,0) = P(y,0) = 0$ for any $x,y$ not equal to zero.\nWe then recursively find $P(4,4) = \\frac{245}{2187}$ so the answer is $245 + 7 = \\boxed{252}$", "Obviously, the only way to reach (0,0) is to get to (1,1) and then have a $\\frac{1}{3}$ chance to get to (0,0). Let x denote a move left 1 unit, y denote a move down 1 unit, and z denote a move left and down one unit each. The possible cases for these moves are $(x,y,z)=(0,0,3),(1,1,2),(2,2,1)$ and $(3,3,0)$ . This gives a probability of $1 \\cdot \\frac{1}{27} + \\frac{4!}{2!} \\cdot \\frac{1}{81} + \\frac{5!}{2! \\cdot 2!} \\cdot \\frac{1}{243} +\\frac{6!}{3! \\cdot 3!} \\cdot \\frac{1}{729}=\\frac{245}{729}$ to get to $(1,1)$ . The probability of reaching $(0,0)$ is $\\frac{245}{3^7}$ . This gives $245+7=\\boxed{252}$", "Since the particle stops at one of the axes, we know that the particle most pass through $(1,1)$ . Thus, it suffices to consider the probability our particle will reach $(1,1)$ . Then the only ways to get to $(1,1)$ from $(4,4)$ are the following:\n(1) 3 moves diagonally\n(2) 2 moves diagonally, 1 move left, 1 move down\n(3) 1 move diagonally, 2 moves left and 2 moves down.\n(4) 3 moves left, 3 moves down.\nThe probability of (1) is $\\frac{1}{3^3}$ . The probability of (2) is $\\frac{\\frac{4!}{2!}}{3^4} = \\frac{12}{3^4}$ . The probability of (3) is $\\frac{\\frac{5!}{2!2!}}{3^5} = \\frac{30}{3^5}$ . The probability of (4) is $\\frac{\\frac{6!}{3!3!}}{3^6} = \\frac{20}{3^6}$ . Adding all of these together, we obtain a total probability of $\\frac{245}{3^6}$ that our particle will hit $(1,1)$ . Trivially, there is a $\\frac{1}{3}$ chance our particle will hit $(0,0)$ from $(1,1)$ . So our final probability will be $\\frac{245}{3^6} \\cdot \\frac{1}{3} = \\frac{245}{3^7} \\implies m = 245, n = 7 \\implies \\boxed{252}$" ]

[ "Let $c$ be the number of questions correct, $w$ be the number of questions wrong, and $b$ be the number of questions left blank. We are given that \\begin{align} c+w+b &= 20 \\\\ 5c-2w &= 48 \\end{align}\nAdding equation $(2)$ to double equation $(1)$ , we get \\[7c+2b=88\\]\nSince we want to maximize the value of $c$ , we try to find the largest multiple of $7$ less than $88$ . This is $84=7\\times 12$ , so let $c=12$ . Then we have \\[7(12)+2b=88\\Rightarrow b=2\\]\nFinally, we have $w=20-12-2=6$ . We want $c$ , so the answer is $12$ , or $\\boxed{12}$", "If John answered 16 questions correctly, then he answered at most 4 questions incorrectly, giving him at least $16 \\cdot 5 - 4 \\cdot 2 = 72$ points. Therefore, John did not answer 16 questions correctly. If he answered 12 questions correctly and 6 questions incorrectly (leaving 2 questions unanswered), then he scored $12 \\cdot 5 - 6 \\cdot 2 = 48$ points. As all other options are less than 12, we conclude that 12 is the most questions John could have answered correctly, and the answer is $\\boxed{12}$" ]

[ "Arranging the nickels in a hexagonal fashion, we see that only $\\boxed{6}$ nickels can be placed around the central nickel." ]

[ "We have $\\log_{10}{(d+1)}-\\log_{10}{d} = \\log_{10}{\\left(\\frac{d+1}{d}\\right)}$ , so the probability of choosing $2$ is $\\log_{10}{\\left(\\frac{3}{2}\\right)}$ . The probability that the digit chosen is in the set must therefore be \\begin{align*}2\\log_{10}{\\left(\\frac{3}{2}\\right)} = &\\log_{10}{\\left(\\left(\\frac{3}{2}\\right)^2\\right)} \\\\ = &\\log_{10}{\\left(\\frac{9}{4}\\right)} \\\\ = &\\log_{10}{9}-\\log_{10}{4} \\\\ = &\\left(\\log_{10}{9}-\\log_{10}{8}\\right)+\\left(\\log_{10}{8}-\\log_{10}{7}\\right)+\\left(\\log_{10}{7}-\\log_{10}{6}\\right) \\\\ &+\\left(\\log_{10}{6}-\\log_{10}{5}\\right)+\\left(\\log_{10}{5}-\\log_{10}{4}\\right),\\end{align*} which, by definition, is the probability that the digit chosen is in the set $\\boxed{4,5,6,7,8}$" ]

[ "A number is divisible by $15$ precisely if it is divisible by $3$ and $5$ . The latter means the last digit must be either $5$ or $0$ , and the former means the sum of the digits must be divisible by $3$ . If the last digit is $0$ , the first digit would be $0$ (because the digits alternate), which is not possible. Hence the last digit must be $5$ , and the number is of the form $5\\square 5\\square 5$ . If the unknown digit is $x$ , we deduce $5+x+5+x+5 \\equiv 0 \\pmod{3} \\Rightarrow 2x \\equiv 0 \\pmod{3}$ . We know $2^{-1}$ exists modulo $3$ because 2 is relatively prime to 3, so we conclude that $x$ (i.e. the second and fourth digit of the number) must be a multiple of $3$ . It can be $0$ $3$ $6$ , or $9$ , so there are $\\boxed{4}$ options: $50505$ $53535$ $56565$ , and $59595$", "As in Solution 1, we find that such numbers must start with $5$ and alternate with $5$ (i.e. must be of the form $5\\square 5\\square 5$ ), where the two digits between the $5$ s need to be the same. Call that digit $x$ . For the number to be divisible by $3$ , the sum of the digits must be divisible by $3$ ; since the sum of the three $5$ s is $15$ , which is already a multiple of $3$ , it must also be the case that $x+x=2x$ is a multiple of $3$ . Thus, the problem reduces to finding the number of digits from $0$ to $9$ for which $2x$ is a multiple of $3$ . This leads to $x=0$ $3$ $6$ , or $9$ , so there are $\\boxed{4}$ possible numbers (namely $50505$ $53535$ $56565$ , and $59595$ ).", "After finding out that the last digit must be $5$ , the number is of the form $5\\square 5\\square 5$ . If the unknown digit is $x$ , we can find that one of the solutions to $x$ is $0$ , since $5+5+5$ is equal to $15$ , which is divisible by $3$ . After trying every one digit number, you'll notice that $x$ must be a multiple of $3$ , meaning that $x=0$ $3$ $6$ , or $9$ $50505$ $53535$ $56565$ , and $59595$ are the $\\boxed{4}$ solutions to this question." ]

[ "assume the number is $ababa$ $10101a+1010b=0 (mod 15)\\newline$ $6a+5b=0 (mod 15)\\newline$ $a=0 (mod 5)\\newline$ $5b=0 (mod 15)\\newline$ $b=0 (mod 3)\\newline$ Solutions: $(5,0),(5,3),(5,6),(5,9)\\newline$ $\\boxed{4}$" ]

[ "The inside diameters of the rings are the positive integers from $1$ to $18$ . The total distance needed is the sum of these values plus $2$ for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series , the answer is $\\frac{18 \\cdot 19}{2} + 2 = \\boxed{173}$", "Alternatively, the sum of the consecutive integers from 3 to 20 is $\\frac{1}{2}(18)(3+20) = 207$ . However, the 17 intersections between the rings must be subtracted, and we also get $207 - 2(17) = \\boxed{173}$" ]

[ "The inside diameters of the rings are the positive integers from $1$ to $18$ . The total distance needed is the sum of these values plus $2$ for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series , the answer is $\\frac{18 \\cdot 19}{2} + 2 = \\boxed{173}$", "Alternatively, the sum of the consecutive integers from 3 to 20 is $\\frac{1}{2}(18)(3+20) = 207$ . However, the 17 intersections between the rings must be subtracted, and we also get $207 - 2(17) = \\boxed{173}$" ]

[ "We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:\n$5:3 = 5(8):3(8) = 40:24$\n$8:5 = 8(5):5(5) = 40:25$\nTherefore, the ratio of 8th graders to 7th graders to 6th graders is $40:25:24$ . Since the ratio is in lowest terms, the smallest number of students participating in the project is $40+25+24 = \\boxed{89}$", "The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). Then there are $40*\\frac{3}{5}=24$ 6th graders and $40*\\frac{5}{8}=25$ 7th graders. The numbers of students is $40+24+25=\\boxed{89}$" ]

[ "If we add $1$ to the number, it becomes divisible by $10, 9, 8, \\cdots, 2, 1$ . The LCM of $1$ through $10$ is $2520$ , therefore the number we want to find is $2520-1=\\boxed{2519}$" ]

[ "The areas of the tops of $A$ $B$ $C$ , and $D$ in the figure formed has sum $1+1+1+1 = 4$ as do the bottoms. Thus, the total so far is $8$ . Now, one of the sides has an area of one, since it combines all of the heights of $A$ $B$ $C$ , and $D$ , which is $1$ . The other side is also the same. Thus the total area now is $10$ . From the front, the surface area is half, because if you looked at it straight from the front it would look exactly like a side of $A$ , with a surface area of half. From the back, it is the same thing. Thus, the total surface area is $10+\\frac{1}{2}+\\frac{1}{2}= 11$ , or $\\boxed{11}$", "The top parts and the bottom parts sum to 8. The sides add on another 2. Therefore, the only logical answer would be $\\boxed{11}$ since it is the only answer greater than 10." ]