problem
string | solution
string | solution_hint
string | hash
int64 |
|---|---|---|---|
In a group of 20 friends, 11 like to ski, 13 like to snowboard, and 3 do not like to do either. How many of the friends like to both ski and snowboard? | \(\boxed{7}\) | Let $x$ be the number of friends who like to both ski and snowboard.
Then $11-x$ of the friends like to ski but not do not like to snowboard, and $13-x$ of the friends like to snowboard but do not like to ski.
Since 3 of the 20 friends do not like to ski or snowboard, then 17 like to either ski or snowboard or both.
Thus, $(11-x)+(13-x)+x=17$ and so $x=7$.
Therefore, 7 of the friends like to both ski and snowboard.
ANSWER: 7 | 8,613,704,553,759,888,000 |
The pair $(x, y)=(2,5)$ is the solution of the system of equations
$$
\begin{aligned}
a x+2 y & =16 \\
3 x-y & =c
\end{aligned}
$$
Determine the value of $\frac{a}{c}$. | \(\boxed{3}\) | Since $(x, y)=(2,5)$ is the solution of the system of equations, then $(x, y)=(2,5)$ satisfies both equations.
Since $(x, y)=(2,5)$ satisfies $a x+2 y=16$, then $2 a+10=16$ or $2 a=6$ and so $a=3$.
Since $(x, y)=(2,5)$ satisfies $3 x-y=c$, then $6-5=c$ or $c=1$.
Therefore, $\frac{a}{c}=3$.
ANSWER: 3 | -3,525,035,608,254,648,300 |
What is the smallest two-digit positive integer $k$ for which the product $45 k$ is a perfect square? | \(\boxed{20}\) | We note that $45=3^{2} \cdot 5$ and so $45 k=3^{2} \cdot 5 \cdot k$.
For $45 k$ to be a perfect square, each prime factor has to occur an even number of times. Therefore, $k$ must be divisible by 5 .
We try the smallest two-digit possibilities for $k$ that are divisible by 5 , namely $k=10,15,20, \ldots$
If $k=10$, then $45 k=450$, which is not a perfect square.
If $k=15$, then $45 k=675$, which is not a perfect square.
If $k=20$, then $45 k=900=30^{2}$, which is a perfect square.
Therefore, the smallest two-digit positive integer $k$ for which $45 k$ is a perfect square is $k=20$. | -96,606,657,541,149,950 |
Each entry in the list below is a positive integer:
$$
a, 8, b, c, d, e, f, g, 2
$$
If the sum of any four consecutive terms in the list is 17 , what is the value of $c+f$ ? | \(\boxed{7}\) | Since the sum of any four consecutive terms is 17 , then $8+b+c+d=17$ or $b+c+d=9$.
Also, $b+c+d+e=17$ and so $9+e=17$ or $e=8$.
Similarly, $e+f+g+2=17$ and $d+e+f+g=17$ tell us that $d=2$.
But $c+d+e+f=17$ and $e=8$ and $d=2$, which gives $c+f=17-8-2=7$.
ANSWER: 7 | -206,071,479,474,804,860 |
In how many different ways can 22 be written as the sum of 3 different prime numbers? That is, determine the number of triples $(a, b, c)$ of prime numbers with $1<a<b<c$ and $a+b+c=22$. | \(\boxed{2}\) | Every prime number other than 2 is odd.
Since $a, b$ and $c$ are all prime and $a+b+c=22$ which is even, it cannot be the case that all of $a, b$ and $c$ are odd (otherwise $a+b+c$ would be odd).
Thus, at least one of $a, b$ and $c$ is even.
Since $1<a<b<c$, then it must be the case that $a=2$ and $b$ and $c$ are odd primes.
Since $a=2$, then $b+c=20$.
Since $b$ and $c$ are primes with $b<c$, then $b=3$ and $c=17$, or $b=7$ and $c=13$.
Therefore, there are two triples $(a, b, c)$ of primes numbers that satisfy the requirements.
ANSWER: 2 | 1,743,037,501,404,293,000 |
For how many one-digit positive integers $k$ is the product $k \cdot 234$ divisible by 12 ? | \(\boxed{4}\) | Note that $234=9 \cdot 26=2 \cdot 3^{2} \cdot 13$.
Thus, $k \cdot 234=k \cdot 2 \cdot 3^{2} \cdot 13$.
This product is divisible by $12=2^{2} \cdot 3$ if and only if $k$ contributes another factor of 2 (that is, if and only if $k$ is even).
Since $k$ is a one-digit positive integer, then $k=2,4,6,8$.
Therefore, there are 4 one-digit positive integers $k$ for which $k \cdot 234$ is divisible by 12 .
ANSWER: 4 | 7,389,902,342,226,603,000 |
The points $A(5,-8), B(9,-30)$ and $C(n, n)$ are collinear (that is, lie on the same straight line). What is the value of $n$ ? | \(\boxed{3}\) | Since $A(5,-8), B(9,-30)$, and $C(n, n)$ lie on the same straight line, then the slopes of $A B$ and $A C$ are equal.
Thus,
$$
\begin{aligned}
\frac{(-30)-(-8)}{9-5} & =\frac{n-(-8)}{n-5} \\
\frac{-22}{4} & =\frac{n+8}{n-5} \\
-22 n+110 & =4 n+32 \\
78 & =26 n \\
n & =3
\end{aligned}
$$
Therefore, $n=3$.
ANSWER: 3 | -1,602,277,047,501,685,000 |
What is the difference between the largest possible three-digit positive integer with no repeated digits and the smallest possible three-digit positive integer with no repeated digits? | \(\boxed{885}\) | To find the largest possible three-digit positive integer with no repeated digits, we put the largest possible digit in each position, starting with the hundreds digit (9), then the tens digit (8), and finally the ones (units) digit (7).
Similarly, to find the smallest possible three-digit positive integer with no repeated digits, we put the smallest possible digit in each position, starting with the hundreds digit (1 because the leading digit cannot be 0), then the tens digit (0), and finally the ones (units) digit (2).
The difference between the numbers 987 and 102 is 885 .
ANSWER: 885 | -7,422,960,132,173,749,000 |
Determine the number of pairs $(x, y)$ of positive integers for which $0<x<y$ and $2 x+3 y=80$. | \(\boxed{5}\) | Since $02 x+3 x=5 x$.
Since $2 x+3 y=80$ and $2 x+3 y>5 x$, then $80>5 x$ or $16>x$.
Since $2 x+3 y=80$, then $2 x=80-3 y$. Since $2 x$ is even, then $80-3 y$ is even. Since 80 is even, then $3 y$ is even, and so $y$ must be even.
Set $y=2 Y$ for some positive integer $Y$.
Then $2 x+3 y=80$ becomes $2 x+6 Y=80$ or $x+3 Y=40$ or $x=40-3 Y$.
Since $x24$ or $Y>8$.
Since $x>0$, then $3 Y<40$ or $Y<13 \frac{1}{3}$.
When $Y=9$ (that is, $y=18$ ), $x=13$.
When $Y=10$ (that is, $y=20$ ), $x=10$.
When $Y=11$ (that is, $y=22$ ), $x=7$.
When $Y=12$ (that is, $y=24$ ), $x=4$.
When $Y=13$ (that is, $y=26$ ), $x=1$.
Therefore, there are 5 pairs that satisfy the given conditions: $(x, y)=(13,18),(10,20),(7,22)$, $(4,24),(1,26)$.
ANSWER: 5 | 3,964,324,921,070,697,000 |
If $a=2^{3}$ and $b=3^{2}$ evaluate $\frac{(a-b)^{2015}+1^{2015}}{(a-b)^{2015}-1^{2015}}$. | \(\boxed{0}\) | Since $a=2^{3}$ and $b=3^{2}$, then $a-b=8-9=-1$.
Therefore,
$$
\frac{(a-b)^{2015}+1^{2015}}{(a-b)^{2015}-1^{2015}}=\frac{(-1)^{2015}+1}{(-1)^{2015}-1}=\frac{-1+1}{-1-1}=0
$$
ANSWER: 0 | -6,227,334,633,082,871,000 |
A moving sidewalk runs from Point $A$ to Point $B$. When the sidewalk is turned off (that is, is not moving) it takes Mario 90 seconds to walk from Point $A$ to Point $B$. It takes Mario 45 seconds to be carried from Point $A$ to Point $B$ by the moving sidewalk when he is not walking. If his walking speed and the speed of the moving sidewalk are constant, how long does it take him to walk from Point $A$ to Point $B$ along the moving sidewalk when it is moving? | \(\boxed{30}\) | Suppose that the distance from Point $A$ to Point $B$ is $d \mathrm{~m}$, that Mario's speed is $w \mathrm{~m} / \mathrm{s}$, and that the speed of the moving sidewalk is $v \mathrm{~m} / \mathrm{s}$.
From the given information, $\frac{d}{w}=45$ and $\frac{d}{v}=90$.
The amount of the time, in seconds, that it takes Mario to walk along the sidewalk when it is moving is $\frac{d}{v+w}$, since his resulting speed is the sum of his walking speed and the speed of the sidewalk.
Here,
$$
\frac{d}{v+w}=\frac{1}{\frac{v+w}{d}}=\frac{1}{\frac{v}{d}+\frac{w}{d}}=\frac{1}{\frac{1}{90}+\frac{1}{45}}=\frac{1}{\frac{3}{90}}=30
$$
Therefore, it takes Mario 30 seconds to walk from Point $A$ to Point $B$ with the sidewalk moving.
ANSWER: 30 seconds | -8,714,924,159,308,996,000 |
What is the measure, in degrees, of the smallest positive angle $x$ for which $4^{\sin ^{2} x} \cdot 2^{\cos ^{2} x}=2 \sqrt[4]{8}$ ? | \(\boxed{60}\) | Using exponent and trigonometric laws,
$$
\begin{aligned}
4^{\sin ^{2} x} \cdot 2^{\cos ^{2} x} & =2 \sqrt[4]{8} \\
\left(2^{2}\right)^{\sin ^{2} x} \cdot 2^{\cos ^{2} x} & =2 \sqrt[4]{2^{3}} \\
2^{2 \sin ^{2} x} \cdot 2^{\cos ^{2} x} & =2^{1} \cdot 2^{3 / 4} \\
2^{2 \sin ^{2} x+\cos ^{2} x} & =2^{7 / 4} \\
2 \sin ^{2} x+\cos ^{2} x & =\frac{7}{4} \\
\sin ^{2} x+\left(\sin ^{2} x+\cos ^{2} x\right) & =\frac{7}{4} \\
\sin ^{2} x+1 & =\frac{7}{4} \\
\sin ^{2} x & =\frac{3}{4}
\end{aligned}
$$
Thus, $\sin x= \pm \frac{\sqrt{3}}{2}$.
The smallest positive angle for which one of these is true is $x=60^{\circ}$ (or $x=\frac{1}{3} \pi$ in radians). | 402,480,323,304,921,900 |
If $\log _{3 n} 675 \sqrt{3}=\log _{n} 75$, determine the value of $n^{5}$. | \(\boxed{5625}\) | Using exponent laws,
$$
\begin{aligned}
\log _{3 n} 675 \sqrt{3} & =\log _{n} 75 \\
(3 n)^{\log _{3 n} 675 \sqrt{3}} & =(3 n)^{\log _{n} 75} \\
675 \sqrt{3} & =3^{\log _{n} 75} \cdot n^{\log _{n} 75} \\
675 \sqrt{3} & =3^{\log _{n} 75} \cdot 75 \\
9 \sqrt{3} & =3^{\log _{n} 75} \\
3^{2} \cdot 3^{1 / 2} & =3^{\log _{n} 75} \\
3^{5 / 2} & =3^{\log _{n} 75} \\
5 / 2 & =\log _{n} 75 \\
n^{5 / 2} & =75 \\
n^{5} & =75^{2}
\end{aligned}
$$
Therefore, $n^{5}=75^{2}=5625$.
ANSWER: 5625 | 8,545,453,531,116,067,000 |
Define $f(x)=\frac{x^{2}}{1+x^{2}}$ and suppose that
$$
\begin{aligned}
A & =f(1)+f(2)+f(3)+\cdots+f(2015) \\
B & =f(1)+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)+\cdots+f\left(\frac{1}{2015}\right)
\end{aligned}
$$
(Each sum contains 2015 terms.) Determine the value of $A+B$. | \(\boxed{2015}\) | If $u \neq 0$, then
$$
f(u)+f\left(\frac{1}{u}\right)=\frac{u^{2}}{1+u^{2}}+\frac{\frac{1}{u^{2}}}{1+\frac{1}{u^{2}}}=\frac{u^{2}}{1+u^{2}}+\frac{\frac{1}{u^{2}}}{\frac{u^{2}}{u^{2}}+\frac{1}{u^{2}}}=\frac{u^{2}}{1+u^{2}}+\frac{1}{u^{2}+1}=\frac{u^{2}+1}{u^{2}+1}=1
$$
Therefore,
$$
A+B=(f(1)+f(1))+\left(f(2)+f\left(\frac{1}{2}\right)\right)+\left(f(3)+f\left(\frac{1}{3}\right)\right)+\cdots+\left(f(2015)+f\left(\frac{1}{2015}\right)\right)
$$
and the right side equals $2015 \cdot 1$.
Therefore, $A+B=2015$.
ANSWER: 2015 | -547,807,735,329,380,200 |
Let $t$ be TNYWR.
The average of the five numbers $12,15,9,14,10$ is $m$.
The average of the four numbers $24, t, 8,12$ is $n$.
What is the value of $n-m$ ? | \(\boxed{1}\) | The average of the numbers in the first list is $m=\frac{12+15+9+14+10}{5}=\frac{60}{5}=12$.
The average of the numbers in the second list is $n=\frac{24+t+8+12}{4}=\frac{44+t}{4}=11+\frac{1}{4} t$.
Therefore, $n-m=\left(11+\frac{1}{4} t\right)-12=\frac{1}{4} t-1$.
Since the answer to (a) is 8 , then $t=8$, and so $n-m=2-1=1$. | -8,951,892,130,542,150,000 |
Let $t$ be TNYWR.
The lines with equations $y=13$ and $y=3 x+t$ intersect at the point $(a, b)$. What is the value of $a$ ? | \(\boxed{4}\) | Since the two lines intersect at $(a, b)$, then these coordinates satisfy the equation of each line.
Therefore, $b=13$ and $b=3 a+t$.
Since $b=13$, then $13=3 a+t$ or $3 a=13-t$, and so $a=\frac{13}{3}-\frac{1}{3} t$.
Since the answer to (b) is 1 , then $t=1$, and so $a=\frac{13}{3}-\frac{1}{3}=4$.
ANSWER: $8,1,4$ | -3,391,327,290,219,310,000 |
Let $t$ be TNYWR.
In last night's 75 minute choir rehearsal, Canada's Totally Musical Choir spent 6 minutes warming up, 30 minutes learning notes, $t$ minutes learning words, and the rest of the rehearsal singing their pieces. If the choir spent $N \%$ of the rehearsal singing their pieces, what is the value of $N$ ? | \(\boxed{32}\) | According to the given information, at last night's rehearsal, Canada's Totally Musical Choir (CTMC) spent $75-6-30-t=39-t$ minutes singing their pieces.
As a percentage of the full 75 minute rehearsal, this is
$$
\frac{39-t}{75} \times 100 \%=\left(\frac{4}{3}(39-t)\right) \%=\left(52-\frac{4}{3} t\right) \%
$$
Since the answer to (a) is 15 , then $t=15$ and so CTMC spent $\left(52-\frac{4}{3} t\right) \%=32 \%$ of their rehearsal singing their pieces, and so $N=32$. | -2,753,005,183,563,200,500 |
What is the smallest positive integer $n$ for which $\sqrt{2019-n}$ is an integer? | \(\boxed{83}\) | In order for $\sqrt{2019-n}$ to be an integer, $2019-n$ must be a perfect square.
Since $n$ is a positive integer, then $2019-n$ is a perfect square less than 2019 .
Since $n$ is to be as small as possible, $2019-n$ must be the largest perfect square less than 2019 . Since $44^{2}=1936$ and $45^{2}=2025$, then 1936 is the largest perfect square less than 2019.
Thus, $2019-n=1936$ and so $n=83$.
ANSWER: 83 | -3,660,672,463,690,432,000 |
When the line with equation $y=-2 x+7$ is reflected across the line with equation $x=3$, the equation of the resulting line is $y=a x+b$. What is the value of $2 a+b$ ? | \(\boxed{-1}\) | When the line with equation $y=-2 x+7$ is reflected across a vertical line, the sign of the slope is reversed, and so becomes 2 .
Since the new line has equation $y=a x+b$, then $a=2$.
The point on the original line that has $x$-coordinate 3 has $y$-coordinate $y=-2(3)+7=1$.
This means that the point $(3,1)$ is on the original line.
When this line is reflected across $x=3$, this point $(3,1)$ must also be on the reflected line.
Substituting into $y=2 x+b$ gives $1=2(3)+b$ and so $b=-5$.
This means that $2 a+b=2(2)+(-5)=-1$. | -1,833,692,455,224,194,800 |
Suppose that
$$
\begin{aligned}
M & =1^{5}+2^{4} \times 3^{3}-4^{2} \div 5^{1} \\
N & =1^{5}-2^{4} \times 3^{3}+4^{2} \div 5^{1}
\end{aligned}
$$
What is the value of $M+N$ ? | \(\boxed{2}\) | Since $M=1^{5}+\left(2^{4} \times 3^{3}\right)-\left(4^{2} \div 5^{1}\right)$ and $N=1^{5}-\left(2^{4} \times 3^{3}\right)+\left(4^{2} \div 5^{1}\right)$, then when $M$ and $N$ are added the terms $\left(2^{4} \times 3^{3}\right)$ and $\left(4^{2} \div 5^{1}\right)$ "cancel" out.
Thus, $M+N=1^{5}+1^{5}=2$.
ANSWER: 2 | -6,652,995,301,049,616,000 |
How many four-digit palindromes $a b b a$ have the property that the two-digit integer $a b$ and the two-digit integer $b a$ are both prime numbers? (For example, 2332 does not have this property, since 23 is prime but 32 is not.) | \(\boxed{9}\) | If $a b$ and $b a$ are both prime numbers, then neither is even which means that neither digit $a$ or $b$ is even and neither equals 5 , otherwise $a b$ or $b a$ would be even or divisible by 5 and so not prime.
Therefore, each of $a$ and $b$ equals $1,3,7$, or 9 .
The two-digit primes using these digits are 11,13,17, 19,31,37,71,73,79, 97 .
If $a b$ equals one of these primes, then $b a$ must be prime as well, which eliminates 19 as a possible value for $a b$, since 91 is not prime.
Therefore, abba could be 1111, 1331, 1771, 3113, 3773, 7117, 7337, 7997, 9779.
There are 9 such palindromes.
ANSWER: 9 | -3,077,758,065,566,318,600 |
Adia writes a list in increasing order of the integers between 1 and 100, inclusive, that cannot be written as the product of two consecutive positive integers. What is the 40th integer in her list? | \(\boxed{46}\) | The integers less than 50 that can be written as a product of two consecutive positive integers are $1 \cdot 2=2,2 \cdot 3=6,3 \cdot 4=12,4 \cdot 5=20,5 \cdot 6=30$, and $6 \cdot 7=42$.
Therefore, there are $50-6=44$ positive integers less than or equal to 50 that cannot be written as the product of two consecutive positive integers.
This means that 50 would be the 44th integer in Adia's list.
Counting backwards from 50, this means that the 40th integer in Adia's list is $50-4=46$. (None of the numbers eliminated are between 46 and 50.)
ANSWER: 46 | -7,666,221,120,331,778,000 |
For how many ordered pairs of positive integers $(a, b)$ is $1<a+b<22$ ? | \(\boxed{210}\) | If $a=1$, we get $1<1+b<22$ or $0<b<21$, which means that $b$ can equal $1,2,3, \ldots, 19,20$.
If $a=2$, we get $1<2+b<22$ or $-1<b<20$.
Since $b$ is positive, then $b$ can equal $1,2,3, \ldots, 18,19$.
In general, $1-a<b<22-a$. Since $b$ is positive, then $b$ satisfies $1 \leq b \leq 21-a$, which means that there are $21-a$ possible values for $b$ for a given $a$.
As $a$ runs from 1 to 20 , there are thus $20,19,18, \ldots, 3,2,1$ possible values for $b$ in these 20 cases. This means that the total number of pairs $(a, b)$ is $20+19+18+\cdots+3+2+1=\frac{1}{2}(20)(21)=210$. | 6,091,911,158,125,323,000 |
Shelly-Ann normally runs along the Laurel Trail at a constant speed of $8 \mathrm{~m} / \mathrm{s}$. One day, onethird of the trail is covered in mud, through which Shelly-Ann can only run one-quarter of her normal speed, and it takes her 12 s to run the entire length of the trail. How long is the trail, in metres? | \(\boxed{48}\) | Suppose that the length of the trail is $d \mathrm{~m}$.
On the muddy day, Shelly-Ann runs $\frac{1}{3} d \mathrm{~m}$ at $2 \mathrm{~m} / \mathrm{s}$ and $\frac{2}{3} d \mathrm{~m}$ at $8 \mathrm{~m} / \mathrm{s}$.
Since this takes 12 seconds in total, then $\frac{\frac{1}{3} d}{2}+\frac{\frac{2}{3} d}{8}=12$.
Multiplying both sides by 8 , we obtain $\frac{4}{3} d+\frac{2}{3} d=96$ which gives $2 d=96$ or $d=48$.
Therefore, the trail is 48 metres long.
ANSWER: 48 metres | -4,915,194,460,954,528,000 |
What is the value of $123456^{2}-123455 \times 123457$ ? | \(\boxed{1}\) | Let $x=123456$.
Thus, $x-1=123455$ and $x+1=123457$.
Therefore,
$$
123456^{2}-123455 \times 123457=x^{2}-(x-1)(x+1)=x^{2}-\left(x^{2}-1\right)=1
$$
ANSWER: 1 | -8,939,354,601,751,332,000 |
Determine the value of $\left(\log _{2} 4\right)\left(\log _{4} 6\right)\left(\log _{6} 8\right)$. | \(\boxed{3}\) | Using the change of base formulas for logarithms,
$$
\left(\log _{2} 4\right)\left(\log _{4} 6\right)\left(\log _{6} 8\right)=\frac{\log 4}{\log 2} \cdot \frac{\log 6}{\log 4} \cdot \frac{\log 8}{\log 6}=\frac{\log 8}{\log 2}=\log _{2} 8=3
$$
ANSWER: 3 | 3,011,282,342,623,619,000 |
The integers $x, y$ and $z$ satisfy $\frac{x}{5}=\frac{6}{y}=\frac{z}{2}$. What is the largest possible value of $x+y+z ?$ | \(\boxed{43}\) | Since $\frac{x}{5}=\frac{6}{y}$, then $x y=30$.
Since $\frac{6}{y}=\frac{z}{2}$, then $y z=12$.
Since we would like the maximum value of $x+y+z$, we may assume that each of $x, y$ and $z$ is positive.
Since $x y=30$ and $y z=12$ and $y$ is a positive integer, then $y$ is a divisor of each of 30 and 12 . This means that $y$ must equal one of $1,2,3$, or 6 .
If $y=1$, then $x=30$ and $z=12$ which gives $x+y+z=43$.
If $y=2$, then $x=15$ and $z=6$ which gives $x+y+z=23$.
If $y=3$, then $x=10$ and $z=4$ which gives $x+y+z=17$.
If $y=6$, then $x=5$ and $z=2$ which gives $x+y+z=13$.
Therefore, the maximum possible value of $x+y+z$ is 43 .
ANSWER: 43 | 4,057,339,326,160,870,400 |
Suppose that $\mathbf{G}=10^{100}$. ( $\mathbf{G}$ is known as a googol.) How many times does the digit 9 occur in the integer equal to $\mathbf{G}-1009^{2}$ ? | \(\boxed{96}\) | We note first that $1009^{2}=(1000+9)^{2}=1000^{2}+2 \cdot 1000 \cdot 9+9^{2}=1018081$. Therefore,
$$
\begin{aligned}
\mathbf{G}-1009^{2} & =10^{100}-1018081 \\
& =\left(10^{100}-1\right)-(1081081-1) \\
& =\underbrace{999 \cdots 999}_{100 \text { digits equal to } 9}-1081080 \\
& =\underbrace{999 \cdots 999}_{93 \text { digits equal to } 9} 9999999-1081080 \\
& =\underbrace{999 \cdots 999}_{93 \text { digits equal to } 9} 8918919
\end{aligned}
$$
and $\mathbf{G}-1009^{2}$ has 96 digits equal to 9 .
ANSWER: 96 | 5,865,918,243,779,671,000 |
The real numbers $x_{1}, x_{2}, x_{3}, \ldots, x_{n}$ are the consecutive terms of an arithmetic sequence. If
$$
\frac{x_{2}}{x_{1}+x_{3}}+\frac{x_{3}}{x_{2}+x_{4}}+\frac{x_{4}}{x_{3}+x_{5}}+\cdots+\frac{x_{n-2}}{x_{n-3}+x_{n-1}}+\frac{x_{n-1}}{x_{n-2}+x_{n}}=1957
$$
what is the value of $n$ ? | \(\boxed{3916}\) | Since the numbers $x_{1}, x_{2}, \ldots, x_{n}$ form an arithmetic sequence, then for each integer $k$ with $2 \leq k \leq n-1$, we have $x_{k}-x_{k-1}=x_{k+1}-x_{k}$.
Rearranging, we obtain $2 x_{k}=x_{k-1}+x_{k+1}$ and so $\frac{x_{k}}{x_{k-1}+x_{k+1}}=\frac{1}{2}$ for each integer $k$ with $2 \leq k \leq n-1$.
We note that there are $(n-1)-2+1=n-2$ integers $k$ in this range.
Therefore, starting with the given equation
$$
\frac{x_{2}}{x_{1}+x_{3}}+\frac{x_{3}}{x_{2}+x_{4}}+\cdots+\frac{x_{n-2}}{x_{n-3}+x_{n-1}}+\frac{x_{n-1}}{x_{n-2}+x_{n}}=1957
$$
we obtain $(n-2) \cdot \frac{1}{2}=1957$ which gives $n-2=3914$ and so $n=3916$.
ANSWER: 3916 | 9,092,677,999,943,281,000 |
Let $t$ be TNYWR.
At the start of 2018, the Canadian Excellent Mathematics Corporation had $t$ employees in its Moose Jaw office, 40 employees in its Okotoks office, and no other employees. During 2018, the number of employees in the Moose Jaw office increased by $25 \%$ and the number of employees in the Okotoks office decreased by 35\%. How many additional employees did the CEMC have at the end of 2018 compared to the beginning of 2018 ? | \(\boxed{16}\) | At the beginning of 2018, there were 40 employees in Okotoks.
At the end of 2018, there were $35 \%$ fewer employees in Okotoks, which is a total of $0.35 \cdot 40=14$ fewer employees.
At the beginning of 2018, there were $t$ employees in Moose Jaw.
At the end of 2018, there were $25 \%$ more employees in Moose Jaw, which is a total of $0.25 t$ more employees.
The net number of additional employees is thus $0.25 t-14$.
Since the answer to (a) is 120 , then $t=120$ and so $0.25 t-14=0.25(120)-14=16$.
Thus, the "CEMC" had 16 more employees at the end of 2018 than it had at the beginning of 2018 . | -6,906,809,629,671,994,000 |
Let $t$ be TNYWR.
Kolapo lists the four-digit positive integers that can be made using the digits 2, 4, 5, and 9, each once. Kolapo lists these integers in increasing order. What is the $t^{\text {th }}$ number in his list? | \(\boxed{5492}\) | There are $4 \cdot 3 \cdot 2 \cdot 1=24$ integers that Kolapo can make using the digits 2, 4, 5, and 9 . These include $3 \times 2 \times 1=6$ integers beginning with 2 , and 6 integers beginning with 4 , and 6 integers beginning with 5 , and 6 integers beginning with 9 .
Since the answer to (b) is 16 , then $t=16$.
The 16th integer in Kolapo's list is in the third group (those beginning with 5), and is the 4 th largest integer in this group.
In increasing order, the integers beginning with 5 in Kolapo's list are 5249, 5294, 5429, $5492,5924,5942$.
Therefore, the 16th number is 5492 .
ANSWER: $120,16,5492$ | 1,752,055,989,713,535,000 |
Let $t$ be TNYWR.
The sum of the even integers from 2 to $2 k$ inclusive equals $t$ for some positive integer $k$. That is,
$$
2+4+6+\cdots+(2 k-2)+2 k=t
$$
What is the value of $k$ ? | \(\boxed{11}\) | Manipulating the left side,
$$
\begin{aligned}
2+4+6+\cdots+(2 k-2)+2 k & =t \\
2(1+2+3+\cdots+(k-1)+k) & =t \\
2\left(\frac{1}{2} k(k+1)\right) & =t \\
k(k+1) & =t
\end{aligned}
$$
Since the answer to (a) is 132 , then $t=132$.
Since $k(k+1)=132$ and $k$ is positive, then $k=11$. | -214,640,839,743,688,500 |
Let $t$ be TNYWR.
Suppose that $O$ is the origin. Points $P(a, b)$ and $Q(c, 1)$ are in the first quadrant with $a=2 c$. If the slope of $O P$ is $t$ and the slope of $O Q$ is 1 , what is the slope of $P Q$ ? | \(\boxed{21}\) | $O$ has coordinates $(0,0)$ and $Q$ has coordinates $(c, 1)$.
Since the slope of $O Q$ is 1 , then $c=1$.
Since $a=2 c$, then $a=2$ which means that the coordinates of $P$ are $(2, b)$.
Since the slope of $O P$ is $t$, then $t=\frac{b-0}{2-0}$, which means that $b=2 t$.
$P$ has coordinates $(2,2 t)$ and $Q$ has coordinates $(1,1)$.
The slope of $P Q$ is thus $\frac{2 t-1}{2-1}$ which equals $2 t-1$.
Since the answer to (b) is 11 , then $t=11$.
This means that the slope of $P Q$ is $2 \cdot 11-1$ which equals 21 .
ANSWER: $132,11,21$ | -2,681,284,194,054,797,300 |
Let $t$ be TNYWR.
The line with equation $y=-2 x+t$ and the parabola with equation $y=(x-1)^{2}+1$ intersect at point $P$ in the first quadrant. What is the $y$-coordinate of $P$ ? | \(\boxed{5}\) | To find the points of intersection of the line with equation $y=-2 x+t$ and the parabola with equation $y=(x-1)^{2}+1$, we equate values of $y$, to obtain
$$
\begin{aligned}
(x-1)^{2}+1 & =-2 x+t \\
x^{2}-2 x+1+1 & =-2 x+t \\
x^{2} & =t-2
\end{aligned}
$$
The points of intersection thus have $x$-coordinates $x=\sqrt{t-2}$ and $x=-\sqrt{t-2}$.
The point $P$ in the first quadrant has a positive $x$-coordinate, and so its $x$-coordinate is $\sqrt{t-2}$.
Thus, the $y$-coordinate of $P$ is $y=-2 \sqrt{t-2}+t$.
Since the answer to (a) is 11 , then $t=11$.
This means that the $y$-coordinate of $P$ is $y=-2 \sqrt{11-2}+11=-2 \cdot 3+11=5$. | -6,603,809,626,981,854,000 |
Determine the largest 6 -digit positive integer that is divisible by 5 . | \(\boxed{999995}\) | The five largest 6-digit numbers are 999 999, 999 998, 999 997, 999 996, and 999995.
In order for an integer to be divisible by 5 , its units digit must be either 0 or 5 .
The largest 6-digit integer that is divisible by 5 is 999995 .
ANSWER: 999995 | 4,226,860,682,381,782,500 |
For each positive integer $n$, the expression $1+2+3+\cdots+(n-1)+n$ represents the sum of all of the integers from 1 to $n$ inclusive. What integer is equal to
$$
(1+2+3+\cdots+2020+2021)-(1+2+3+\cdots+2018+2019) ?
$$ | \(\boxed{4041}\) | By regrouping terms, we have
$$
\begin{aligned}
& (1+2+3+\cdots+2020+2021)-(1+2+3+\cdots+2018+2019) \\
= & (1+2+3+\cdots+2018+2019)+2020+2021-(1+2+3+\cdots+2018+2019) \\
= & 2020+2021 \\
= & 4041 .
\end{aligned}
$$
ANSWER: 4041 | -2,965,456,545,020,012,000 |
The three scales shown below are balanced. The mass of $\boldsymbol{\lambda}$ is $1 \mathrm{~kg}$. Which of the other objects, (circle, square and triangle), has a mass of $1 \mathrm{~kg}$ ?
 | \(\boxed{1}\) | Suppose the mass of $\boldsymbol{\Delta}$ is $x \mathrm{~kg}$, the mass of $\square$ is $y \mathrm{~kg}$, and the mass of $\square$ is $z \mathrm{~kg}$.
From the first scale, we have that $3 y=2 x$ or $y=\frac{2}{3} x$.
From the third scale, we have that $2 y=x+1$, so we can substitute $y=\frac{2}{3} x$ into this equation
to get $\frac{4}{3} x=x+1$. This can be solved for $x$ to get $x=3$.
Substituting $x=3$ into $3 y=2 x$ gives $3 y=6$ so $y=2$.
From the third scale, we have that $5 z=x+y=5$, and substituting $x=3$ and $y=2$ gives $5 z=5$ or $z=1$.
ANSWER: | -1,183,085,144,841,975,600 |
The diagram shows the first four levels of a school's emergency telephone tree. In the case of an emergency, the principal calls two students. The first level consists of just the principal, and in the second level, two students are contacted. In the next level, each of these two students contacts two students who have not been contacted, so after the third level, a total of 6 students have been contacted. This continues so that each student contacts two students who have not yet been contacted. After the $8^{\text {th }}$ level, how many students in total have been contacted?
 | \(\boxed{254}\) | The number of students contacted in level 1 is 0 .
The number of students contacted in level 2 is $2=2^{1}$.
The number of students contacted in level 3 is $4=2^{2}$.
The number of students contacted in level $n$ is twice the number of students that were contacted in the previous level.
This means the number of students contacted in level 4 is $2\left(2^{2}\right)=2^{3}$, the number of students contacted in level 5 is $2\left(2^{3}\right)=2^{4}$, and so on.
The number of students contacted after 8 levels is therefore
$$
\begin{aligned}
2^{1}+2^{2}+2^{3}+2^{4}+2^{5}+2^{6}+2^{7} & =2+4+8+16+32+64+128 \\
& =254 .
\end{aligned}
$$
# | -4,408,662,720,158,511,000 |
Nabil has a tablet that starts with its battery fully charged to $100 \%$. The battery life decreases at a constant rate as the tablet is being used. He uses the tablet for exactly 60 minutes, after which $68 \%$ of the battery life remains. For how many more minutes can Nabil use the tablet before the battery is at $0 \%$ ? | \(\boxed{127.5}\) | Since the battery loses $100 \%-68 \%=32 \%$ every 60 minutes, the percentage of battery life after $t$ minutes is $P=100-\frac{32}{60} t$.
We can solve this equation for $P=0$ to get $t=\frac{100 \times 60}{32}=\frac{375}{2}$ minutes.
Since we are asked how long after the first hour ( 60 minutes) the battery will be at $0 \%$, the answer is $\frac{375}{2}-60=127.5$ minutes.
## Solution 2
The battery loses its charge at a rate of $100 \%-68 \%=32 \%$ every 60 minutes.
For the remaining $68 \%$ to disappear, it will take $\frac{68 \%}{32 \%} \times 60$ minutes, or $\frac{255}{2}=127.5$ minutes.
ANSWER: 127.5 | -8,681,472,763,345,261,000 |
Suppose $f$ is a function that satisfies $f(2)=20$ and $f(2 n)+n f(2)=f(2 n+2)$ for all positive integers $n$. What is the value of $f(10)$ ? | \(\boxed{220}\) | With $n=1$, we have $f(2 \times 1)+1 \times f(2)=f(2 \times 1+2)$ which simplifies to $f(2)+f(2)=f(4)$, so $f(4)=20+20=40$.
With $n=2$, we have $f(2 \times 2)+2 \times f(2)=f(2 \times 2+2)$ which simplifies to $f(4)+2 f(2)=f(6)$.
Since $f(4)=40$ and $f(2)=20$, we have $f(6)=40+2(20)=80$.
Continuing with $n=3$, we have $f(6)+3 f(2)=f(8)$, so $f(8)=80+3(20)=140$.
With $n=4$, we have $f(8)+4 f(2)=f(10)$, so $f(10)=140+4(20)=220$.
ANSWER: 220 | -8,426,902,112,854,534,000 |
If $n$ is a positive integer, the symbol $n$ ! (read " $n$ factorial") represents the product of the integers from 1 to $n$. For example, $4 !=(1)(2)(3)(4)$ or $4 !=24$. Determine
$$
\frac{1}{\log _{2} 100 !}+\frac{1}{\log _{3} 100 !}+\frac{1}{\log _{4} 100 !}+\cdots+\frac{1}{\log _{99} 100 !}+\frac{1}{\log _{100} 100 !}
$$ | \(\boxed{1}\) | By the change of base formula for logarithms, if $x$ and $y$ are positive real numbers, then
$$
\frac{1}{\log _{x} y}=\frac{1}{\frac{\log _{10} y}{\log _{10} x}}=\frac{\log _{10} x}{\log _{10} y}=\log _{y} x
$$
By this and the identity $\log x+\log y=\log x y$, we have
$$
\begin{aligned}
\frac{1}{\log _{2} 100 !}+\frac{1}{\log _{3} 100 !}+\cdots+\frac{1}{\log _{100} 100 !} & =\log _{100 !} 2+\log _{100 !} 3+\cdots+\log _{100 !} 99+\log _{100 !} 100 \\
& =\log _{100 !}(2 \times 3 \times \cdots \times 99 \times 100) \\
& =\log _{100 !}(1 \times 2 \times 3 \times \cdots \times 99 \times 100) \\
& =\log _{100 !}(100 !) \\
& =1
\end{aligned}
$$
ANSWER: 1 | -8,718,183,081,330,353,000 |
A line passing through $(-5, k)$ and $(13,-7)$ has a slope of $-\frac{1}{2}$. What is the value of $k$ ?
## | \(\boxed{2}\) | The slope of the line is $\frac{-7-k}{13-(-5)}=-\frac{7+k}{18}$.
Therefore, $-\frac{7+k}{18}=-\frac{1}{2}=-\frac{9}{18}$, so $7+k=9$ from which it follows that $k=2$. | 7,163,357,625,400,953,000 |
Let $t$ be TNYWR
Three buckets, labelled $A, B$, and $C$, are filled with water.
The amount of water in bucket $A$ is 6 litres more than half of the amount in bucket $C$.
The amount of water in bucket $B$ is the average (mean) of the amounts in buckets $A$ and $C$.
The amount of water in bucket $C$ is $18 t+8$ litres.
In total, how many litres of water are there in the three buckets?
## | \(\boxed{108}\) | Let $a$ be the number of litres of water in bucket $A, b$ be the number of litres in bucket $B$, and $c$ be the number of litres in bucket $C$.
The information given translates in to the equtions $a=\frac{1}{2} c+6, b=\frac{a+c}{2}$, and $c=18 t+8$.
Substituting $c=18 t+8$ into $a=\frac{1}{2} c+6$ gives $a=\frac{1}{2}(18 t+8)+6=9 t+10$.
Substituting the values of $a$ and $c$ in terms of $t$ into $b=\frac{a+c}{2}$, we have
$$
b=\frac{(9 t+10)+(18 t+8)}{2}=\frac{27 t}{2}+9 .
$$
Therefore,
$$
a+b+c=(9 t+10)+\left(\frac{27 t}{2}+9\right)+(18 t+8)=\frac{81 t}{2}+27 .
$$
Using that $t=2$, we have $a+b+c=81+27=108$. | 8,582,441,271,386,433,000 |
Let $d$ be TNYWR.
Lawrence runs $\frac{d}{2} \mathrm{~km}$ at an average speed of 8 minutes per kilometre.
George runs $\frac{d}{2} \mathrm{~km}$ at an average speed of 12 minutes per kilometre.
How many minutes more did George run than Lawrence?
## | \(\boxed{104}\) | Observe that the time it takes to run a given distance (in minutes) is the product of the distance (in kilometres) and the rate (in minutes per kilometre).
The time it took Lawrence to complete the run was $8 \times \frac{d}{2}=4 d$ minutes.
The time it took George to complete the run was $12 \times \frac{d}{2}=6 d$ minutes.
Therefore, it took George $6 d-4 d=2 d$ minutes longer than Lawrence to complete the run.
Substituting $d=52$, it took George 104 minutes longer than Lawrence to complete the run. | 2,056,039,383,685,833,500 |
Let $t$ be TNYWR.
The sum of two numbers is $t$ and the positive difference between the squares of these two numbers is 208. What is the larger of the two numbers? | \(\boxed{53}\) | Let the numbers be $a$ and $b$.
We have that $a+b=t$ and $a^{2}-b^{2}=208$.
Factoring the difference of squares, we have $(a-b)(a+b)=208$, so $t(a-b)=208$, which can be solved for $a-b$ to get $a-b=\frac{208}{t}$.
Adding this equation to $a+b=t$, we get $2 a=t+\frac{208}{t}$ so $a=\frac{t}{2}+\frac{104}{t}$.
Substituting $t=104$, we get $a=\frac{104}{2}+\frac{104}{104}=53$. Observe that since $(a-b)(a+b)=208$ and $a+b$ and 208 are positive, we have that $a-b>0$ so $a>b$.
Therefore, $a$ is the largest of the two numbers so the answer is 53 .
ANSWER: $(52,104,53)$ | 8,593,034,133,313,098,000 |
Integers $x$ and $y$, with $x>y$, satisfy $x+y=7$ and $x y=12$.
Integers $m$ and $n$, with $m>n$, satisty $m+n=13$ and $m^{2}+n^{2}=97$.
If $A=x-y$ and $B=m-n$, determine the value of $A-B$. | \(\boxed{-4}\) | If $x$ and $y$ are integers with $x>y$ and $x+y=7$ and $x y=12$, then $x=4$ and $y=3$, so $A=x-y=1$. (We can check that there are no other solutions.)
If $m$ and $n$ are integers with $m>n$ and $m+n=13$ and $m^{2}+n^{2}=97$, then $m=9$ and $n=4$, so $B=m-n=5$. (We can check that there are no other solutions.)
Therefore, $A-B=1-5=-4$.
ANSWER: -4 | -5,459,214,244,739,503,000 |
Robert was born in the year $n^{2}$. On his birthday in the year $(n+1)^{2}$, he will be 89 years old. In what year was he born? | \(\boxed{1936}\) | Robert was born in the year $n^{2}$ and will be 89 years old in the year $(n+1)^{2}$.
Therefore, $(n+1)^{2}-n^{2}=89$ or $n^{2}+2 n+1-n^{2}=89$ and so $2 n+1=89$ or $n=44$.
Thus, $n^{2}=1936$, so Robert was born in the year 1936 .
ANSWER: 1936 | -3,132,980,729,012,162,000 |
Suppose that
$$
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right) \cdots\left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right)=2014
$$
for some positive integer $k$. (There are $k$ factors in the product.) What is the value of $k$ ? | \(\boxed{4026}\) | Simplifying,
$$
\begin{aligned}
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right) \cdots\left(1+\frac{1}{k}\right)\left(1+\frac{1}{k+1}\right) & =2014 \\
\left(\frac{3}{2}\right)\left(\frac{4}{3}\right)\left(\frac{5}{4}\right) \cdots\left(\frac{k+1}{k}\right)\left(\frac{k+2}{k+1}\right) & =2014 \\
\frac{3 \cdot 4 \cdot 5 \cdots(k+1) \cdot(k+2)}{2 \cdot 3 \cdot 4 \cdots \cdots \cdot(k+1)} & =2014 \\
\frac{k+2}{2} & =2014 \\
k+2 & =4028 \\
k & =4026
\end{aligned}
$$
Therefore, $k=4026$.
ANSWER: 4026 | 1,864,895,293,534,006,300 |
In an arithmetic sequence with 20 terms, the sum of the first three terms is 15 and the sum of the last three terms is 12 . What is the sum of the 20 terms in the sequence?
(An arithmetic sequence is a sequence in which each term after the first is obtained from the previous term by adding a constant. For example, $3,5,7,9$ are the first four terms of an arithmetic sequence.) | \(\boxed{90}\) | Suppose that the sequence has first term $a$ and common difference $d$.
Then the 20 terms are $a, a+d, a+2 d, \ldots, a+17 d, a+18 d, a+19 d$.
The sum of an arithmetic sequence with first term $a$, common difference $d$, and $n$ terms is $\frac{n}{2}\left(a+(a+(n-1) d)\right.$, so the sum of this sequence is $\frac{20}{2}(a+(a+19 d))=10(2 a+19 d)$.
Since the sum of the first three terms is 15 , then $a+(a+d)+(a+2 d)=15$ or $3 a+3 d=15$ and so $a+d=5$.
Since the sum of the last three terms is 12 , then $(a+17 d)+(a+18 d)+(a+19 d)=12$ or $3 a+54 d=12$ and so $a+18 d=4$.
Therefore, the sum of all 20 terms is $10(2 a+19 d)=10((a+d)+(a+18 d))=10(5+4)=90$. | 8,969,352,444,655,341,000 |
A parabola has equation $y=k^{2}-x^{2}$, for some positive number $k$. Rectangle $A B C D$ is drawn with sides parallel to the axes so that $A$ and $D$ are the points where the parabola intersects the $x$-axis and so that the vertex, $V$, of the parabola is the midpoint of $B C$. If the perimeter of the rectangle is 48 , what is the value of $k$ ?
 | \(\boxed{4}\) | Since the parabola has equation $y=k^{2}-x^{2}$, then its $y$-intercept is $k^{2}$, and so the height of rectangle $A B C D$ is $k^{2}$.
Since the parabola has equation $y=k^{2}-x^{2}=(k-x)(k+x)$, then its $x$-intercepts are $-k$ and $k$, so the width of rectangle is $k-(-k)=2 k$.
Since the perimeter is 48 , then $2 k^{2}+2(2 k)=48$ or $k^{2}+2 k-24=0$.
Factoring, we obtain $(k+6)(k-4)=0$.
Since $k>0$, then $k=4$.
ANSWER: 4 | 8,239,171,587,699,116,000 |
A quantity of grey paint has a mass of $12 \mathrm{~kg}$. The grey paint is a mixture of black paint and white paint, of which $80 \%$ by mass is white paint. More white paint is added until the mixture is $90 \%$ white paint by mass. What is the resulting total mass of the paint, in $\mathrm{kg}$ ? | \(\boxed{24}\) | Since the grey paint has mass $12 \mathrm{~kg}$ and is $80 \%$ white paint, then the amount of white paint is $0.8 \times 12=9.6 \mathrm{~kg}$ and so the amount of black paint is $12-9.6=2.4 \mathrm{~kg}$.
For the new mixture to be $90 \%$ white paint and $10 \%$ black paint, the mass of white paint should be 9 times the mass of black paint, or $9 \times 2.4=21.6 \mathrm{~kg}$.
In total, the new mixture has mass $2.4+21.6=24 \mathrm{~kg}$.
ANSWER: 24 | -2,936,921,051,170,650,000 |
In a regular polygon, a diagonal is a line segment joining a vertex to any non-neighbouring vertex. For example, a regular hexagon has 9 diagonals. If a regular polygon with $n$ sides has 90 diagonals, what is the value of $n$ ? | \(\boxed{15}\) | Consider one of the $n$ vertices of a regular polygon with $n$ sides.
This vertex will be part of $(n-3)$ diagonals - one joining to every vertex other than itself and its two neighbours.
This means that there are $\frac{1}{2} n(n-3)$ diagonals in total, since each of $n$ vertices is part of $n-3$ diagonals, and each diagonal is counted twice in this way.
For there to be 90 diagonals, then $\frac{1}{2} n(n-3)=90$ or $n^{2}-3 n-180=0$.
Thus, $(n-15)(n+12)=0$.
Since $n>0$, then $n=15$. | -6,045,856,306,182,962,000 |
The sum of 8 one-digit positive integers is 68 . If seven of these integers are equal, determine the other integer. | \(\boxed{5}\) | Let $x$ be the one-digit integer that is included 7 times.
Since $x$ is at most 9 , then $7 x$ is at most 63 .
If $x=9$, then the other integer is $68-7 x=68-63=5$.
If $x \leq 8$, then $7 x$ is at most 56 , and so the other integer would have to be at least 12 , which is not possible since it is a one-digit integer.
Thus, the other integer is 5 .
ANSWER: 5 | 2,518,543,931,719,851,500 |
The five-digit positive integer $15 A B 9$ is a perfect square for some digits $A$ and $B$. What is the value of $A+B$ ? | \(\boxed{3}\) | Since $120^{2}=14400$ and $130^{2}=16900$, then $15 A B 9$ is the square of an integer between 120 and 130 .
Since $15 A B 9$ has a units digit of 9 , then it is the square of an integer ending in 3 or 7 .
Note that $123^{2}=15129$ and $127^{2}=16129$, then $A=1$ and $B=2$, so $A+B=3$.
ANSWER: 3 | -3,614,397,945,444,163,000 |
The numbers $36,27,42,32,28,31,23,17$ are grouped in pairs so that the sum of each pair is the same. Which number is paired with 32 ? | \(\boxed{27}\) | The 8 numbers in the list have a sum of 236 .
When these 8 numbers are grouped into 4 pairs with equal sums, this sum must be $\frac{236}{4}=59$.
Therefore, 32 must be paired with $59-32=27$.
(Note that the pairings $36+23$ and $27+32$ and $42+17$ and $28+31$ satisfy the requirements.)
ANSWER: 27 | 7,243,114,708,334,919,000 |
In the diagram, a positive integer is hidden behind each star. The integer shown on each star is the product of the integers hidden behind the other three stars. What is the product of all four hidden integers?
 | \(\boxed{840}\) | Suppose that the numbers hidden behind the four stars, in order, are $w, x, y, z$.
Since the number on the first star is 280 , then the product of the numbers behind the other three stars is 280 , or $x y z=280$.
Similarly, the numbers on the other three stars give the equations $w y z=168, w x z=105$, and $w x y=120$.
Multiplying these four equations together, we obtain
$$
(x y z)(w y z)(w x z)(w x y)=(280)(168)(105)(120)
$$
or
$$
w^{3} x^{3} y^{3} z^{3}=\left(2^{3} \cdot 5 \cdot 7\right)\left(2^{3} \cdot 3 \cdot 7\right)(3 \cdot 5 \cdot 7)\left(2^{3} \cdot 3 \cdot 5\right)
$$
and so
$$
(w x y z)^{3}=2^{9} \cdot 3^{3} \cdot 5^{3} \cdot 7^{3}
$$
Therefore, the product of the four hidden integers, or $w x y z$, equals $2^{3} \cdot 3 \cdot 5 \cdot 7=840$. | 961,116,710,073,590,100 |
Let $t$ be TNYWR.
The average of the list of five numbers $13,16,10,15,11$ is $m$.
The average of the list of four numbers $16, t, 3,13$ is $n$.
What is the value of $m-n$ ? | \(\boxed{4}\) | The average of the first list is $m=\frac{13+16+10+15+11}{5}=\frac{65}{5}=13$.
The average of the second list is $n=\frac{16+t+3+13}{4}=\frac{32+t}{4}=8+\frac{1}{4} t$.
Therefore, $m-n=13-\left(8+\frac{1}{4} t\right)=5-\frac{1}{4} t$.
Since the answer to (a) is 4 , then $t=4$, and so $m-n=5-1=4$. | 2,492,813,287,661,434,400 |
Let $t$ be TNYWR.
The lines with equations $y=12$ and $y=2 x+t$ intersect at the point $(a, b)$. What is the value of $a$ ? | \(\boxed{4}\) | Since the two lines intersect at $(a, b)$, then these coordinates satisfy the equation of each line.
Therefore, $b=12$ and $b=2 a+t$.
Since $b=12$, then $12=2 a+t$ or $2 a=12-t$, and so $a=6-\frac{1}{2} t$.
Since the answer to (b) is 4 , then $t=4$, and so $a=6-2=4$.
ANSWER: $4,4,4$ | 4,700,693,722,585,222,000 |
Suppose that $a$ and $b$ are positive integers with $2^{a} \times 3^{b}=324$.
Evaluate $2^{b} \times 3^{a}$. | \(\boxed{144}\) | Repeatedly dividing, $324=3 \times 108=3^{2} \times 36=3^{2} \times 3^{2} \times 2^{2}=2^{2} \times 3^{4}$.
Therefore, $a=2$ and $b=4$, so $2^{b} \times 3^{a}=2^{4} \times 3^{2}=16 \times 9=144$. | -9,076,334,833,522,835,000 |
Let $t$ be TNYWR.
Three siblings share a box of chocolates that contains $t$ pieces. Sarah eats $\frac{1}{3}$ of the total number of chocolates and Andrew eats $\frac{3}{8}$ of the total number of chocolates. Cecily eats the remaining chocolates in the box. How many more chocolates does Sarah eat than Cecily eats? | \(\boxed{6}\) | Sarah eats $\frac{1}{3} t$ chocolates and Andrew eats $\frac{3}{8} t$ chocolates.
Since Cecily eats the rest of the chocolates, she eats $t-\frac{1}{3} t-\frac{3}{8} t=\frac{24-8-9}{24} t=\frac{7}{24} t$.
Therefore, Sarah eats $\frac{1}{3} t-\frac{7}{24} t=\frac{1}{24} t$ more chocolates than Cecily.
Since the answer to (a) is 144, then $t=144$ and so Sarah eats 6 more chocolates than Cecily. | -5,291,330,526,551,558,000 |
What is the greatest common divisor of the three integers 36,45 and 495 ? | \(\boxed{9}\) | Since $36=2^{2} \times 3^{2}$ and $45=3^{2} \times 5$, then the greatest common divisor of 36 and 45 is 9 . Since 9 is also a divisor of 495, then it is the greatest common divisor of the three integers. | -7,727,246,545,221,513,000 |
Let $t$ be TNYWR.
The expression $(t x+3)^{3}$ can be re-written in the form $a x^{3}+b x^{2}+c x+d$ for some positive integers $a, b, c, d$. Determine the value of the largest of $a, b, c$, and $d$. | \(\boxed{324}\) | Expanding,
$$
(t x+3)^{3}=(t x+3)(t x+3)^{2}=(t x+3)\left(t^{2} x^{2}+6 t x+9\right)=t^{3} x^{3}+9 t^{2} x^{2}+27 t x+27
$$
In terms of $t$, the coefficients are $a=t^{3}, b=9 t^{2}, c=27 t$, and $d=27$.
Since the answer to (b) is 6 , then $t=6$, and so the coefficients are $a=6^{3}=216$, $b=9(36)=324, c=27(6)=162$, and $d=27$.
Therefore, the largest of the values of $a, b, c, d$ is 324 .
ANswer: 9, 6,324 | 2,313,709,297,763,269,600 |
A car and a minivan drive from Alphaville to Betatown. The car travels at a constant speed of $40 \mathrm{~km} / \mathrm{h}$ and the minivan travels at a constant speed of $50 \mathrm{~km} / \mathrm{h}$. The minivan passes the car 10 minutes before the car arrives at Betatown. How many minutes pass between the time at which the minivan arrives in Betatown and the time at which the car arrives in Betatown? | \(\boxed{2}\) | The car takes 10 minutes to travel from the point at which the minivan passes it until it arrives in Betatown.
Since the car drives at $40 \mathrm{~km} / \mathrm{h}$ and since 10 minutes equals $\frac{1}{6}$ hour, then the car travels $40 \mathrm{~km} / \mathrm{h} \cdot \frac{1}{6} \mathrm{~h}=\frac{20}{3} \mathrm{~km}$ in these 10 minutes.
Thus, the distance between the point where the vehicles pass and Betatown is $\frac{20}{3} \mathrm{~km}$.
Since the minivan travels at $50 \mathrm{~km} / \mathrm{h}$, it covers this distance in $\frac{203 \mathrm{~km}}{50 \mathrm{~km} / \mathrm{h}}=\frac{2}{15} \mathrm{~h}$.
Now $\frac{2}{15} \mathrm{~h}=\frac{8}{60} \mathrm{~h}$ which equals 8 minutes, and so the minivan arrives in Betatown $10-8=2$ minutes before the car.
ANSWER: 2 minutes | 4,246,390,666,086,851,600 |
Bethany is told to create an expression from $2 \square 0 \square 1 \square 7$ by putting a + in one box, a in another, and $a x$ in the remaining box. There are 6 ways in which she can do this. She calculates the value of each expression and obtains a maximum value of $M$ and a minimum value of $m$. What is $M-m$ ? | \(\boxed{15}\) | The six expressions that Bethany creates are
$$
\begin{aligned}
& 2+0-1 \times 7=2+0-7=-5 \\
& 2+0 \times 1-7=2+0-7=-5 \\
& 2-0+1 \times 7=2-0+7=9 \\
& 2-0 \times 1+7=2-0+7=9 \\
& 2 \times 0+1-7=0+1-7=-6 \\
& 2 \times 0-1+7=0-1+7=6
\end{aligned}
$$
Of these, the maximum value is $M=9$ and the minimum value is $m=-6$.
Thus, $M-m=9-(-6)=15$.
ANSWER: 15 | -5,321,250,837,250,721,000 |
If $n$ is the largest positive integer with $n^{2}<2018$ and $m$ is the smallest positive integer with $2018<m^{2}$, what is $m^{2}-n^{2}$ ? | \(\boxed{89}\) | Since $\sqrt{2018} \approx 44.92$, the largest perfect square less than 2018 is $44^{2}=1936$ and the smallest perfect square greater than 2018 is $45^{2}=2025$.
Therefore, $m^{2}=2025$ and $n^{2}=1936$, which gives $m^{2}-n^{2}=2025-1936=89$.
ANSWER: 89 | 4,800,741,562,767,904,000 |
If $N$ is a positive integer with $\sqrt{12}+\sqrt{108}=\sqrt{N}$, determine the value of $N$. | \(\boxed{192}\) | Since $\sqrt{12}+\sqrt{108}=\sqrt{N}$, then $\sqrt{2^{2} \cdot 3}+\sqrt{6^{2} \cdot 3}=\sqrt{N}$ or $2 \sqrt{3}+6 \sqrt{3}=\sqrt{N}$.
This means that $\sqrt{N}=8 \sqrt{3}=\sqrt{8^{2} \cdot 3}=\sqrt{192}$ and so $N=192$.
ANSWER: 192 | 33,342,527,400,418,696 |
Determine the value of the expression
$$
1+2-3+4+5-6+7+8-9+10+11-12+\cdots+94+95-96+97+98-99
$$
(The expression consists of 99 terms. The operations alternate between two additions and one subtraction.) | \(\boxed{1584}\) | Removing the initial 0 , the remaining 99 terms can be written in groups of the form $(3 k-2)+$ $(3 k-1)-3 k$ for each $k$ from 1 to 33 .
The expression $(3 k-2)+(3 k-1)-3 k$ simplifies to $3 k-3$.
Therefore, the given sum equals
$$
0+3+6+\cdots+93+96
$$
Since $k$ ran from 1 to 33 , then this sum includes 33 terms and so equals
$$
\frac{33}{2}(0+96)=33(48)=33(50)-33(2)=1650-66=1584
$$
ANSWER: 1584 | 7,425,024,150,924,961,000 |
The product of the roots of the quadratic equation $2 x^{2}+p x-p+4=0$ is 9 . What is the sum of the roots of this equation? | \(\boxed{7}\) | The product of the roots of the quadratic equation $a x^{2}+b x+c=0$ is $\frac{c}{a}$ and the sum of the roots is $-\frac{b}{a}$.
Since the product of the roots of $2 x^{2}+p x-p+4=0$ is 9 , then $\frac{-p+4}{2}=9$ and so $-p+4=18$, which gives $p=-14$.
Therefore, the quadratic equation is $2 x^{2}-14 x+18=0$ and the sum of its roots is $-\frac{(-14)}{2}=7$. ANSWER: 7 | -34,751,841,917,970,708 |
Let $t$ be TNYWR.
What is the area of a triangle with base $2 t$ and height $3 t-1$ ? | \(\boxed{70}\) | The area of a triangle with base $2 t$ and height $3 t-1$ is $\frac{1}{2}(2 t)(3 t-1)$ or $t(3 t-1)$. Since the answer to (a) is 5 , then $t=5$, and so $t(3 t-1)=5(14)=70$. | -4,814,035,576,617,072,000 |
When the integer $300^{8}$ is written out, it has $d$ digits. What is the value of $d$ ? | \(\boxed{20}\) | We note that $300^{8}=3^{8} \cdot 100^{8}=3^{8} \cdot\left(10^{2}\right)^{8}=6561 \cdot 10^{16}$.
Multiplying 6561 by $10^{16}$ is equivalent to appending 16 zeroes to the right end of 6561 , creating an integer with 20 digits. | -6,496,079,938,409,626,000 |
Let $t$ be TNYWR.
If the graph of $y=2 \sqrt{2 t} \sqrt{x}-2 t$ passes through the point $(a, a)$, what is the value of $a$ ? | \(\boxed{6}\) | Since the graph of $y=2 \sqrt{2 t} \sqrt{x}-2 t$ passes through the point ( $a$, a), then $a=2 \sqrt{2 t} \sqrt{a}-2 t$. Rearranging, we obtain $a-2 \sqrt{2 t} \sqrt{a}+2 t=0$. We re-write as $(\sqrt{a})^{2}-2 \sqrt{a} \sqrt{2 t}+(\sqrt{2 t})^{2}=0$ or $(\sqrt{a}-\sqrt{2 t})^{2}=0$.
Therefore, $\sqrt{a}=\sqrt{2 t}$ or $a=2 t$.
Since the answer to (a) is 3 , then $t=3$ and so $a=6$. | 5,735,786,366,525,975,000 |
Let $t$ be TNYWR.
Suppose that
$$
\frac{1}{2^{12}}+\frac{1}{2^{11}}+\frac{1}{2^{10}}+\cdots+\frac{1}{2^{t+1}}+\frac{1}{2^{t}}=\frac{n}{2^{12}}
$$
(The sum on the left side consists of $13-t$ terms.)
What is the value of $n$ ? | \(\boxed{127}\) | Multiplying both sides of the given equation by $2^{12}$, we obtain
$$
1+2^{1}+2^{2}+\cdots+2^{12-(t+1)}+2^{12-t}=n
$$
Since the answer to (b) is 6 , then $t=6$ and so we have
$$
1+2^{1}+2^{2}+2^{3}+2^{4}+2^{5}+2^{6}=n
$$
Therefore, $n=1+2+4+8+16+32+64=127$.
ANSwer: $3,6,127$ | 8,648,372,534,184,894,000 |
The series below includes the consecutive even integers from 2 to 2022 inclusive, where the signs of the terms alternate between positive and negative:
$$
S=2-4+6-8+10-\cdots-2016+2018-2020+2022
$$
What is the value of $S$ ? | \(\boxed{1012}\) | The terms being added and subtracted are the integers 2(1), 2(2), 2(3), and so on up to 2(1011). This means there are 1011 terms in total, which is an odd number of terms.
We will now insert parentheses to group the terms in pairs starting with the first two terms, then the second two terms, and so on.
Since the total number of terms is odd, 2022 will not be grouped with another term:
$$
S=(2-4)+(6-8)+(10-12)+\cdots+(2014-2016)+(2018-2020)+2022
$$
Each of the parenthetical expressions is equal to -2 , and the number of parenthetical expressions is $\frac{1011-1}{2}=505$.
Therefore,
$$
S=505(-2)+2022=-1010+2022=1012
$$
ANSWER: 1012 | -4,278,565,434,116,826,000 |
What is the largest integer $n$ with the properties that $200<n<250$ and that $12 n$ is a perfect square? | \(\boxed{243}\) | Suppose $12 n=k^{2}$ for some integer $k$. Then $k^{2}$ is even and so $k$ must be even, which means $\frac{k}{2}$ is an integer.
Dividing both sides of $12 n=k^{2}$ by 4 gives $3 n=\left(\frac{k}{2}\right)^{2}$, and since $\frac{k}{2}$ is an integer, this means $3 n$ is a perfect square.
We are given that $200<n<250$, which implies $600<3 n<750$.
The perfect squares between 600 and 750 are 625,676 , and 729 , among which 729 is the only multiple of 3 , so $3 n=729$ or $n=243$.
This value of $n$ satisfies $200<n<250$, and $12 n=2916=54^{2}$.
We have shown that $n=243$ is the only $n$ that satisfies the conditions, so it must be the largest. | -1,636,397,179,737,286,100 |
What is the sum of the digits of the integer equal to $3 \times 10^{500}-2022 \times 10^{497}-2022$ ? | \(\boxed{4491}\) | We can rewrite the given expression as follows:
$$
\begin{aligned}
3 \times 10^{500}-2022 \times 10^{497}-2022 & =3000 \times 10^{497}-2022 \times 10^{497}-2022 \\
& =(3000-2022) \times 10^{497}-2022 \\
& =978 \times 10^{497}-2022 \\
& =\left(978 \times 10^{497}-1\right)-2021
\end{aligned}
$$
The integer $978 \times 10^{497}-1$ is the 500 -digit integer with leading digits 977 followed by 4979 's. Thus, the digits of the given integer, from left to right, are 9, 7 , and 7 followed by $497-4=493$ 9 's, and the final four digits are $9-2=7,9-0=9,9-2=7$, and $9-1=8$.
Therefore, the sum of the digits of the integer is
$$
9+7+7+(493 \times 9)+7+9+7+8=4491
$$
ANSWER: 4491 | 377,207,393,070,194,940 |
The integer 2022 is positive and has four digits. Three of its digits are 2 and one of its digits is 0 . What is the difference between the largest and smallest four-digit integers that can be made using three 2 's and one 0 as digits? | \(\boxed{198}\) | Since the first digit of a four-digit integer cannot be 0 , there are three integers with the given property: 2022, 2202, and 2220 .
The largest is 2220 and the smallest is 2022 , so the answer is $2220-2022=198$.
ANSWER: 198 | 1,594,164,799,171,400,200 |
A total of $\$ 425$ was invested in three different accounts, Account A, Account B and Account C. After one year, the amount in Account A had increased by 5\%, the amount in Account B had increased by 8\%, and the amount in Account C had increased by 10\%. The increase in dollars was the same in each of the three accounts. How much money was originally invested in Account C? | \(\boxed{100}\) | Let $a$ be the amount in dollars initially invested in Account $\mathrm{A}$, let $b$ be the amount in dollars initially invested in Account B, and let $c$ be the amount in dollars initially invested in Account C. The given information leads to the equations $a+b+c=425,0.05 a=0.08 b$, and $0.08 b=0.1 c$. The second and third equations can be solved for $a$ and $b$, respectively, to get $a=\frac{8}{5} b$ and
$b=\frac{5}{4} c$.
Substituting $b=\frac{5}{4} c$ into the equation $a=\frac{8}{5} b$, we get $a=2 c$.
Substituting $a=2 c$ and $b=\frac{5}{4} c$ into the equation $a+b+c=425$ gives
$$
425=a+b+c=2 c+\frac{5}{4} c+c=\frac{17}{4} c
$$
Therefore, the amount in dollars initially invested in Account C is $c=\frac{4}{17} \times 425=100$.
ANSWER: $\$ 100$
# | 4,486,426,697,718,415,000 |
A list of integers consists of $(m+1)$ ones, $(m+2)$ twos, $(m+3)$ threes, $(m+4)$ fours, and $(m+5)$ fives. The average (mean) of the list of integers is $\frac{19}{6}$. What is $m$ ? | \(\boxed{9}\) | The sum of the integers in the list is
$$
(m+1)+2(m+2)+3(m+3)+4(m+4)+5(m+5)=15 m+1+4+9+16+25=15 m+55
$$
The number of integers in the list is
$$
(m+1)+(m+2)+(m+3)+(m+4)+(m+5)=5 m+15
$$
The average of the integers in the list is $\frac{19}{6}$, so this means
$$
\frac{19}{6}=\frac{15 m+55}{5 m+15}=\frac{3 m+11}{m+3}
$$
The above equation is equivalent to $19(m+3)=6(3 m+11)$, or $19 m+57=18 m+66$, which can be solved for $m$ to get $m=9$. | 4,826,081,159,483,003,000 |
The equation $x^{2}-7 x+k=0$ has solutions $x=3$ and $x=a$. The equation $x^{2}-8 x+k+1=0$ has solutions $x=b$ and $x=c$. What is the value of $a+b c$ ? | \(\boxed{17}\) | Using that 3 is a solution to $x^{2}-7 x+k=0$ we have that $3^{2}-7(3)+k=0$ or $k=21-9=12$. This means $x^{2}-7 x+k$ is $x^{2}-7 x+12=(x-3)(x-4)$, the roots of which are 3 and 4 , which means $a=4$.
The polynomial $x^{2}-8 x+k+1$ factors as $(x-b)(x-c)$, so upon expanding the latter expression, we have $b c=k+1=12+1=13$.
Therefore, $a+b c=4+13=17$.
ANSWER: 17 | 3,731,170,067,351,035,000 |
The functions $f(x)$ and $g(x)$ are defined by $f(x)=9^{x}$ and $g(x)=\log _{3}(9 x)$. The real number $x$ satisfies $g(f(x))=f(g(2))$. What is the value of $x$ ? | \(\boxed{161}\) | For positive $x$,
$$
\begin{aligned}
g(f(x)) & =\log _{3}(9 f(x)) \\
& =\log _{3}\left(9 \times 9^{x}\right) \\
& =\log _{3}\left(9^{x+1}\right) \\
& =\log _{3}\left(\left(3^{2}\right)^{x+1}\right) \\
& =\log _{3}\left(3^{2 x+2}\right) \\
& =(2 x+2) \log _{3}(3) \\
& =2 x+2
\end{aligned}
$$
A similar calculation shows that
$$
\begin{aligned}
f(g(2)) & =9^{g(2)} \\
& =9^{\log _{3}(18)} \\
& =3^{2 \log _{3}(18)} \\
& =\left(3^{\log _{3}(18)}\right)^{2} \\
& =18^{2} \\
& =324
\end{aligned}
$$
Therefore, $2 x+2=324$ so $x=\frac{324-2}{2}=161$. | -4,398,049,934,303,167,500 |
The real number $\theta$ is an angle measure in degrees that satisfies $0^{\circ}<\theta<360^{\circ}$ and
$$
2022^{2 \sin ^{2} \theta-3 \sin \theta+1}=1
$$
The sum of the possible values of $\theta$ is $k^{\circ}$. What is the value of $k$ ? | \(\boxed{270}\) | If $2022^{x}=1$, then $x=0$. This means we have $2 \sin ^{2} \theta-3 \sin \theta+1=0$.
The expression on the left can be factored to get $(\sin \theta-1)(2 \sin \theta-1)=0$.
Therefore, the given equation is true exactly when $\sin \theta=1$ or $\sin \theta=\frac{1}{2}$.
Since $0^{\circ}<\theta<360^{\circ}$, the only value of $\theta$ with $\sin \theta=1$ is $\theta=90^{\circ}$.
The only values of $\theta$ with $\sin \theta=\frac{1}{2}$ are $\theta=30^{\circ}$ and $\theta=150^{\circ}$.
Therefore, the answer is $90^{\circ}+30^{\circ}+150^{\circ}=270^{\circ}$.
ANSWER: $270^{\circ}$ | -3,226,312,287,835,982,000 |
What is the largest integer that can be placed in the box so that $\frac{\square}{11}<\frac{2}{3}$ ?
## | \(\boxed{7}\) | Since 33 is positive, $\frac{\square}{11}<\frac{2}{3}$ implies
$$
33\left(\frac{\square}{11}\right)<33\left(\frac{2}{3}\right)
$$
which simplifies to $3 \times \square<22$.
The largest multiple of 3 that is less than 22 is $3 \times 7$, so this means the number in the box cannot be larger than 7 .
Indeed, $\frac{7}{11}=\frac{21}{33}$ is less than $\frac{2}{3}=\frac{22}{33}$, so the answer is 7 . | -5,620,472,674,965,968,000 |
Let $t$ be TNYWR.
If $6 x+t=4 x-9$, what is the value of $x+4$ ?
## | \(\boxed{-4}\) | Rearranging the equation, we have $2 x=-t-9$ or $x=\frac{-t-9}{2}$, so
$$
x+4=\frac{-t-9}{2}+4=\frac{-t-9+8}{2}=\frac{-t-1}{2}
$$
Substituting $t=7$ into this equation gives $x+4=\frac{-7-1}{2}=-4$. | -2,590,005,726,978,281,000 |
Let $x$ be the number of prime numbers between 10 and 30 .
What is the number equal to $\frac{x^{2}-4}{x+2}$ ?
## | \(\boxed{4}\) | The prime numbers between 10 and 30 are $11,13,17,19,23$, and 29 , which means $x=6$. Factoring the numerator in $\frac{x^{2}-4}{x+2}$ gives $\frac{(x-2)(x+2)}{x+2}$ which is equal to $x-2$ as long as $x \neq 2$.
Since $x=6$, the answer is $6-2=4$. | 1,402,752,544,444,760,000 |
Let $t$ be TNYWR.
Alida, Bono, and Cate each have some jelly beans.
The number of jelly beans that Alida and Bono have combined is $6 t+3$.
The number of jelly beans that Alida and Cate have combined is $4 t+5$.
The number of jelly beans that Bono and Cate have combined is $6 t$.
How many jelly beans does Bono have?
## | \(\boxed{15}\) | Let $A$ be the number of beans that Alida has, $B$ be the number of beans that Bono has, and $C$ be the number of beans that Cate has.
The given information translates to
$$
\begin{aligned}
& A+B=6 t+3 \\
& A+C=4 t+5 \\
& B+C=6 t
\end{aligned}
$$
Adding these three equations together gives $2(A+B+C)=16 t+8$, which implies $A+B+C=8 t+4$.
Subtracting $A+C=4 t+5$ from $A+B+C=8 t+4$ gives
$$
B=(A+B+C)-(A+C)=(8 t+4)-(4 t+5)=4 t-1
$$
Substituting $t=4$ into $B=4 t-1$ gives $B=4(4)-1=15$. | -2,226,317,289,558,774,500 |
Let $t$ be TNYWR.
There is exactly one real number $x$ with the property that both $x^{2}-t x+36=0$ and $x^{2}-8 x+t=0$.
What is the value of $x$ ?
## The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 2022 Canadian Team Mathematics Contest | \(\boxed{3}\) | Suppose $x$ is the real number such that $x^{2}-t x+36=0$ and $x^{2}-8 x+t=0$.
Therefore, we get that $x^{2}-t x+36=x^{2}-8 x+t$ which can be rearranged to get $36-t=$ $t x-8 x=(t-8) x$.
Dividing by $t-8$ gives $x=\frac{36-t}{t-8}$.
Substituting $t=15$ gives $x=\frac{36-15}{15-8}=\frac{21}{7}=3$.
ANSWER: $(4,15,3)$ | -4,412,793,734,571,553,300 |
Let $t$ be TNYWR.
If $\frac{5+x}{t+x}=\frac{2}{3}$, what is the value of $x$ ?
## | \(\boxed{11}\) | Multiplying through by $3(t+x)$ gives $3(5+x)=2(t+x)$.
Expanding gives $15+3 x=2 t+2 x$ which can be rearranged to $x=2 t-15$.
Substituting $t=13$ gives $x=2(13)-15=11$. | 289,433,724,064,737,340 |
Zeljko travelled at $30 \mathrm{~km} / \mathrm{h}$ for 20 minutes and then travelled at $20 \mathrm{~km} / \mathrm{h}$ for 30 minutes. How far did he travel, in kilometres? | \(\boxed{20}\) | Since 20 minutes is $\frac{1}{3}$ of an hour and Zeljko travelled at $30 \mathrm{~km} / \mathrm{h}$ for $\frac{1}{3}$ of an hour, then Zeljko travelled $30 \cdot \frac{1}{3}=10 \mathrm{~km}$ during this portion of the trip.
Since 30 minutes is $\frac{1}{2}$ of an hour and Zeljko travelled at $20 \mathrm{~km} / \mathrm{h}$ for $\frac{1}{2}$ of an hour, then Zeljko travelled $20 \cdot \frac{1}{2}=10 \mathrm{~km}$ during this portion of the trip.
In total, Zeljko travelled $10+10=20 \mathrm{~km}$.
ANSWER: 20 | -3,186,899,363,056,116,700 |
The average (mean) of $3,5,6,8$, and $x$ is 7 . What is the value of $x$ ? | \(\boxed{13}\) | Since the average of 5 numbers is 7 , then the sum of the 5 numbers is $5 \times 7=35$.
Therefore, $3+5+6+8+x=35$ or $22+x=35$ and so $x=13$.
ANSWER: 13 | 5,951,098,517,390,366,000 |
A parabola has equation $y=a x^{2}+b x+c$ and passes through the points $(-3,50),(-1,20)$ and $(1,2)$. Determine the value of $a+b+c$. | \(\boxed{2}\) | Since $(1,2)$ lies on the parabola with equation $y=a x^{2}+b x+c$, then the coordinates of the point satisfy the equation of the parabola.
Thus, $2=a\left(1^{2}\right)+b(1)+c$ or $a+b+c=2$.
ANSWER: 2 | 7,334,825,061,399,834,000 |
For some positive integers $m$ and $n, 2^{m}-2^{n}=1792$. Determine the value of $m^{2}+n^{2}$. | \(\boxed{185}\) | Since $2^{11}=2048$ and $2^{8}=256$, then $2^{11}-2^{8}=2048-256=1792$.
Therefore, $m=11$ and $n=8$, which gives $m^{2}+n^{2}=11^{2}+8^{2}=121+64=185$.
Alternatively, we can factor the left side of the equation $2^{m}-2^{n}=1792$ to obtain $2^{n}\left(2^{m-n}-1\right)=$ 1792 .
Since $m>n$, then $2^{m-n}-1$ is an integer.
Now, $1792=2^{8} \cdot 7$.
Since $2^{n}\left(2^{m-n}-1\right)=2^{8} \cdot 7$, then $2^{n}=2^{8}$ (which gives $n=8$ ) and $2^{m-n}-1=7$ (which gives $m-n=3$ and so $m=11$ ).
ANSWER: 185 | -5,206,973,383,419,482,000 |
Let $t$ be TNYWR.
If $\frac{3(x+5)}{4}=t+\frac{3-3 x}{2}$, what is the value of $x$ ? | \(\boxed{3}\) | If $\frac{3(x+5)}{4}=t+\frac{3-3 x}{2}$, then $3(x+5)=4 t+2(3-3 x)$ or $3 x+15=4 t+6-6 x$, which gives $9 x=4 t-9$ or $x=\frac{4}{9} t-1$.
Since the answer to (a) is 9 , then $t=9$ and so $x=\frac{4}{9} t-1=4-1=3$. | 1,769,776,142,299,515,100 |
Let $t$ be TNYWR.
The $y$-coordinate of the vertex of the parabola with equation $y=3 x^{2}+6 \sqrt{m} x+36$ is $t$. What is the value of $m$ ? | \(\boxed{11}\) | We start with the given equation and complete the square:
$y=3 x^{2}+6 \sqrt{m} x+36=3\left(x^{2}+2 \sqrt{m} x+12\right)=3\left((x+\sqrt{m})^{2}-m+12\right)=3(x+\sqrt{m})^{2}+(36-3 m)$
Therefore, the coordinates of the vertex of this parabola are $(-\sqrt{m}, 36-3 m)$.
Since the $y$-coordinate of the vertex is $t$, then $36-3 m=t$ or $3 m=36-t$ and so $m=12-\frac{1}{3} t$.
Since the answer to (b) is 3 , then $t=3$, and so $m=12-\frac{1}{3} t=12-1=11$.
ANSWER: 9,3,11 | -5,602,205,319,712,801,000 |
What is the sum of the $x$-intercept of the line with equation $20 x+16 y-40=0$ and the $y$-intercept of the line with equation $20 x+16 y-64=0$ ? | \(\boxed{6}\) | To find the $x$-intercept of the line with equation $20 x+16 y-40=0$, we set $y=0$ and get $20 x-40=0$ or $x=2$.
To find the $y$-intercept of the line with equation $20 x+16 y-64=0$, we set $x=0$ and get $16 y-64=0$ or $y=4$.
The sum of the intercepts is $2+4=6$. | 8,570,926,631,750,459,000 |
Let $t$ be TNYWR.
Over the winter, Oscar counted the birds in his backyard. He counted three different types of birds: sparrows, finches and cardinals. Three-fifths of the birds that he counted were sparrows. One-quarter of the birds that he counted were finches. If Oscar counted exactly $10 t$ cardinals, how many sparrows did he count? | \(\boxed{240}\) | Suppose that Oscar saw $N$ birds in total.
From the given information, he saw $\frac{3}{5} N$ sparrows and $\frac{1}{4} N$ finches.
Therefore, he saw $N-\frac{3}{5} N-\frac{1}{4} N=N-\frac{12}{20} N-\frac{5}{20} N=\frac{3}{20} N$ cardinals.
Note that Oscar saw $\frac{3}{5} N$ sparrows and $\frac{3}{20} N$ cardinals. Since $\frac{3 / 5}{3 / 20}=4$, then he saw four times as many sparrows as cardinals.
Since he saw $10 t$ cardinals, then he saw $4 \times 10 t=40 t$ sparrows.
Since the answer to (a) is 6 , then $t=6$, and so Oscar saw $40 t=240$ sparrows. | 4,520,884,204,399,049,000 |
Ingrid starts with $n$ chocolates, while Jin and Brian each start with 0 chocolates. Ingrid gives one third of her chocolates to Jin. Jin gives 8 chocolates to Brian and then Jin eats half of her remaining chocolates. Jin now has 5 chocolates. What is the value of $n$ ? | \(\boxed{54}\) | Working backwards, since Jin had 5 chocolates left after eating half of them, she had $2 \times 5=10$ chocolates after giving chocolates to Brian.
Since Jin gave 8 chocolates to Brian and had 10 chocolates after doing this, Jin was given $10+8=18$ chocolates by Ingrid.
Ingrid gave one third of her chocolates to Jin, which means Ingrid started with $n=3 \times 18=54$ chocolates.
ANSWER: 54 | 4,901,243,508,915,007,000 |
For what value of $k$ is $k \%$ of 25 equal to $20 \%$ of 30 ? | \(\boxed{24}\) | Since $20 \%$ of 30 is $0.2 \times 30=6$, we want to find $k$ so that $k \%$ of 25 equals 6 .
This means $\frac{k}{100} \times 25=6$ or $\frac{k}{4}=6$ and so $k=24$.
ANSWER: 24 | 8,896,349,845,487,437,000 |
A group of eight students have lockers that are arranged as shown, in two rows of four lockers with one row directly on top of the other. The students are allowed to paint their lockers either blue or red according to two rules. The first rule is that there must be two blue lockers and two red lockers in each row. The second rule is that lockers in the same column must have different colours. How many ways are there for the students to paint their lockers according to the rules?
 | \(\boxed{6}\) | As soon as two lockers are painted blue in the top row, the other two lockers in the top row must be painted red.
Once the top row is painted, the colours of the lockers in the bottom row are determined.
If the lockers in the top row are numbered $1,2,3$, and 4 , then there are six possibilities for the two blue lockers.
They are 1 and 2, 1 and 3,1 and 4, 2 and 3,2 and 4, and 3 and 4.
ANSWER: 6 | 5,794,140,377,997,781,000 |
A fish and chips truck sells three kinds of fish: cod, haddock, and halibut. During a music festival, the truck sold 220 pieces of fish, of which $40 \%$ were haddock and $40 \%$ were halibut. How many pieces of cod did the truck sell? | \(\boxed{44}\) | The cod accounts for $100 \%-40 \%-40 \%=20 \%$ of the total pieces of fish sold.
Therefore, the number of pieces of cod sold was $0.2 \times 220=44$.
ANSWER: 44 | -9,066,581,155,734,483,000 |
If $\frac{x}{2}-5=9$, what is the value of $\sqrt{7 x}$ ? | \(\boxed{14}\) | Rearranging the given equation, we get $\frac{x}{2}=14$ which implies that $x=28=2^{2} \times 7$.
Therefore, $\sqrt{7 x}=\sqrt{7 \times 2^{2} \times 7}=\sqrt{2^{2} \times 7^{2}}=2 \times 7=14$.
ANSWER: 14 | -3,747,760,393,875,721,700 |
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