problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
The line with equation $y=5 x+a$ passes through the point $\left(a, a^{2}\right)$. If $a \neq 0$, what is the value of $a$ ? | Since $\left(a, a^{2}\right)$ lies on the line with equation $y=5 x+a$, then $a^{2}=5 a+a$ or $a^{2}=6 a$.
Since $a \neq 0$, then $a=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What is the value of $\frac{24+12}{4^{2}-4}$ ? | Solution 1
The numerator is equal to $24+12=36$. The denominator is equal to $4^{2}-4=16-4=12$, so
$$
\frac{24+12}{4^{2}-4}=\frac{36}{12}=3
$$
## Solution 2
By factoring 4 out of both the numerator and denominator, we get
$$
\frac{24+12}{4^{2}-4}=\frac{4(6+3)}{4(4-1)}=\frac{6+3}{4-1}=\frac{9}{3}=3 .
$$
# | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If $3 k=10$, what is the value of $\frac{6}{5} k-2$ ? | Solution 1
By factoring and substituting, we get
$$
\frac{6}{5} k-2=\frac{2}{5}(3 k)-2=\frac{2}{5}(10)-2=4-2=2 .
$$
## Solution 2
Solving for $k$, we have $k=\frac{10}{3}$, so
$$
\frac{6}{5} k-2=\frac{6}{5} \times \frac{10}{3}-2=\frac{60}{15}-2=4-2=2 .
$$
ANSWER: 2
# | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Segment $A B$ is reflected in the $y$-axis to obtain $E D$. Segment $B C$ is reflected in the $y$-axis to obtain $D C$. Determine the sum of the slopes of $D C$ and $E D$.
 | Solution 1
The slope of segment $A B$ is $\frac{0-6}{-2-(-4)}=-3$.
Since $E D$ is the reflection of $A B$ in the $y$-axis, its slope is the negative of the slope of $A B$. This means the slope of $E D$ is 3 .
Similarly, the slope of segment $B C$ is $\frac{4-0}{0-(-2)}=2$, so the slope of $D C$ is -2 .
Therefore, t... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Maggie graphs the six possible lines of the form $y=m x+b$ where $m$ is either 1 or -2 , and $b$ is either 0,1 or 2 . For example, one of the lines is $y=x+2$. The lines are all graphed on the same axes. There are exactly $n$ distinct points, each of which lies on two or more of these lines. What is the value of $n$ ? | If two lines are parallel, then they will not intersect unless they are the same line. This means two lines that have the same slope but different $y$-intercepts cannot intersect.
On the other hand, two lines that have different slopes always intersect at exactly one point.
The six lines that Maggie graphs can be bro... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Twenty-seven unit cubes are each coloured completely black or completely red. The unit cubes are assembled into a larger cube. If $\frac{1}{3}$ of the surface area of the larger cube is red, what is the smallest number of unit cubes that could have been coloured red? | Since $\sqrt[3]{27}=3$, the dimensions of the larger cube must be $3 \times 3 \times 3$.
Therefore, each side of the larger cube has area $3 \times 3=9$.
A cube has 6 faces, so the total surface of the cube is made up of $9 \times 6=54$ of the $1 \times 1$ squares from the faces of the unit cubes.
Since $\frac{1}{3}... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
What is the integer that is greater than $\sqrt{11}$ but less than $\sqrt{19}$ ? | Since $3^{2}=9<11<16=4^{2}$, it must be that $3<\sqrt{11}<4$.
Similarly, $4^{2}=16<19<25=5^{2}$, so $4<\sqrt{19}<5$.
Therefore, $\sqrt{11}<4<\sqrt{19}$.
ANSWER: 4 | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the value of $\frac{3^{5}-3^{4}}{3^{3}}$ ? | By factoring $3^{3}$ out of the numerator, we have
$$
\frac{3^{5}-3^{4}}{3^{3}}=\frac{3^{3}\left(3^{2}-3\right)}{3^{3}}=3^{2}-3=9-3=6 .
$$
ANSWER: 6 | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
On Fridays, the price of a ticket to a museum is $\$ 9$. On one particular Saturday, there were 200 visitors to the museum, which was twice as many visitors as there were the day before. The total money collected from ticket sales on that particular Saturday was $\frac{4}{3}$ as much as the day before. The price of tic... | There were 200 visitors on Saturday, so there were 100 visitors the day before. Since tickets cost $\$ 9$ on Fridays, the total money collected on Friday was $\$ 900$.
Therefore, the amount of money collected from ticket sales on the Saturday was $\$ 900 \times \frac{4}{3}=\$ 1200$.
Since there were 200 visitors on S... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
How many times does the digit 0 appear in the integer equal to $20^{10}$ ? | By factoring and using exponent rules, we have $20^{10}=(2 \times 10)^{10}=2^{10} \times 10^{10}$.
Therefore, $20^{10}=1024 \times 10^{10}$, which is the integer 1024 followed by ten zeros.
Thus, $20^{10}$ has eleven digits that are 0 . That is, 10 zeros at the end and one coming from the 1024 at the beginning.
ANSW... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Suppose that $f(x)=2 \sin ^{2}(\log x)+\cos \left(\log x^{2}\right)-5$ for each $x>0$. What is the value of $f(\pi)$ ? | Using a logarithm rule, we have that $\log x^{2}=2 \log x$, so
$$
f(x)=2 \sin ^{2}(\log x)+\cos (2 \log x)-5 .
$$
Using the trigonometric identity $\cos 2 x=\cos ^{2} x-\sin ^{2} x$, we get
$$
f(x)=2 \sin ^{2}(\log x)+\cos ^{2}(\log x)-\sin ^{2}(\log x)-5
$$
which simplifies to
$$
f(x)=\sin ^{2}(\log x)+\cos ^{2}(... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine the sum of the real numbers $x$ for which $\frac{2 x}{x^{2}+5 x+3}+\frac{3 x}{x^{2}+x+3}=1$. | By clearing denominators, the equation becomes
$$
2 x\left(x^{2}+x+3\right)+3 x\left(x^{2}+5 x+3\right)=\left(x^{2}+5 x+3\right)\left(x^{2}+x+3\right)
$$
and after expanding the left side, we get
$$
2 x^{3}+2 x^{2}+6 x+3 x^{3}+15 x^{2}+9 x=\left(x^{2}+5 x+3\right)\left(x^{2}+x+3\right) \text {. }
$$
There are many ... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Evaluate $\frac{2+5 \times 5}{3}$.
## | Evaluating, $\frac{2+5 \times 5}{3}=\frac{2+25}{3}=\frac{27}{3}=9$. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
An equilateral triangle has sides of length $x+5, y+11$, and 14 . What is the value of $x+y$ ?
## | In an equilateral triangle, all sides are of equal length. This means $x+5=14$ and $y+11=14$.
Solving these equations for $x$ and $y$, respectively, we get $x=14-5=9$ and $y=14-11=3$. Therefore, $x+y=9+3=12$. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $t$ be TNYWR.
Gray has $t$ dollars consisting of $\$ 1$ and $\$ 2$ coins. If she has the same number of $\$ 1$ and $\$ 2$ coins, how many $\$ 1$ coins does she have?
## | Let $k$ be the number of $\$ 1$ coins that Gray has. It is given that $k$ is also the number of $\$ 2$ coins, which means the total amount of money in dollars that Gray has is
$$
k(1)+k(2)=k(1+2)=k(3) .
$$
The answer to (a) is 12 . Therefore, $\$ 3 k=\$ t$ so $k=\frac{t}{3}=\frac{12}{3}=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The numbers $x+5,14, x$, and 5 have an average of 9 . What is the value of $x$ ?
## | There are four numbers in total, so their average is
$$
\frac{(x+5)+14+x+5}{4}=\frac{2 x+24}{4}=\frac{x+12}{2} .
$$
The average is 9 , so $\frac{x+12}{2}=9$ or $x+12=18$ which means $x=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $t$ be TNYWR.
Each of the three lines having equations $x+t y+8=0,5 x-t y+4=0$, and $3 x-k y+1=0$ passes through the same point. What is the value of $k$ ?
## | Adding the equations $x+t y+8=0$ and $5 x-t y+4=0$, we get $x+t y+8+5 x-t y+4=0+0$ or $6 x+12=0$.
Solving for $x$ gives $x=-2$. Note that we have found the value of $x$ without knowing the value of $t$.
Substituting into the first equation gives $-2+t y+8=0$ or $t y=-6$, which means $y=-\frac{6}{t}$. Substituting the... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In a group of 20 friends, 11 like to ski, 13 like to snowboard, and 3 do not like to do either. How many of the friends like to both ski and snowboard? | Let $x$ be the number of friends who like to both ski and snowboard.
Then $11-x$ of the friends like to ski but not do not like to snowboard, and $13-x$ of the friends like to snowboard but do not like to ski.
Since 3 of the 20 friends do not like to ski or snowboard, then 17 like to either ski or snowboard or both.
... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
The pair $(x, y)=(2,5)$ is the solution of the system of equations
$$
\begin{aligned}
a x+2 y & =16 \\
3 x-y & =c
\end{aligned}
$$
Determine the value of $\frac{a}{c}$. | Since $(x, y)=(2,5)$ is the solution of the system of equations, then $(x, y)=(2,5)$ satisfies both equations.
Since $(x, y)=(2,5)$ satisfies $a x+2 y=16$, then $2 a+10=16$ or $2 a=6$ and so $a=3$.
Since $(x, y)=(2,5)$ satisfies $3 x-y=c$, then $6-5=c$ or $c=1$.
Therefore, $\frac{a}{c}=3$.
ANSWER: 3 | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Each entry in the list below is a positive integer:
$$
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$ ? | 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 | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The sum of the lengths of all of the edges of rectangular prism $A B C D E F G H$ is 24 . If the total surface area of the prism is 11, determine the length of the diagonal $A H$.
 | Suppose that the prism has $A B=x, A D=y$, and $A F=z$.
Since the sum of all of the edge lengths is 24 , then $4 x+4 y+4 z=24$ or $x+y+z=6$.
(The prism has 4 edges of each length.)
Since the surface area is 11 , then $2 x y+2 x z+2 y z=11$.
(The prism has 2 faces that are $x$ by $y, 2$ faces that are $x$ by $z$, an... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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$. | 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 od... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
For how many one-digit positive integers $k$ is the product $k \cdot 234$ divisible by 12 ? | 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, the... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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$ ? | 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}
$$
Therefore, $n=3$.
ANSWER: 3 | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine the number of pairs $(x, y)$ of positive integers for which $0<x<y$ and $2 x+3 y=80$. | Since $02 x+3 x=5 x$.
Since $2 x+3 y=80$ and $2 x+3 y>5 x$, then $80>5 x$ or $16>x$.
Since $2 x+3 y=80$, then $2 x=80-3 y$. Since $2 x$ is even, then $80-3 y$ is even. Since 80 is even, then $3 y$ is even, and so $y$ must be even.
Set $y=2 Y$ for some positive integer $Y$.
Then $2 x+3 y=80$ becomes $2 x+6 Y=80$ or ... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If $a=2^{3}$ and $b=3^{2}$ evaluate $\frac{(a-b)^{2015}+1^{2015}}{(a-b)^{2015}-1^{2015}}$. | Since $a=2^{3}$ and $b=3^{2}$, then $a-b=8-9=-1$.
Therefore,
$$
\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 | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For each positive integer $n$, define the point $P_{n}$ to have coordinates $\left((n-1)^{2}, n(n-1)\right)$ and the point $Q_{n}$ to have coordinates $\left((n-1)^{2}, 0\right)$. For how many integers $n$ with $2 \leq n \leq 99$ is the area of trapezoid $Q_{n} P_{n} P_{n+1} Q_{n+1}$ a perfect square? | We first note that the coordinates of $P_{n+1}$ are $\left(((n+1)-1)^{2},(n+1)((n+1)-1)\right)=\left(n^{2}, n(n+1)\right)$ and that the coordinates of $Q_{n+1}$ are $\left(((n+1)-1)^{2}, 0\right)=\left(n^{2}, 0\right)$.
Since $P_{n}$ and $Q_{n}$ have the same $x$-coordinate, then $P_{n} Q_{n}$ is vertical.
Since $P_{... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Suppose that $n$ is a positive integer and that the set $S$ contains exactly $n$ distinct positive integers. If the mean of the elements of $S$ is equal to $\frac{2}{5}$ of the largest element of $S$ and is also equal to $\frac{7}{4}$ of the smallest element of $S$, determine the minimum possible value of $n$. | Suppose that the largest integer in $S$ is $L$, the smallest integer in $S$ is $P$, and the mean of the elements of $S$ is $m$.
We are told that $m=\frac{2}{5} L$ and $m=\frac{7}{4} P$.
Thus, $\frac{7}{4} P=\frac{2}{5} L$ or $P=\frac{8}{35} L$.
Since $P$ and $L$ are positive integers, then $L$ must be divisible by 3... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Evaluate $2+0+1+5$. | Evaluating, $2+0+1+5=8$. | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
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$ ? | 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 $... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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$ ? | 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$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If $2^{k+4}=1024$, what is the value of $k$ ? | Since $1024=2^{10}$ and $2^{k+4}=1024$, then $k+4=10$ or $k=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $t$ be TNYWR.
If $2 t+2 x-t-3 x+4 x+2 t=30$, what is the value of $x$ ? | If $2 t+2 x-t-3 x+4 x+2 t=30$, then $3 t+3 x=30$ or $t+x=10$ and so $x=10-t$. Since the answer to (a) is 6 , then $t=6$ and so $x=10-t=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $t$ be TNYWR.
In the diagram, $\angle B A E=\angle C B E=\angle D C E=90^{\circ}$. If $A E=\sqrt{5}, A B=\sqrt{4}, B C=\sqrt{3}$, and $C D=\sqrt{t}$, what is the length of $D E$ ?
 | We use the Pythagorean Theorem repeatedly and obtain
$$
D E^{2}=C D^{2}+C E^{2}=C D^{2}+\left(B C^{2}+B E^{2}\right)=C D^{2}+B C^{2}+\left(A B^{2}+A E^{2}\right)
$$
Since $A E=\sqrt{5}, A B=\sqrt{4}, B C=\sqrt{3}$, and $C D=\sqrt{t}$, then
$$
D E^{2}=t+3+4+5=t+12
$$
Since the answer to (b) is 4 , then $t=4$, and so... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Let $t$ be TNYWR.
Determine the number of integers $b>0$ for which $30 t$ is divisible by $b!$.
(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$.) | We do some preliminary work assuming that the number $t$ to be received is a positive integer. (This turns out to be the case; if it were not the case, we might have to change strategies.)
Since $1!=1$, then $30 t$ is divisible by 1 ! regardless of the value of $t$.
Since $2!=2$, then $30 t=2(15 t)$ is divisible by 2... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the diagram, points $A, B, C$, and $D$ are on a circle. Philippe uses a ruler to connect each pair of these points with a line segment. How many line segments does he draw?
 | Philippe connects $A B, A C, A D, B C, B D$, and $C D$.
He draws 6 line segments.
ANSWER: 6 | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Suppose that $a, b, c$, and $d$ are positive integers with $0<a<b<c<d<10$. What is the maximum possible value of $\frac{a-b}{c-d}$ ? | First, we note that $\frac{a-b}{c-d}=\frac{b-a}{d-c}$ and each of $b-a$ and $d-c$ is positive.
For $\frac{b-a}{d-c}$ to be as large as possible, we want $b-a$ to be as large as possible and $d-c$ to be as small as possible.
Since $d$ and $c$ are integers with $d>c$, then the smallest possible value of $d-c$ is 1 .
F... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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$ ? | 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 t... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Suppose that $\left(2^{3}\right)^{x}=4096$ and that $y=x^{3}$. What is the ones (units) digit of the integer equal to $3^{y}$ ? | We note that $\left(2^{3}\right)^{x}=2^{3 x}$ and that $4096=4 \cdot 41024=4 \cdot 4 \cdot 256=4 \cdot 4 \cdot 16 \cdot 16=2^{2} \cdot 2^{2} \cdot 2^{4} \cdot 2^{4}=2^{12}$.
Thus, $2^{3 x}=2^{12}$ which gives $3 x=12$ and so $x=4$.
Since $y=x^{3}$, then $y=4^{3}=64$.
We need to consider the integer $3^{64}$.
The fi... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For what value of $x$ is $4 x-8+3 x=12+5 x$ ? | Simplifying, we obtain $7 x-8=12+5 x$ and so $2 x=20$ or $x=10$.
ANSWER: 10 | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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}
$$
What is the value of $M+N$ ? | Since $M=1^{5}+\left(2^{4} \times 3^{3}\right)-\left(4^{2} \div 5^{1}\right)$ and $N=1^{5}-\left(2^{4} \times 3^{3}\right)+\left(4^{2} \div 5^{1}\right)$, then when $M$ and $N$ are added the terms $\left(2^{4} \times 3^{3}\right)$ and $\left(4^{2} \div 5^{1}\right)$ "cancel" out.
Thus, $M+N=1^{5}+1^{5}=2$.
ANSWER: 2 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
How many four-digit palindromes $a b b a$ have the property that the two-digit integer $a b$ and the two-digit integer $b a$ are both prime numbers? (For example, 2332 does not have this property, since 23 is prime but 32 is not.) | If $a b$ and $b a$ are both prime numbers, then neither is even which means that neither digit $a$ or $b$ is even and neither equals 5 , otherwise $a b$ or $b a$ would be even or divisible by 5 and so not prime.
Therefore, each of $a$ and $b$ equals $1,3,7$, or 9 .
The two-digit primes using these digits are 11,13,17... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the value of $123456^{2}-123455 \times 123457$ ? | Let $x=123456$.
Thus, $x-1=123455$ and $x+1=123457$.
Therefore,
$$
123456^{2}-123455 \times 123457=x^{2}-(x-1)(x+1)=x^{2}-\left(x^{2}-1\right)=1
$$
ANSWER: 1 | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine the value of $\left(\log _{2} 4\right)\left(\log _{4} 6\right)\left(\log _{6} 8\right)$. | Using the change of base formulas for logarithms,
$$
\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
$$
ANSWER: 3 | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $t$ be TNYWR.
The sum of the even integers from 2 to $2 k$ inclusive equals $t$ for some positive integer $k$. That is,
$$
2+4+6+\cdots+(2 k-2)+2 k=t
$$
What is the value of $k$ ? | Manipulating the left side,
$$
\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}
$$
Since the answer to (a) is 132 , then $t=132$.
Since $k(k+1)=132$ and $k$ is positive, then $k=11$. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
How many perfect squares are there between 2 and 150 ? | Since $1^{2}=1,2^{2}=4,12^{2}=144$, and $13^{2}=169$, then the perfect squares between 2 and 150 are $2^{2}$ through $12^{2}$, of which there are 11 . | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Let $t$ be TNYWR.
The line with equation $y=-2 x+t$ and the parabola with equation $y=(x-1)^{2}+1$ intersect at point $P$ in the first quadrant. What is the $y$-coordinate of $P$ ? | To find the points of intersection of the line with equation $y=-2 x+t$ and the parabola with equation $y=(x-1)^{2}+1$, we equate values of $y$, to obtain
$$
\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{... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The equations $x-2 y-3=0$ and $18 x-k^{2} y-9 k=0$ represent two lines. For some real number $k$, these two lines are distinct and parallel. Determine the value of $k$. | We can re-write the equation of the first line in slope-intercept form as $y=\frac{1}{2} x-\frac{3}{2}$.
If $k=0$, then the second line has equation $18 x=0$ or $x=0$ which is vertical and cannot be parallel to the first line which has slope $\frac{1}{2}$.
This means $k \neq 0$, so the equation of the second line can... | -6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine the number of ordered pairs of positive integers $(a, b)$ for which $20 a+21 b=2021$. | Since $a$ is a positive integer, the units digit of $20 a$ is 0 .
In order for $20 a+21 b=2021$, the units digit of $21 b$ must be 1 , and this implies that the units digit of $b$ must be 1 .
If $b \geq 100$, then $21 b \geq 2100>2021$, which would mean $20 a$ is negative.
Since $a$ must be positive, we must have $b... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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}$ ?
 | 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 equ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
$A C$ and $B D$ are two perpendicular chords in a circle. The chords intersect at $E$, as shown, such that $B E=3, E C=2$, and $A E=6$. The exact perimeter of the quadrilateral $A B C D$ may be written in the form $m \sqrt{n}+p \sqrt{q}$, where $m, n, p$, and $q$ are positive integers, $q>n$, and neither $q$ nor $n$ ha... | If $A B$ and $D C$ are drawn, then $\angle A B D=\angle A C D$ since they are subtended by the same arc.
As well, we are given that $\angle A E B=\angle D E C=90^{\circ}$, so $\triangle A E... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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... | 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
$$
\begin{aligned}
\frac{1}{\log _{2} 100 !}+\f... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If $(y-5)^{2}=(y-9)^{2}$, what is the value of $y$ ?
## | Expanding both sides, we have $y^{2}-10 y+25=y^{2}-18 y+81$.
Rearranging, we have $8 y=56$, so $y=7$. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
A line passing through $(-5, k)$ and $(13,-7)$ has a slope of $-\frac{1}{2}$. What is the value of $k$ ?
## | The slope of the line is $\frac{-7-k}{13-(-5)}=-\frac{7+k}{18}$.
Therefore, $-\frac{7+k}{18}=-\frac{1}{2}=-\frac{9}{18}$, so $7+k=9$ from which it follows that $k=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Beginning at 100, David counts down by 11 at a time. What is the first number less than 0 that David will count? | David counts 100, 89, 78, 67, 56, 45, 34, 23, 12, 1, - 10 .
The first number less than 0 that he counts is -10 .
ANSWER: -10 | -10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
In the diagram, $\triangle A B C$ is right-angled at $B$ with $A B=24$ and $B C=10$. If $A B$ and $B C$ are each increased by 6 , by how much does $A C$ increase?
 | Using the Pythagorean Theorem in the given triangle, we obtain
$$
A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{24^{2}+10^{2}}=\sqrt{676}=26
$$
since $A C>0$.
If $A B$ and $B C$ are each increased by 6 , they become 30 and 16 , so the hypotenuse of the new triangle is $\sqrt{30^{2}+16^{2}}=\sqrt{1156}=34$.
Therefore, $A C$ incr... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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$. | 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.)
Therefo... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Line segment $P Q$ has endpoints at $P(6,-2)$ and $Q(-3,10)$. Point $R(a, b)$ is a point on $P Q$ such that the distance from $P$ to $R$ is one-third the distance from $P$ to $Q$. What is $b-a$ ? | To get from $P(6,-2)$ to $Q(-3,10)$, we go $6-(-3)=9$ units to the left and $10-(-2)=12$ units up.
One-third of this difference would be 3 units to the left and 4 units up, which means that the coordinates of $R$ are $(a, b)=(6-3,-2+4)=(3,2)$.
Finally, $b-a=2-3=-1$.
ANSWER: -1 | -1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
In how many ways can 105 be expressed as the sum of at least two consecutive positive integers? | In any sequence of consecutive integers, the average, $a$, is either an integer (if there is an odd number of integers in the sequence) or halfway between two integers (if there is an even number of integers in the sequence).
Furthermore, the sum of this sequence is the product of the average $a$ with the number of in... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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... | 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 ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine the number of triples $(a, b, c)$ of three positive integers with $a<b<c$ whose sum is 100 and whose product is 18018 . | We note that $18018=18 \times 1001=2 \times 3^{2} \times 11 \times 91=2 \times 3^{2} \times 7 \times 11 \times 13$.
We want to find all triples $(a, b, c)$ of positive integers with $a+b+c=100$ and $a b c=18018$.
First, we note that none of $a, b, c$ can be a multiple of more than one of 7,11 or 13:
If one was a mul... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
The sum of 8 one-digit positive integers is 68 . If seven of these integers are equal, determine the other integer. | 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.... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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$ ? | 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 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Evaluate $2 \times 0+1 \times 4$. | Evaluating, $2 \times 0+1 \times 4=0+4=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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$ ? | 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$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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$ ? | 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 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Evaluate $\frac{1}{2}\left(\frac{1}{\frac{1}{9}}+\frac{1}{\frac{1}{6}}-\frac{1}{\frac{1}{5}}\right)$. | Evaluating, $\frac{1}{2}\left(\frac{1}{\frac{1}{9}}+\frac{1}{\frac{1}{6}}-\frac{1}{\frac{1}{5}}\right)=\frac{1}{2}(9+6-5)=\frac{1}{2}(10)=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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? | 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 answe... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What is the greatest common divisor of the three integers 36,45 and 495 ? | 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. | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the value of $x$ so that $\frac{8}{x}+6=8$ ? | If $\frac{8}{x}+6=8$, then $\frac{8}{x}=2$ and so $x=4$.
ANSWER: 4 | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 ... | 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... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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? | 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... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Evaluate $\frac{9+2 \times 3}{3}$. | Evaluating, $\frac{9+2 \times 3}{3}=\frac{9+6}{3}=\frac{15}{3}=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If $w$ is a positive integer with $w^{2}-5 w=0$, what is the value of $w$ ? | Since $w^{2}-5 w=0$, then $w(w-5)=0$ and so $w=0$ or $w=5$.
Since $w$ is positive, then $w=5$. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Suppose that $x=\sqrt{20-17-2 \times 0-1+7}$. What is the value of $x$ ? | Evaluating, $x=\sqrt{20-17-2 \times 0-1+7}=\sqrt{20-17-0-1+7}=\sqrt{9}=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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$ ? | 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... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Line segment $A D$ is divided into three segments by points $B$ and $C$, so that $A B: B C=1: 2$ and $B C: C D=6: 5$. The length of $A D$ is 56 units. What is the length of $A B$ ?
 | Let $x$ the real number satisfying $B C=6 x$.
From $A B: B C=1: 2$ we get that $A B=3 x$.
From $B C: C D=6: 5$ we get that $C D=5 x$.
This means $A D=A B+B C+C D=3 x+6 x+5 x=14 x$.
We are given that $A D=56$, so $56=14 x$ or $x=4$.
Therefore, $A B=3 x=3(4)=12$.
ANSWER: 12 | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The integers $a$ and $b$ have the property that the expression
$$
\frac{2 n^{3}+3 n^{2}+a n+b}{n^{2}+1}
$$
is an integer for every integer $n$. What is the value of the expression above when $n=4$ ? | By rearranging, we get
$$
\begin{aligned}
\frac{2 n^{3}+3 n^{2}+a n+b}{n^{2}+1} & =\frac{2 n^{3}+2 n+3 n^{2}+3+(a-2) n+(b-3)}{n^{2}+1} \\
& =\frac{2 n\left(n^{2}+1\right)+3\left(n^{2}+1\right)+(a-2) n+(b-3)}{n^{2}+1} \\
& =2 n+3+\frac{(a-2) n+(b-3)}{n^{2}+1}
\end{aligned}
$$
The quantity $2 n+3$ is an integer if $n$ ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
In the diagram, there are exactly nine $1 \times 1$ squares. What is the largest number of $1 \times 1$ squares that can be shaded so that no two shaded squares share a side?
 | In the diagram below, there are 6 squares shaded in such a way that no two shaded squares share an edge, which shows that the answer is at least 6 .
We will now argue that it is impossible ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Consider the lines with equations $y=m x$ and $y=-2 x+28$ where $m$ is some real number. The area enclosed by the two lines and the $x$-axis in the first quadrant is equal to 98 . What is the value of $m$ ? | The area of interest is a triangle with vertices at the origin, the $x$-intercept of $y=-2 x+28$, and the point of intersection of the lines with equations $y=m x$ and $y=-2 x+28$.
Setting ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
For some integers $n$, the expression $\frac{8(n-1)}{(n-1)(n-2)}$ is equal to an integer $M$. What is the sum of all possible values of $M$ ? | The expression $\frac{8(n-1)}{(n-1)(n-2)}$ is undefined when $n=1$. For every integer $n \neq 1$, the expression is equal to $\frac{8}{n-2}$.
Thus, if $n$ has the property that $\frac{8(n-1)}{(n-1)(n-2)}$ is also an integer, then $n$ must be an integer different from 1 with the property that $n-2$ is a factor of 8 .
... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
All the points on the line with equation $y=3 x+6$ are translated up 3 units, then translated left 4 units, and then reflected in the line with equation $y=x$. Determine the $y$-intercept of the resulting line. | Solution 1
The original line contains the points $(0,6)$ and $(1,9)$.
Translating these points 3 units up gives the points $(0,9)$ and $(1,12)$.
Translating these points 4 units to the left gives $(-4,9)$ and $(-3,12)$.
Reflecting these points in the line $y=x$ gives $(9,-4)$ and $(12,-3)$.
The slope of the line t... | -7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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$ ? | 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}... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What is the largest integer that can be placed in the box so that $\frac{\square}{11}<\frac{2}{3}$ ?
## | 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}... | 7 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Let $t$ be TNYWR.
If $6 x+t=4 x-9$, what is the value of $x+4$ ?
## | 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 $x+4=\frac{-7-1}{2}=-4$. | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $x$ be the number of prime numbers between 10 and 30 .
What is the number equal to $\frac{x^{2}-4}{x+2}$ ?
## | 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 $6-2=4$. | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 | 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: $(4,15,3)$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $t$ be TNYWR.
If $\frac{5+x}{t+x}=\frac{2}{3}$, what is the value of $x$ ?
## | 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 $x=2(13)-15=11$. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What is the value of $2^{0}+20^{0}+201^{0}+2016^{0}$ ? | Evaluating, $2^{0}+20^{0}+201^{0}+2016^{0}=1+1+1+1=4$.
ANSWER: 4 | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
What is the value of $2+0-1 \times 6$ ? | Evaluating, $2+0-1 \times 6=2+0-6=-4$.
ANSwER: -4 | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 $a+b+c$. | 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 | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Determine the value of the following sum:
$$
\log _{3}\left(1-\frac{1}{15}\right)+\log _{3}\left(1-\frac{1}{14}\right)+\log _{3}\left(1-\frac{1}{13}\right)+\cdots+\log _{3}\left(1-\frac{1}{8}\right)+\log _{3}\left(1-\frac{1}{7}\right)+\log _{3}\left(1-\frac{1}{6}\right)
$$
(Note that the sum includes a total of 10 te... | Using logarithm rules,
$$
\begin{aligned}
\log _{3}(1- & \left.\frac{1}{15}\right)+\log _{3}\left(1-\frac{1}{14}\right)+\log _{3}\left(1-\frac{1}{13}\right)+\cdots+\log _{3}\left(1-\frac{1}{8}\right)+\log _{3}\left(1-\frac{1}{7}\right)+\log _{3}\left(1-\frac{1}{6}\right) \\
& =\log _{3}\left(\frac{14}{15}\right)+\log ... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Evaluate $10-2 \times 3$. | Evaluating, $10-2 \times 3=10-6=4$. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
If $x: 6=15: 10$, what is the value of $x$ ? | Since $x: 6=15: 10$, then $\frac{x}{6}=\frac{15}{10}$ which gives $x=\frac{6 \cdot 15}{10}=9$. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $t$ be TNYWR.
If $\frac{3(x+5)}{4}=t+\frac{3-3 x}{2}$, what is the value of $x$ ? | 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 $x=\frac{4}{9} t-1=4-1=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 $m$ ? | 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 $t$, ... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 $20 x+16 y-64=0$ ? | 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 $2+4=6$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $a$ be the largest positive integer so that $a^{3}$ is less than 999.
Let $b$ be the smallest positive integer so that $b^{5}$ is greater than 99 .
What is the value of $a-b$ ? | Since $9^{3}=729$ and $10^{3}=1000$, then the largest positive integer $a$ with $a^{3}99$ is $b=3$.
Therefore, $a-b=6$. | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
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 ... | 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 lock... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
In $\triangle A B C, \tan \angle B C A=1$ and $\tan \angle B A C=\frac{1}{7}$. The perimeter of $\triangle A B C$ is $24+18 \sqrt{2}$. The altitude from $B$ to $A C$ has length $h$ and intersects $A C$ at $D$. What is the value of $h$ ?
![](https://cdn.mathpix.com/cropped/2024_04_17_587bc2a8915214d42f9eg-2.jpg?height=... | We are given that $\tan \angle B A C=\frac{1}{7}$, but $\angle B A C=\angle B A D$ and $\triangle B A D$ has a right angle at $D$, so $\tan \angle B A D=\frac{B D}{A D}=\frac{1}{7}$. Since $B D=h, A D=7 \cdot B D=7 h$.
Using that $\tan \angle B C D=\tan \angle B C A=1$, we get $C D=h$ by similar reasoning. Therefore, ... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.