id int64 | question string | solution list | final_answer list | context string | image_1 image | image_2 image | image_3 image | image_4 image | image_5 image | modality string | difficulty string | is_multiple_answer bool | unit string | answer_type string | error string | question_type string | subfield string | subject string | language string |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
2,464 | A multiplicative partition of a positive integer $n \geq 2$ is a way of writing $n$ as a product of one or more integers, each greater than 1. Note that we consider a positive integer to be a multiplicative partition of itself. Also, the order of the factors in a partition does not matter; for example, $2 \times 3 \times 5$ and $2 \times 5 \times 3$ are considered to be the same partition of 30 . For each positive integer $n \geq 2$, define $P(n)$ to be the number of multiplicative partitions of $n$. We also define $P(1)=1$. Note that $P(40)=7$, since the multiplicative partitions of 40 are $40,2 \times 20,4 \times 10$, $5 \times 8,2 \times 2 \times 10,2 \times 4 \times 5$, and $2 \times 2 \times 2 \times 5$.
(In each part, we use "partition" to mean "multiplicative partition". We also call the numbers being multiplied together in a given partition the "parts" of the partition.)
Determine the value of $P(1000)$. | [
"First, we note that $1000=10^{3}=(2 \\cdot 5)^{3}=2^{3} 5^{3}$.\n\nWe calculate the value of $P\\left(p^{3} q^{3}\\right)$ for two distinct prime numbers $p$ and $q$. It will turn out that this value does not depend on $p$ and $q$. This value will be the value of $P(1000)$, since 1000 has this form of prime factor... | [
"31"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,466 | What are all values of $x$ such that
$$
\log _{5}(x+3)+\log _{5}(x-1)=1 ?
$$ | [
"Combining the logarithms,\n\n$$\n\\begin{aligned}\n\\log _{5}(x+3)+\\log _{5}(x-1) & =1 \\\\\n\\log _{5}((x+3)(x-1)) & =1 \\\\\n\\log _{5}\\left(x^{2}+2 x-3\\right) & =1 \\\\\nx^{2}+2 x-3 & =5 \\\\\nx^{2}+2 x-8 & =0 \\\\\n(x+4)(x-2) & =0\n\\end{aligned}\n$$\n\nTherefore, $x=-4$ or $x=2$. Substituting the two value... | [
"2"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,467 | A chef aboard a luxury liner wants to cook a goose. The time $t$ in hours to cook a goose at $180^{\circ} \mathrm{C}$ depends on the mass of the goose $m$ in kilograms according to the formula
$$
t=a m^{b}
$$
where $a$ and $b$ are constants. The table below gives the times observed to cook a goose at $180^{\circ} \mathrm{C}$.
| Mass, $m(\mathrm{~kg})$ | Time, $t(\mathrm{~h})$ |
| :---: | :---: |
| 3.00 | 2.75 |
| 6.00 | 3.75 |
Using the data in the table, determine both $a$ and $b$ to two decimal places. | [
"From the table we have two pieces of information, so we substitute both of these into the given formula.\n\n$$\n\\begin{aligned}\n& 2.75=a(3.00)^{b} \\\\\n& 3.75=a(6.00)^{b}\n\\end{aligned}\n$$\n\nWe can now proceed in either of two ways to solve for $b$.\n\nMethod 1 to find $b$\n\nDividing the second equation by ... | [
"$1.68,0.45$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,469 | A circle passes through the origin and the points of intersection of the parabolas $y=x^{2}-3$ and $y=-x^{2}-2 x+9$. Determine the coordinates of the centre of this circle. | [
"We first determine the three points through which the circle passes.\n\nThe first point is the origin $(0,0)$.\n\nThe second and third points are found by determining the points of intersection of the two parabolas $y=x^{2}-3$ and $y=-x^{2}-2 x+9$. We do this by setting the $y$ values equal.\n\n$$\nx^{2}-3=-x^{2}-... | [
"$(-\\frac{1}{2}, \\frac{7}{2})$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Geometry | Math | English |
2,472 | In a soccer league with 5 teams, each team plays 20 games(that is, 5 games with each of the other 4 teams). For each team, every game ends in a win (W), a loss (L), or a tie (T). The numbers of wins, losses and ties for each team at the end of the season are shown in the table. Determine the values of $x, y$ and $z$.
| Team | W | L | T |
| :---: | ---: | ---: | ---: |
| A | 2 | 15 | 3 |
| B | 7 | 9 | 4 |
| C | 6 | 12 | 2 |
| D | 10 | 8 | 2 |
| E | $x$ | $y$ | $z$ | | [
"In total, there are $\\frac{1}{2} \\times 5 \\times 20=50$ games played, since each of 5 teams plays 20 games (we divide by 2 since each game is double-counted).\n\nIn each game, there is either a loss or a tie.\n\nThe number of games with a loss is $44+y$ from the second column, and the number of games with a tie... | [
"19,0,1"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,475 | Three thin metal rods of lengths 9,12 and 15 are welded together to form a right-angled triangle, which is held in a horizontal position. A solid sphere of radius 5 rests in the triangle so that it is tangent to each of the three sides. Assuming that the thickness of the rods can be neglected, how high above the plane of the triangle is the top of the sphere? | [
"Consider the cross-section of the sphere in the plane defined by the triangle. This crosssection will be a circle, since any cross-section of a sphere is a circle. This circle will be tangent to the three sides of the triangle, ie. will be the inscribed circle (or incircle) of the triangle. Let the centre of this ... | [
"5"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,479 | Triangle $A B C$ has vertices $A(0,5), B(3,0)$ and $C(8,3)$. Determine the measure of $\angle A C B$. | [
"First, we calculate the side lengths of $\\triangle A B C$ :\n\n$$\n\\begin{aligned}\n& A B=\\sqrt{(0-3)^{2}+(5-0)^{2}}=\\sqrt{34} \\\\\n& B C=\\sqrt{(3-8)^{2}+(0-3)^{2}}=\\sqrt{34} \\\\\n& A C=\\sqrt{(0-8)^{2}+(5-3)^{2}}=\\sqrt{68}\n\\end{aligned}\n$$\n\nSince $A B=B C$ and $A C=\\sqrt{2} A B=\\sqrt{2} B C$, then... | [
"$45^{\\circ}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,481 | Blaise and Pierre will play 6 games of squash. Since they are equally skilled, each is equally likely to win any given game. (In squash, there are no ties.) The probability that each of them will win 3 of the 6 games is $\frac{5}{16}$. What is the probability that Blaise will win more games than Pierre? | [
"There are two possibilities: either each player wins three games or one player wins more games than the other.\n\nSince the probability that each player wins three games is $\\frac{5}{16}$, then the probability that any one player wins more games than the other is $1-\\frac{5}{16}=\\frac{11}{16}$.\n\nSince each of... | [
"$\\frac{11}{32}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,482 | Determine all real values of $x$ for which
$$
3^{x+2}+2^{x+2}+2^{x}=2^{x+5}+3^{x}
$$ | [
"Using exponent rules and arithmetic, we manipulate the given equation:\n\n$$\n\\begin{aligned}\n3^{x+2}+2^{x+2}+2^{x} & =2^{x+5}+3^{x} \\\\\n3^{x} 3^{2}+2^{x} 2^{2}+2^{x} & =2^{x} 2^{5}+3^{x} \\\\\n9\\left(3^{x}\\right)+4\\left(2^{x}\\right)+2^{x} & =32\\left(2^{x}\\right)+3^{x} \\\\\n8\\left(3^{x}\\right) & =27\\... | [
"3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,486 | Determine all real values of $x$ such that
$$
\log _{5 x+9}\left(x^{2}+6 x+9\right)+\log _{x+3}\left(5 x^{2}+24 x+27\right)=4
$$ | [
"We manipulate the given equation into a sequence of equivalent equations:\n\n$$\n\\begin{array}{rll}\n\\log _{5 x+9}\\left(x^{2}+6 x+9\\right)+\\log _{x+3}\\left(5 x^{2}+24 x+27\\right) & =4 & \\\\\n\\frac{\\log \\left(x^{2}+6 x+9\\right)}{\\log (5 x+9)}+\\frac{\\log \\left(5 x^{2}+24 x+27\\right)}{\\log (x+3)} & ... | [
"$0,-1,-\\frac{3}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,489 | For each positive integer $N$, an Eden sequence from $\{1,2,3, \ldots, N\}$ is defined to be a sequence that satisfies the following conditions:
(i) each of its terms is an element of the set of consecutive integers $\{1,2,3, \ldots, N\}$,
(ii) the sequence is increasing, and
(iii) the terms in odd numbered positions are odd and the terms in even numbered positions are even.
For example, the four Eden sequences from $\{1,2,3\}$ are
$$
\begin{array}{llll}
1 & 3 & 1,2 & 1,2,3
\end{array}
$$
Determine the number of Eden sequences from $\{1,2,3,4,5\}$. | [
"The Eden sequences from $\\{1,2,3,4,5\\}$ are\n\n$$\n135 \\quad 5 \\quad 1,2 \\quad 1,4 \\quad 3,4 \\quad 1,2,3 \\quad 1,2,5 \\quad 1,4,5 \\quad 3,4,5 \\quad 1,2,3,4 \\quad 1,2,3,4,5\n$$\n\nThere are 12 such sequences.\n\nWe present a brief justification of why these are all of the sequences.\n\n* An Eden sequence... | [
"12"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,490 | For each positive integer $N$, an Eden sequence from $\{1,2,3, \ldots, N\}$ is defined to be a sequence that satisfies the following conditions:
(i) each of its terms is an element of the set of consecutive integers $\{1,2,3, \ldots, N\}$,
(ii) the sequence is increasing, and
(iii) the terms in odd numbered positions are odd and the terms in even numbered positions are even.
For example, the four Eden sequences from $\{1,2,3\}$ are
$$
\begin{array}{llll}
1 & 3 & 1,2 & 1,2,3
\end{array}
$$
For each positive integer $N$, define $e(N)$ to be the number of Eden sequences from $\{1,2,3, \ldots, N\}$. If $e(17)=4180$ and $e(20)=17710$, determine $e(18)$ and $e(19)$. | [
"We will prove that, for all positive integers $n \\geq 3$, we have $e(n)=e(n-1)+e(n-2)+1$. Thus, if $e(18)=m$, then $e(19)=e(18)+e(17)+1=m+4181$ and\n\n$$\ne(20)=e(19)+e(18)+1=(m+4181)+m+1\n$$\n\nSince $e(20)=17710$, then $17710=2 m+4182$ or $2 m=13528$ and so $m=6764$.\n\nTherefore, $e(18)=6764$ and $e(19)=6764+4... | [
"$6764$,$10945$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,492 | If $a$ is chosen randomly from the set $\{1,2,3,4,5\}$ and $b$ is chosen randomly from the set $\{6,7,8\}$, what is the probability that $a^{b}$ is an even number? | [
"Since there are 5 choices for $a$ and 3 choices for $b$, there are fifteen possible ways of choosing $a$ and $b$.\n\nIf $a$ is even, $a^{b}$ is even; if $a$ is odd, $a^{b}$ is odd.\n\nSo the choices of $a$ and $b$ which give an even value for $a^{b}$ are those where $a$ is even, or 6 of the choices (since there ar... | [
"$\\frac{2}{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,493 | A bag contains some blue and some green hats. On each turn, Julia removes one hat without looking, with each hat in the bag being equally likely to be chosen. If it is green, she adds a blue hat into the bag from her supply of extra hats, and if it is blue, she adds a green hat to the bag. The bag initially contains 4 blue hats and 2 green hats. What is the probability that the bag again contains 4 blue hats and 2 green hats after two turns? | [
"Starting with 4 blue hats and 2 green hats, the probability that Julia removes a blue hat is $\\frac{4}{6}=\\frac{2}{3}$. The result would be 3 blue hats and 3 green hats, since a blue hat is replaced with a green hat.\n\nIn order to return to 4 blue hats and 2 green hats from 3 blue and 3 green, Julia would need ... | [
"$\\frac{11}{18}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,494 | Suppose that, for some angles $x$ and $y$,
$$
\begin{aligned}
& \sin ^{2} x+\cos ^{2} y=\frac{3}{2} a \\
& \cos ^{2} x+\sin ^{2} y=\frac{1}{2} a^{2}
\end{aligned}
$$
Determine the possible value(s) of $a$. | [
"Adding the two equations, we obtain\n\n$$\n\\begin{aligned}\n\\sin ^{2} x+\\cos ^{2} x+\\sin ^{2} y+\\cos ^{2} y & =\\frac{3}{2} a+\\frac{1}{2} a^{2} \\\\\n2 & =\\frac{3}{2} a+\\frac{1}{2} a^{2} \\\\\n4 & =3 a+a^{2} \\\\\n0 & =a^{2}+3 a-4 \\\\\n0 & =(a+4)(a-1)\n\\end{aligned}\n$$\n\nand so $a=-4$ or $a=1$.\n\nHowe... | [
"1"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,496 | The sequence $2,5,10,50,500, \ldots$ is formed so that each term after the second is the product of the two previous terms. The 15 th term ends with exactly $k$ zeroes. What is the value of $k$ ? | [
"We calculate the first 15 terms, writing each as an integer times a power of 10:\n\n$$\n\\begin{gathered}\n2,5,10,5 \\times 10,5 \\times 10^{2}, 5^{2} \\times 10^{3}, 5^{3} \\times 10^{5}, 5^{5} \\times 10^{8}, 5^{8} \\times 10^{13}, 5^{13} \\times 10^{21}, 5^{21} \\times 10^{34} \\\\\n5^{34} \\times 10^{55}, 5^{5... | [
"233"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,498 | If $\log _{2} x-2 \log _{2} y=2$, determine $y$, as a function of $x$ | [
"We use logarithm rules to rearrange the equation to solve for $y$ :\n\n$$\n\\begin{aligned}\n\\log _{2} x-2 \\log _{2} y & =2 \\\\\n\\log _{2} x-\\log _{2}\\left(y^{2}\\right) & =2 \\\\\n\\log _{2}\\left(\\frac{x}{y^{2}}\\right) & =2 \\\\\n\\frac{x}{y^{2}} & =2^{2} \\\\\n\\frac{1}{4} x & =y^{2} \\\\\ny & = \\pm \\... | [
"$\\frac{1}{2},\\sqrt{x}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Expression | null | Open-ended | Algebra | Math | English |
2,500 | Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$.
Determine all real numbers $k$ for which $f(x)$ is constant for all values of $x$. | [
"Since $\\sin ^{2} x+\\cos ^{2} x=1$, then $\\cos ^{2} x=1-\\sin ^{2} x$, so\n\n$$\n\\begin{aligned}\nf(x) & =\\sin ^{6} x+\\left(1-\\sin ^{2} x\\right)^{3}+k\\left(\\sin ^{4} x+\\left(1-\\sin ^{2} x\\right)^{2}\\right) \\\\\n& =\\sin ^{6} x+1-3 \\sin ^{2} x+3 \\sin ^{4} x-\\sin ^{6} x+k\\left(\\sin ^{4} x+1-2 \\si... | [
"$-\\frac{3}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,501 | Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$.
If $k=-0.7$, determine all solutions to the equation $f(x)=0$. | [
"Since $\\sin ^{2} x+\\cos ^{2} x=1$, then $\\cos ^{2} x=1-\\sin ^{2} x$, so\n\n$$\n\\begin{aligned}\nf(x) & =\\sin ^{6} x+\\left(1-\\sin ^{2} x\\right)^{3}+k\\left(\\sin ^{4} x+\\left(1-\\sin ^{2} x\\right)^{2}\\right) \\\\\n& =\\sin ^{6} x+1-3 \\sin ^{2} x+3 \\sin ^{4} x-\\sin ^{6} x+k\\left(\\sin ^{4} x+1-2 \\si... | [
"$x=\\frac{1}{6} \\pi+\\pi k, \\frac{1}{3} \\pi+\\pi k, \\frac{2}{3} \\pi+\\pi k, \\frac{5}{6} \\pi+\\pi k$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Expression | null | Open-ended | Algebra | Math | English |
2,502 | Define $f(x)=\sin ^{6} x+\cos ^{6} x+k\left(\sin ^{4} x+\cos ^{4} x\right)$ for some real number $k$.
Determine all real numbers $k$ for which there exists a real number $c$ such that $f(c)=0$. | [
"Since $\\sin ^{2} x+\\cos ^{2} x=1$, then $\\cos ^{2} x=1-\\sin ^{2} x$, so\n\n$$\n\\begin{aligned}\nf(x) & =\\sin ^{6} x+\\left(1-\\sin ^{2} x\\right)^{3}+k\\left(\\sin ^{4} x+\\left(1-\\sin ^{2} x\\right)^{2}\\right) \\\\\n& =\\sin ^{6} x+1-3 \\sin ^{2} x+3 \\sin ^{4} x-\\sin ^{6} x+k\\left(\\sin ^{4} x+1-2 \\si... | [
"$[-1,-\\frac{1}{2}]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Algebra | Math | English |
2,504 | Hexagon $A B C D E F$ has vertices $A(0,0), B(4,0), C(7,2), D(7,5), E(3,5)$, $F(0,3)$. What is the area of hexagon $A B C D E F$ ? | [
"Let $P$ be the point with coordinates $(7,0)$ and let $Q$ be the point with coordinates $(0,5)$.\n\n<img_4025>\n\nThen $A P D Q$ is a rectangle with width 7 and height 5 , and so it has area $7 \\cdot 5=35$.\n\nHexagon $A B C D E F$ is formed by removing two triangles from rectangle $A P D Q$, namely $\\triangle B... | [
"29"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,506 | A list $a_{1}, a_{2}, a_{3}, a_{4}$ of rational numbers is defined so that if one term is equal to $r$, then the next term is equal to $1+\frac{1}{1+r}$. For example, if $a_{3}=\frac{41}{29}$, then $a_{4}=1+\frac{1}{1+(41 / 29)}=\frac{99}{70}$. If $a_{3}=\frac{41}{29}$, what is the value of $a_{1} ?$ | [
"If $r$ is a term in the sequence and $s$ is the next term, then $s=1+\\frac{1}{1+r}$.\n\nThis means that $s-1=\\frac{1}{1+r}$ and so $\\frac{1}{s-1}=1+r$ which gives $r=\\frac{1}{s-1}-1$.\n\nTherefore, since $a_{3}=\\frac{41}{29}$, then\n\n$$\na_{2}=\\frac{1}{a_{3}-1}-1=\\frac{1}{(41 / 29)-1}-1=\\frac{1}{12 / 29}-... | [
"$\\frac{7}{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,507 | A hollow cylindrical tube has a radius of $10 \mathrm{~mm}$ and a height of $100 \mathrm{~mm}$. The tube sits flat on one of its circular faces on a horizontal table. The tube is filled with water to a depth of $h \mathrm{~mm}$. A solid cylindrical rod has a radius of $2.5 \mathrm{~mm}$ and a height of $150 \mathrm{~mm}$. The rod is inserted into the tube so that one of its circular faces sits flat on the bottom of the tube. The height of the water in the tube is now $64 \mathrm{~mm}$. Determine the value of $h$. | [
"Initially, the water in the hollow tube forms a cylinder with radius $10 \\mathrm{~mm}$ and height $h \\mathrm{~mm}$. Thus, the volume of the water is $\\pi(10 \\mathrm{~mm})^{2}(h \\mathrm{~mm})=100 \\pi h \\mathrm{~mm}^{3}$.\n\nAfter the rod is inserted, the level of the water rises to $64 \\mathrm{~mm}$. Note t... | [
"60"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,508 | A function $f$ has the property that $f\left(\frac{2 x+1}{x}\right)=x+6$ for all real values of $x \neq 0$. What is the value of $f(4) ?$ | [
"We note that $\\frac{2 x+1}{x}=\\frac{2 x}{x}+\\frac{1}{x}=2+\\frac{1}{x}$.\n\nTherefore, $\\frac{2 x+1}{x}=4$ exactly when $2+\\frac{1}{x}=4$ or $\\frac{1}{x}=2$ and so $x=\\frac{1}{2}$.\n\nAlternatively, we could solve $\\frac{2 x+1}{x}=4$ directly to obtain $2 x+1=4 x$, which gives $2 x=1$ and so $x=\\frac{1}{2... | [
"$\\frac{13}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,509 | Determine all real numbers $a, b$ and $c$ for which the graph of the function $y=\log _{a}(x+b)+c$ passes through the points $P(3,5), Q(5,4)$ and $R(11,3)$. | [
"Since the graph passes through $(3,5),(5,4)$ and $(11,3)$, we can substitute these three points and obtain the following three equations:\n\n$$\n\\begin{aligned}\n& 5=\\log _{a}(3+b)+c \\\\\n& 4=\\log _{a}(5+b)+c \\\\\n& 3=\\log _{a}(11+b)+c\n\\end{aligned}\n$$\n\nSubtracting the second equation from the first and... | [
"$\\frac{1}{3},-2,5$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,510 | A computer is programmed to choose an integer between 1 and 99, inclusive, so that the probability that it selects the integer $x$ is equal to $\log _{100}\left(1+\frac{1}{x}\right)$. Suppose that the probability that $81 \leq x \leq 99$ is equal to 2 times the probability that $x=n$ for some integer $n$. What is the value of $n$ ? | [
"The probability that the integer $n$ is chosen is $\\log _{100}\\left(1+\\frac{1}{n}\\right)$.\n\nThe probability that an integer between 81 and 99 , inclusive, is chosen equals the sum of the probabilities that the integers $81,82, \\ldots, 98,99$ are selected, which equals\n\n$$\n\\log _{100}\\left(1+\\frac{1}{8... | [
"9"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,520 | What is the smallest positive integer $x$ for which $\frac{1}{32}=\frac{x}{10^{y}}$ for some positive integer $y$ ? | [
"Since $10^{y} \\neq 0$, the equation $\\frac{1}{32}=\\frac{x}{10^{y}}$ is equivalent to $10^{y}=32 x$.\n\nSo the given question is equivalent to asking for the smallest positive integer $x$ for which $32 x$ equals a positive integer power of 10 .\n\nNow $32=2^{5}$ and so $32 x=2^{5} x$.\n\nFor $32 x$ to equal a po... | [
"3125"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,521 | Determine all possible values for the area of a right-angled triangle with one side length equal to 60 and with the property that its side lengths form an arithmetic 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.) | [
"Since the three side lengths of a right-angled triangle form an arithemetic sequence and must include 60 , then the three side lengths are $60,60+d, 60+2 d$ or $60-d, 60,60+d$ or $60-2 d, 60-d, 60$, for some $d \\geq 0$.\n\nFor a triangle with sides of length $60,60+d, 60+2 d$ to be right-angled, by the Pythagorea... | [
"2400, 1350, 864"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,522 | Amrita and Zhang cross a lake in a straight line with the help of a one-seat kayak. Each can paddle the kayak at $7 \mathrm{~km} / \mathrm{h}$ and swim at $2 \mathrm{~km} / \mathrm{h}$. They start from the same point at the same time with Amrita paddling and Zhang swimming. After a while, Amrita stops the kayak and immediately starts swimming. Upon reaching the kayak (which has not moved since Amrita started swimming), Zhang gets in and immediately starts paddling. They arrive on the far side of the lake at the same time, 90 minutes after they began. Determine the amount of time during these 90 minutes that the kayak was not being paddled. | [
"Suppose that Amrita paddles the kayak for $p \\mathrm{~km}$ and swims for $s \\mathrm{~km}$.\n\nSince Amrita leaves the kayak in the lake and it does not move, then Zhang swims $p \\mathrm{~km}$ and paddles the kayak for $s \\mathrm{~km}$.\n\nNote that each paddles at $7 \\mathrm{~km} / \\mathrm{h}$ and each swims... | [
"50"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | minutes | Numerical | null | Open-ended | Algebra | Math | English |
2,523 | Determine all pairs $(x, y)$ of real numbers that satisfy the system of equations
$$
\begin{aligned}
x\left(\frac{1}{2}+y-2 x^{2}\right) & =0 \\
y\left(\frac{5}{2}+x-y\right) & =0
\end{aligned}
$$ | [
"From the first equation, $x\\left(\\frac{1}{2}+y-2 x^{2}\\right)=0$, we obtain $x=0$ or $\\frac{1}{2}+y-2 x^{2}=0$.\n\nFrom the second equation, $y\\left(\\frac{5}{2}+x-y\\right)=0$, we obtain $y=0$ or $\\frac{5}{2}+x-y=0$.\n\nIf $x=0$, the first equation is satisified.\n\nFor the second equation to be true in thi... | [
"$(0,0),(0, \\frac{5}{2}),(\\frac{1}{2}, 0),(-\\frac{1}{2}, 0),(\\frac{3}{2}, 4),(-1, \\frac{3}{2})$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Algebra | Math | English |
2,524 | Determine all real numbers $x>0$ for which
$$
\log _{4} x-\log _{x} 16=\frac{7}{6}-\log _{x} 8
$$ | [
"Note that $x \\neq 1$ since 1 cannot be the base of a logarithm. This tells us that $\\log x \\neq 0$. Using the fact that $\\log _{a} b=\\frac{\\log b}{\\log a}$ and then using other logarithm laws, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n\\log _{4} x-\\log _{x} 16 & =\\frac{7}{6}-\... | [
"$2^{-2 / 3}$, $8$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,525 | The string $A A A B B B A A B B$ is a string of ten letters, each of which is $A$ or $B$, that does not include the consecutive letters $A B B A$.
The string $A A A B B A A A B B$ is a string of ten letters, each of which is $A$ or $B$, that does include the consecutive letters $A B B A$.
Determine, with justification, the total number of strings of ten letters, each of which is $A$ or $B$, that do not include the consecutive letters $A B B A$. | [
"There are $2^{10}=1024$ strings of ten letters, each of which is $A$ or $B$, because there are 2 choices for each of the 10 positions in the string.\n\nWe determine the number of these strings that do not include the \"substring\" $A B B A$ (that is, that do not include consecutive letters $A B B A$ ) by counting ... | [
"631"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,527 | Let $k$ be a positive integer with $k \geq 2$. Two bags each contain $k$ balls, labelled with the positive integers from 1 to $k$. André removes one ball from each bag. (In each bag, each ball is equally likely to be chosen.) Define $P(k)$ to be the probability that the product of the numbers on the two balls that he chooses is divisible by $k$.
Calculate $P(10)$. | [
"Here, $k=10$ and so there are 10 balls in each bag.\n\nSince there are 10 balls in each bag, there are $10 \\cdot 10=100$ pairs of balls that can be chosen.\n\nLet $a$ be the number on the first ball chosen and $b$ be the number on the second ball chosen. To determine $P(10)$, we count the number of pairs $(a, b)$... | [
"$\\frac{27}{100}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,531 | In an arithmetic sequence, the first term is 1 and the last term is 19 . The sum of all the terms in the sequence is 70 . How many terms does the sequence have? (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 is an arithmetic sequence with four terms.) | [
"The sum of the terms in an arithmetic sequence is equal to the average of the first and last terms times the number of terms.\n\nIf $n$ is the number of terms in the sequence, then $\\frac{1}{2}(1+19) n=70$ or $10 n=70$ and so $n=7$.",
"Let $n$ be the number of terms in the sequence and $d$ the common difference... | [
"7"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,532 | Suppose that $a(x+b(x+3))=2(x+6)$ for all values of $x$. Determine $a$ and $b$. | [
"Since the given equation is true for all values of $x$, then it is true for any particular value of $x$ that we try.\n\nIf $x=-3$, the equation becomes $a(-3+b(0))=2(3)$ or $-3 a=6$ and so $a=-2$.\n\nIf $x=0$, the equation becomes $-2(0+b(3))=2(6)$ or $-6 b=12$ and so $b=-2$.\n\nTherefore, $a=-2$ and $b=-2$.",
"... | [
"-2,-2"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,534 | An integer $n$, with $100 \leq n \leq 999$, is chosen at random. What is the probability that the sum of the digits of $n$ is 24 ? | [
"The number of integers between 100 and 999 inclusive is $999-100+1=900$.\n\nAn integer $n$ in this range has three digits, say $a, b$ and $c$, with the hundreds digit equal to $a$.\n\nNote that $0 \\leq b \\leq 9$ and $0 \\leq c \\leq 9$ and $1 \\leq a \\leq 9$.\n\nTo have $a+b+c=24$, then the possible triples for... | [
"$\\frac{1}{90}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,536 | The parabola $y=x^{2}-2 x+4$ is translated $p$ units to the right and $q$ units down. The $x$-intercepts of the resulting parabola are 3 and 5 . What are the values of $p$ and $q$ ? | [
"Completing the square on the original parabola, we obtain\n\n$$\ny=x^{2}-2 x+4=x^{2}-2 x+1-1+4=(x-1)^{2}+3\n$$\n\nTherefore, the vertex of the original parabola is $(1,3)$.\n\nSince the new parabola is a translation of the original parabola and has $x$-intercepts 3 and 5 , then its equation is $y=1(x-3)(x-5)=x^{2}... | [
"3,4"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Geometry | Math | English |
2,539 | If $\log _{2} x,\left(1+\log _{4} x\right)$, and $\log _{8} 4 x$ are consecutive terms of a geometric sequence, determine the possible values of $x$.
(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.) | [
"First, we convert each of the logarithms to a logarithm with base 2:\n\n$$\n\\begin{aligned}\n1+\\log _{4} x & =1+\\frac{\\log _{2} x}{\\log _{2} 4}=1+\\frac{\\log _{2} x}{2}=1+\\frac{1}{2} \\log _{2} x \\\\\n\\log _{8} 4 x & =\\frac{\\log _{2} 4 x}{\\log _{2} 8}=\\frac{\\log _{2} 4+\\log _{2} x}{3}=\\frac{2}{3}+\... | [
"$64,\\frac{1}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,544 | Determine the two pairs of positive integers $(a, b)$ with $a<b$ that satisfy the equation $\sqrt{a}+\sqrt{b}=\sqrt{50}$. | [
"First, we note that $\\sqrt{50}=5 \\sqrt{2}$.\n\nNext, we note that $\\sqrt{2}+4 \\sqrt{2}=5 \\sqrt{2}$ and $2 \\sqrt{2}+3 \\sqrt{2}=5 \\sqrt{2}$.\n\nFrom the first of these, we obtain $\\sqrt{2}+\\sqrt{32}=\\sqrt{50}$.\n\nFrom the second of these, we obtain $\\sqrt{8}+\\sqrt{18}=\\sqrt{50}$.\n\nThus, $(a, b)=(2,3... | [
"(2,32), (8,18)"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Number Theory | Math | English |
2,545 | Consider the system of equations:
$$
\begin{aligned}
c+d & =2000 \\
\frac{c}{d} & =k
\end{aligned}
$$
Determine the number of integers $k$ with $k \geq 0$ for which there is at least one pair of integers $(c, d)$ that is a solution to the system. | [
"From the second equation, we note that $d \\neq 0$.\n\nRearranging this second equation, we obtain $c=k d$.\n\nSubstituting into the first equation, we obtain $k d+d=2000$ or $(k+1) d=2000$.\n\nSince $k \\geq 0$, note that $k+1 \\geq 1$.\n\nThis means that if $(c, d)$ is a solution, then $k+1$ is a divisor of 2000... | [
"20"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,548 | Determine all real numbers $x$ for which $2 \log _{2}(x-1)=1-\log _{2}(x+2)$. | [
"Using logarithm and exponent laws, we obtain the following equivalent equations:\n\n$$\n\\begin{aligned}\n2 \\log _{2}(x-1) & =1-\\log _{2}(x+2) \\\\\n2 \\log _{2}(x-1)+\\log _{2}(x+2) & =1 \\\\\n\\log _{2}\\left((x-1)^{2}\\right)+\\log _{2}(x+2) & =1 \\\\\n\\log _{2}\\left((x-1)^{2}(x+2)\\right) & =1 \\\\\n(x-1)^... | [
"$\\sqrt{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,549 | Consider the function $f(x)=x^{2}-2 x$. Determine all real numbers $x$ that satisfy the equation $f(f(f(x)))=3$. | [
"Let $a=f(f(x))$.\n\nThus, the equation $f(f(f(x)))=3$ is equivalent to $f(a)=3$.\n\nSince $f(a)=a^{2}-2 a$, then we obtain the equation $a^{2}-2 a=3$ which gives $a^{2}-2 a-3=0$ and $(a-3)(a+1)=0$.\n\nThus, $a=3$ or $a=-1$ which means that $f(f(x))=3$ or $f(f(x))=-1$.\n\nLet $b=f(x)$.\n\nThus, the equations $f(f(x... | [
"$3,1,-1,1+\\sqrt{2}, 1-\\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,551 | Suppose that $x$ satisfies $0<x<\frac{\pi}{2}$ and $\cos \left(\frac{3}{2} \cos x\right)=\sin \left(\frac{3}{2} \sin x\right)$.
Determine all possible values of $\sin 2 x$, expressing your answers in the form $\frac{a \pi^{2}+b \pi+c}{d}$ where $a, b, c, d$ are integers. | [
"Since $0<x<\\frac{\\pi}{2}$, then $0<\\cos x<1$ and $0<\\sin x<1$.\n\nThis means that $0<\\frac{3}{2} \\cos x<\\frac{3}{2}$ and $0<\\frac{3}{2} \\sin x<\\frac{3}{2}$. Since $3<\\pi$, then $0<\\frac{3}{2} \\cos x<\\frac{\\pi}{2}$ and $0<\\frac{3}{2} \\sin x<\\frac{\\pi}{2}$.\n\nIf $Y$ and $Z$ are angles with $0<Y<\... | [
"$\\frac{\\pi^{2}-9}{9}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,552 | For positive integers $a$ and $b$, define $f(a, b)=\frac{a}{b}+\frac{b}{a}+\frac{1}{a b}$.
For example, the value of $f(1,2)$ is 3 .
Determine the value of $f(2,5)$. | [
"By definition, $f(2,5)=\\frac{2}{5}+\\frac{5}{2}+\\frac{1}{2 \\cdot 5}=\\frac{2 \\cdot 2+5 \\cdot 5+1}{2 \\cdot 5}=\\frac{4+25+1}{10}=\\frac{30}{10}=3$."
] | [
"3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,553 | For positive integers $a$ and $b$, define $f(a, b)=\frac{a}{b}+\frac{b}{a}+\frac{1}{a b}$.
For example, the value of $f(1,2)$ is 3 .
Determine all positive integers $a$ for which $f(a, a)$ is an integer. | [
"By definition, $f(a, a)=\\frac{a}{a}+\\frac{a}{a}+\\frac{1}{a^{2}}=2+\\frac{1}{a^{2}}$.\n\nFor $2+\\frac{1}{a^{2}}$ to be an integer, it must be the case that $\\frac{1}{a^{2}}$ is an integer.\n\nFor $\\frac{1}{a^{2}}$ to be an integer and since $a^{2}$ is an integer, $a^{2}$ needs to be a divisor of 1 .\n\nSince ... | [
"1"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,555 | Amir and Brigitte play a card game. Amir starts with a hand of 6 cards: 2 red, 2 yellow and 2 green. Brigitte starts with a hand of 4 cards: 2 purple and 2 white. Amir plays first. Amir and Brigitte alternate turns. On each turn, the current player chooses one of their own cards at random and places it on the table. The cards remain on the table for the rest of the game. A player wins and the game ends when they have placed two cards of the same colour on the table. Determine the probability that Amir wins the game. | [
"On her first two turns, Brigitte either chooses two cards of the same colour or two cards of different colours. If she chooses two cards of different colours, then on her third turn, she must choose a card that matches one of the cards that she already has.\n\nTherefore, the game ends on or before Brigitte's third... | [
"$\\frac{7}{15}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,558 | Consider the sequence $t_{1}=1, t_{2}=-1$ and $t_{n}=\left(\frac{n-3}{n-1}\right) t_{n-2}$ where $n \geq 3$. What is the value of $t_{1998}$ ? | [
"Calculating some terms, $t_{1}=1, t_{2}=-1, t_{3}=0, t_{4}=\\frac{-1}{3}, t_{5}=0, t_{6}=\\frac{-1}{5}$ etc.\n\nBy pattern recognition, $t_{1998}=\\frac{-1}{1997}$.",
"$$\n\\begin{aligned}\nt_{1998} & =\\frac{1995}{1997} t_{1996}=\\frac{1995}{1997} \\times \\frac{1993}{1995} t_{1994} \\\\\n& =\\frac{1995}{1997} ... | [
"$\\frac{-1}{1997}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,559 | The $n$th term of an arithmetic sequence is given by $t_{n}=555-7 n$.
If $S_{n}=t_{1}+t_{2}+\ldots+t_{n}$, determine the smallest value of $n$ for which $S_{n}<0$. | [
"This is an arithmetic sequence in which $a=548$ and $d=-7$.\n\nTherefore, $S_{n}=\\frac{n}{2}[2(548)+(n-1)(-7)]=\\frac{n}{2}[-7 n+1103]$.\n\nWe now want $\\frac{n}{2}(-7 n+1103)<0$.\n\nSince $n>0,-7 n+1103<0$\n\n$$\nn>157 \\frac{4}{7}\n$$\n\nTherefore the smallest value of $n$ is 158 .",
"For this series we want... | [
"158"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,560 | If $x$ and $y$ are real numbers, determine all solutions $(x, y)$ of the system of equations
$$
\begin{aligned}
& x^{2}-x y+8=0 \\
& x^{2}-8 x+y=0
\end{aligned}
$$ | [
"Subtracting,\n\n$$\n\\begin{array}{r}\nx^{2}-x y+8=0 \\\\\nx^{2}-8 x+y=0 \\\\\n\\hline-x y+8 x+8-y=0 \\\\\n8(1+x)-y(1+x)=0 \\\\\n(8-y)(1+x)=0 \\\\\ny=8 \\text { or } x=-1\n\\end{array}\n$$\n\n\n\nIf $y=8$, both equations become $x^{2}-8 x+8=0, x=4 \\pm 2 \\sqrt{2}$.\n\nIf $x=-1$ both equations become $y+9=0, y=-9$... | [
"$(-1,-9),(4+2 \\sqrt{2}, 8),(4-2 \\sqrt{2}, 8)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Algebra | Math | English |
2,565 | The equations $x^{2}+5 x+6=0$ and $x^{2}+5 x-6=0$ each have integer solutions whereas only one of the equations in the pair $x^{2}+4 x+5=0$ and $x^{2}+4 x-5=0$ has integer solutions.
Determine $q$ in terms of $a$ and $b$. | [
"We have that $x^{2}+p x+q=0$ and $x^{2}+p x-q=0$ both have integer solutions.\n\nFor $x^{2}+p x+q=0$, its roots are $\\frac{-p \\pm \\sqrt{p^{2}-4 q}}{2}$.\n\nIn order that these roots be integers, $p^{2}-4 q$ must be a perfect square.\n\nTherefore, $p^{2}-4 q=m^{2}$ for some positive integer $m$.\n\nSimilarly for... | [
"$\\frac{a b}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Algebra | Math | English |
2,568 | Determine all values of $k$, with $k \neq 0$, for which the parabola
$$
y=k x^{2}+(5 k+3) x+(6 k+5)
$$
has its vertex on the $x$-axis. | [
"For the parabola to have its vertex on the $x$-axis, the equation\n\n$$\ny=k x^{2}+(5 k+3) x+(6 k+5)=0\n$$\n\nmust have two equal real roots.\n\nThat is, its discriminant must equal 0 , and so\n\n$$\n\\begin{aligned}\n(5 k+3)^{2}-4 k(6 k+5) & =0 \\\\\n25 k^{2}+30 k+9-24 k^{2}-20 k & =0 \\\\\nk^{2}+10 k+9 & =0 \\\\... | [
"$-1,-9$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,569 | The function $f(x)$ satisfies the equation $f(x)=f(x-1)+f(x+1)$ for all values of $x$. If $f(1)=1$ and $f(2)=3$, what is the value of $f(2008)$ ? | [
"Since $f(x)=f(x-1)+f(x+1)$, then $f(x+1)=f(x)-f(x-1)$, and so\n\n$$\n\\begin{aligned}\n& f(1)=1 \\\\\n& f(2)=3 \\\\\n& f(3)=f(2)-f(1)=3-1=2 \\\\\n& f(4)=f(3)-f(2)=2-3=-1 \\\\\n& f(5)=f(4)-f(3)=-1-2=-3 \\\\\n& f(6)=f(5)-f(4)=-3-(-1)=-2 \\\\\n& f(7)=f(6)-f(5)=-2-(-3)=1=f(1) \\\\\n& f(8)=f(7)-f(6)=1-(-2)=3=f(2)\n\\en... | [
"-1"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,570 | The numbers $a, b, c$, in that order, form a three term arithmetic sequence (see below) and $a+b+c=60$.
The numbers $a-2, b, c+3$, in that order, form a three term geometric sequence. Determine all possible values of $a, b$ and $c$.
(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, $3,5,7$ is an arithmetic sequence with three terms.
A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a constant. For example, $3,6,12$ is a geometric sequence with three terms.)
Present your answer in the form of coordinates (e.g. (1, 2, 3) for a=1, b=2, c=3). | [
"Since $a, b, c$ form an arithmetic sequence, then we can write $a=b-d$ and $c=b+d$ for some real number $d$.\n\nSince $a+b+c=60$, then $(b-d)+b+(b+d)=60$ or $3 b=60$ or $b=20$.\n\nTherefore, we can write $a, b, c$ as $20-d, 20,20+d$.\n\n(We could have written $a, b, c$ instead as $a, a+d, a+2 d$ and arrived at the... | [
"$(27,20,13), (18,20,22)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Algebra | Math | English |
2,571 | The average of three consecutive multiples of 3 is $a$.
The average of four consecutive multiples of 4 is $a+27$.
The average of the smallest and largest of these seven integers is 42 .
Determine the value of $a$. | [
"Since the average of three consecutive multiples of 3 is $a$, then $a$ is the middle of these three integers, so the integers are $a-3, a, a+3$.\n\nSince the average of four consecutive multiples of 4 is $a+27$, then $a+27$ is halfway in between the second and third of these multiples (which differ by 4), so the s... | [
"27"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,572 | Billy and Crystal each have a bag of 9 balls. The balls in each bag are numbered from 1 to 9. Billy and Crystal each remove one ball from their own bag. Let $b$ be the sum of the numbers on the balls remaining in Billy's bag. Let $c$ be the sum of the numbers on the balls remaining in Crystal's bag. Determine the probability that $b$ and $c$ differ by a multiple of 4 . | [
"Suppose that Billy removes the ball numbered $x$ from his bag and that Crystal removes the ball numbered $y$ from her bag.\n\nThen $b=1+2+3+4+5+6+7+8+9-x=45-x$.\n\nAlso, $c=1+2+3+4+5+6+7+8+9-y=45-y$.\n\nHence, $b-c=(45-x)-(45-y)=y-x$.\n\nSince $1 \\leq x \\leq 9$ and $1 \\leq y \\leq 9$, then $-8 \\leq y-x \\leq 8... | [
"$\\frac{7}{27}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,574 | The equation $2^{x+2} 5^{6-x}=10^{x^{2}}$ has two real solutions. Determine these two solutions. | [
"Rewriting the equation, we obtain\n\n$$\n\\begin{aligned}\n2^{x+2} 5^{6-x} & =2^{x^{2}} 5^{x^{2}} \\\\\n1 & =2^{x^{2}} 2^{-2-x} 5^{x^{2}} 5^{x-6} \\\\\n1 & =2^{x^{2}-x-2} 5^{x^{2}+x-6} \\\\\n0 & =\\left(x^{2}-x-2\\right) \\log _{10} 2+\\left(x^{2}+x-6\\right) \\log _{10} 5 \\\\\n0 & =(x-2)(x+1) \\log _{10} 2+(x-2)... | [
"$2,-\\log _{10} 250$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,575 | Determine all real solutions to the system of equations
$$
\begin{aligned}
& x+\log _{10} x=y-1 \\
& y+\log _{10}(y-1)=z-1 \\
& z+\log _{10}(z-2)=x+2
\end{aligned}
$$
and prove that there are no more solutions. | [
"First, we rewrite the system as\n\n$$\n\\begin{aligned}\n& x+\\log _{10} x=y-1 \\\\\n& (y-1)+\\log _{10}(y-1)=z-2 \\\\\n& (z-2)+\\log _{10}(z-2)=x\n\\end{aligned}\n$$\n\nSecond, we make the substitution $a=x, b=y-1$ and $c=z-2$, allowing us to rewrite\n\n\n\nthe system as\n\n$$\n\\begin{aligned}\na+\\log _{10} a &... | [
"$1,2,3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English |
2,576 | The positive integers 34 and 80 have exactly two positive common divisors, namely 1 and 2. How many positive integers $n$ with $1 \leq n \leq 30$ have the property that $n$ and 80 have exactly two positive common divisors? | [
"Since $80=2^{4} \\cdot 5$, its positive divisors are $1,2,4,5,8,10,16,20,40,80$.\n\nFor an integer $n$ to share exactly two positive common divisors with 80, these divisors must be either 1 and 2 or 1 and 5 . ( 1 is a common divisor of any two integers. The second common divisor must be a prime number since any co... | [
"9"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,577 | A function $f$ is defined so that
- $f(1)=1$,
- if $n$ is an even positive integer, then $f(n)=f\left(\frac{1}{2} n\right)$, and
- if $n$ is an odd positive integer with $n>1$, then $f(n)=f(n-1)+1$.
For example, $f(34)=f(17)$ and $f(17)=f(16)+1$.
Determine the value of $f(50)$. | [
"We start with $f(50)$ and apply the given rules for the function until we reach $f(1)$ :\n\n$$\n\\begin{aligned}\nf(50) & =f(25) \\\\\n& =f(24)+1 \\\\\n& =f(12)+1 \\\\\n& =f(6)+1 \\\\\n& =f(3)+1 \\\\\n& =(f(2)+1)+1 \\\\\n& =f(1)+1+1 \\\\\n& =1+1+1 \\\\\n& =3\n\\end{aligned}\n$$\n\n(since 50 is even and $\\frac{1}{... | [
"3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,578 | The perimeter of equilateral $\triangle P Q R$ is 12. The perimeter of regular hexagon $S T U V W X$ is also 12. What is the ratio of the area of $\triangle P Q R$ to the area of $S T U V W X$ ? | [
"Since the hexagon has perimeter 12 and has 6 sides, then each side has length 2 .\n\nSince equilateral $\\triangle P Q R$ has perimeter 12 , then its side length is 4 .\n\nConsider equilateral triangles with side length 2.\n\nSix of these triangles can be combined to form a regular hexagon with side length 2 and f... | [
"$\\frac{2}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,580 | For how many integers $k$ with $0<k<18$ is $\frac{5 \sin \left(10 k^{\circ}\right)-2}{\sin ^{2}\left(10 k^{\circ}\right)} \geq 2$ ? | [
"Let $\\theta=10 k^{\\circ}$.\n\nThe given inequalities become $0^{\\circ}<\\theta<180^{\\circ}$ and $\\frac{5 \\sin \\theta-2}{\\sin ^{2} \\theta} \\geq 2$.\n\nWhen $0^{\\circ}<\\theta<180^{\\circ}, \\sin \\theta \\neq 0$.\n\nThis means that we can can multiply both sides by $\\sin ^{2} \\theta>0$ and obtain the e... | [
"13"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,582 | Eight people, including triplets Barry, Carrie and Mary, are going for a trip in four canoes. Each canoe seats two people. The eight people are to be randomly assigned to the four canoes in pairs. What is the probability that no two of Barry, Carrie and Mary will be in the same canoe? | [
"Among a group of $n$ people, there are $\\frac{n(n-1)}{2}$ ways of choosing a pair of these people:\n\nThere are $n$ people that can be chosen first.\n\nFor each of these $n$ people, there are $n-1$ people that can be chosen second.\n\nThis gives $n(n-1)$ orderings of two people.\n\nEach pair is counted twice (giv... | [
"$\\frac{4}{7}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,583 | Diagonal $W Y$ of square $W X Y Z$ has slope 2. Determine the sum of the slopes of $W X$ and $X Y$. | [
"Suppose that $W Y$ makes an angle of $\\theta$ with the horizontal.\n\n<img_3532>\n\nSince the slope of $W Y$ is 2 , then $\\tan \\theta=2$, since the tangent of an angle equals the slope of a line that makes this angle with the horizontal.\n\nSince $\\tan \\theta=2>1=\\tan 45^{\\circ}$, then $\\theta>45^{\\circ}$... | [
"$-\\frac{8}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,584 | Determine all values of $x$ such that $\log _{2 x}(48 \sqrt[3]{3})=\log _{3 x}(162 \sqrt[3]{2})$. | [
"Since the base of a logarithm must be positive and cannot equal 1 , then $x>0$ and $x \\neq \\frac{1}{2}$ and $x \\neq \\frac{1}{3}$.\n\nThis tells us that $\\log 2 x$ and $\\log 3 x$ exist and do not equal 0 , which we will need shortly when we apply the change of base formula.\n\nWe note further that $48=2^{4} \... | [
"$\\sqrt{6}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,588 | In an infinite array with two rows, the numbers in the top row are denoted $\ldots, A_{-2}, A_{-1}, A_{0}, A_{1}, A_{2}, \ldots$ and the numbers in the bottom row are denoted $\ldots, B_{-2}, B_{-1}, B_{0}, B_{1}, B_{2}, \ldots$ For each integer $k$, the entry $A_{k}$ is directly above the entry $B_{k}$ in the array, as shown:
| $\ldots$ | $A_{-2}$ | $A_{-1}$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $\ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $\ldots$ | $B_{-2}$ | $B_{-1}$ | $B_{0}$ | $B_{1}$ | $B_{2}$ | $\ldots$ |
For each integer $k, A_{k}$ is the average of the entry to its left, the entry to its right, and the entry below it; similarly, each entry $B_{k}$ is the average of the entry to its left, the entry to its right, and the entry above it.
In one such array, $A_{0}=A_{1}=A_{2}=0$ and $A_{3}=1$.
Determine the value of $A_{4}$. | [
"We draw part of the array using the information that $A_{0}=A_{1}=A_{2}=0$ and $A_{3}=1$ :\n\n$$\n\\begin{array}{l|l|l|l|l|l|l|lll|c|c|c|c|c|c}\n\\cdots & A_{0} & A_{1} & A_{2} & A_{3} & A_{4} & A_{5} & \\cdots & \\cdots & 0 & 0 & 0 & 1 & A_{4} & A_{5} & \\cdots \\\\\n\\hline \\cdots & B_{0} & B_{1} & B_{2} & B_{3... | [
"6"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,592 | The populations of Alphaville and Betaville were equal at the end of 1995. The population of Alphaville decreased by $2.9 \%$ during 1996, then increased by $8.9 \%$ during 1997 , and then increased by $6.9 \%$ during 1998 . The population of Betaville increased by $r \%$ in each of the three years. If the populations of the towns are equal at the end of 1998, determine the value of $r$ correct to one decimal place. | [
"If $P$ is the original population of Alphaville and Betaville,\n\n$$\n\\begin{aligned}\nP(.971)(1.089)(1.069) & =P\\left(1+\\frac{r}{100}\\right)^{3} \\\\\n1.1303 & =\\left(1+\\frac{r}{100}\\right)^{3}\n\\end{aligned}\n$$\n\nFrom here,\n\nPossibility 1\n\n$$\n\\begin{aligned}\n1+\\frac{r}{100} & =(1.1303)^{\\frac{... | [
"4.2"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | % | Numerical | 1e-1 | Open-ended | Algebra | Math | English |
2,596 | Determine the coordinates of the points of intersection of the graphs of $y=\log _{10}(x-2)$ and $y=1-\log _{10}(x+1)$. | [
"The intersection takes place where,\n\n$$\n\\begin{aligned}\n& \\log _{10}(x-2)=1-\\log _{10}(x+1) \\\\\n& \\log _{10}(x-2)+\\log _{10}(x+1)=1 \\\\\n& \\log _{10}\\left(x^{2}-x-2\\right)=1\n\\end{aligned}\n$$\n\n\n\n$$\n\\begin{aligned}\n& x^{2}-x-2=10 \\\\\n& x^{2}-x-12=0 \\\\\n& (x-4)(x+3)=0 \\\\\n& x=4 \\text {... | [
"$(4, \\log _{10} 2)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Tuple | null | Open-ended | Geometry | Math | English |
2,598 | Charlie was born in the twentieth century. On his birthday in the present year (2014), he notices that his current age is twice the number formed by the rightmost two digits of the year in which he was born. Compute the four-digit year in which Charlie was born. | [
"Let $N$ be the number formed by the rightmost two digits of the year in which Charlie was born. Then his current age is $100-N+14=114-N$. Setting this equal to $2 N$ and solving yields $N=38$, hence the answer is 1938 .",
"Let $N$ be the number formed by the rightmost two digits of the year in which Charlie was ... | [
"1938"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,599 | Let $A, B$, and $C$ be randomly chosen (not necessarily distinct) integers between 0 and 4 inclusive. Pat and Chris compute the value of $A+B \cdot C$ by two different methods. Pat follows the proper order of operations, computing $A+(B \cdot C)$. Chris ignores order of operations, choosing instead to compute $(A+B) \cdot C$. Compute the probability that Pat and Chris get the same answer. | [
"If Pat and Chris get the same answer, then $A+(B \\cdot C)=(A+B) \\cdot C$, or $A+B C=A C+B C$, or $A=A C$. This equation is true if $A=0$ or $C=1$; the equation places no restrictions on $B$. There are 25 triples $(A, B, C)$ where $A=0,25$ triples where $C=1$, and 5 triples where $A=0$ and $C=1$. As all triples a... | [
"$\\frac{9}{25}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,600 | Bobby, Peter, Greg, Cindy, Jan, and Marcia line up for ice cream. In an acceptable lineup, Greg is ahead of Peter, Peter is ahead of Bobby, Marcia is ahead of Jan, and Jan is ahead of Cindy. For example, the lineup with Greg in front, followed by Peter, Marcia, Jan, Cindy, and Bobby, in that order, is an acceptable lineup. Compute the number of acceptable lineups. | [
"There are 6 people, so there are $6 !=720$ permutations. However, for each arrangement of the boys, there are $3 !=6$ permutations of the girls, of which only one yields an acceptable lineup. The same logic holds for the boys. Thus the total number of permutations must be divided by $3 ! \\cdot 3 !=36$, yielding $... | [
"20"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,601 | In triangle $A B C, a=12, b=17$, and $c=13$. Compute $b \cos C-c \cos B$. | [
"Using the Law of Cosines, $a^{2}+b^{2}-2 a b \\cos C=c^{2}$ implies\n\n$$\nb \\cos C=\\frac{a^{2}+b^{2}-c^{2}}{2 a}\n$$\n\nSimilarly,\n\n$$\nc \\cos B=\\frac{a^{2}-b^{2}+c^{2}}{2 a}\n$$\n\nThus\n\n$$\n\\begin{aligned}\nb \\cos C-c \\cos B & =\\frac{a^{2}+b^{2}-c^{2}}{2 a}-\\frac{a^{2}-b^{2}+c^{2}}{2 a} \\\\\n& =\\... | [
"10"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,602 | The sequence of words $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=X, a_{2}=O$, and for $n \geq 3, a_{n}$ is $a_{n-1}$ followed by the reverse of $a_{n-2}$. For example, $a_{3}=O X, a_{4}=O X O, a_{5}=O X O X O$, and $a_{6}=O X O X O O X O$. Compute the number of palindromes in the first 1000 terms of this sequence. | [
"Let $P$ denote a palindromic word, let $Q$ denote any word, and let $\\bar{R}$ denote the reverse of word $R$. Note that if two consecutive terms of the sequence are $a_{n}=P, a_{n+1}=Q$, then $a_{n+2}=Q \\bar{P}=Q P$ and $a_{n+3}=Q P \\bar{Q}$. Thus if $a_{n}$ is a palindrome, so is $a_{n+3}$. Because $a_{1}$ and... | [
"667"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,603 | Compute the smallest positive integer $n$ such that $214 \cdot n$ and $2014 \cdot n$ have the same number of divisors. | [
"Let $D(n)$ be the number of divisors of the integer $n$. Note that if $D(214 n)=D(2014 n)$ and if some $p$ divides $n$ and is relatively prime to both 214 and 2014 , then $D\\left(\\frac{214 n}{p}\\right)=D\\left(\\frac{2014 n}{p}\\right)$. Thus any prime divisor of the smallest possible positive $n$ will be a div... | [
"19133"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,604 | Let $N$ be the least integer greater than 20 that is a palindrome in both base 20 and base 14 . For example, the three-digit base-14 numeral (13)5(13) ${ }_{14}$ (representing $13 \cdot 14^{2}+5 \cdot 14^{1}+13 \cdot 14^{0}$ ) is a palindrome in base 14 , but not in base 20 , and the three-digit base-14 numeral (13)31 14 is not a palindrome in base 14 . Compute the base-10 representation of $N$. | [
"Because $N$ is greater than 20, the base-20 and base-14 representations of $N$ must be at least two digits long. The smallest possible case is that $N$ is a two-digit palindrome in both bases. Then $N=20 a+a=21 a$, where $1 \\leq a \\leq 19$. Similarly, in order to be a two-digit palindrome in base $14, N=14 b+b=1... | [
"105"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,606 | $\quad$ Compute the greatest integer $k \leq 1000$ such that $\left(\begin{array}{c}1000 \\ k\end{array}\right)$ is a multiple of 7 . | [
"The ratio of binomial coefficients $\\left(\\begin{array}{c}1000 \\\\ k\\end{array}\\right) /\\left(\\begin{array}{c}1000 \\\\ k+1\\end{array}\\right)=\\frac{k+1}{1000-k}$. Because 1000 is 1 less than a multiple of 7 , namely $1001=7 \\cdot 11 \\cdot 13$, either $1000-k$ and $k+1$ are both multiples of 7 or neithe... | [
"979"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,607 | An integer-valued function $f$ is called tenuous if $f(x)+f(y)>x^{2}$ for all positive integers $x$ and $y$. Let $g$ be a tenuous function such that $g(1)+g(2)+\cdots+g(20)$ is as small as possible. Compute the minimum possible value for $g(14)$. | [
"For a tenuous function $g$, let $S_{g}=g(1)+g(2)+\\cdots+g(20)$. Then:\n\n$$\n\\begin{aligned}\nS_{g} & =(g(1)+g(20))+(g(2)+g(19))+\\cdots+(g(10)+g(11)) \\\\\n& \\geq\\left(20^{2}+1\\right)+\\left(19^{2}+1\\right)+\\cdots+\\left(11^{2}+1\\right) \\\\\n& =10+\\sum_{k=11}^{20} k^{2} \\\\\n& =2495 .\n\\end{aligned}\n... | [
"136"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,608 | Let $T=(0,0), N=(2,0), Y=(6,6), W=(2,6)$, and $R=(0,2)$. Compute the area of pentagon $T N Y W R$. | [
"Pentagon $T N Y W R$ fits inside square $T A Y B$, where $A=(6,0)$ and $B=(0,6)$. The region of $T A Y B$ not in $T N Y W R$ consists of triangles $\\triangle N A Y$ and $\\triangle W B R$, as shown below.\n\n<img_3654>\n\nThus\n\n$$\n\\begin{aligned}\n{[T N Y W R] } & =[T A Y B]-[N A Y]-[W B R] \\\\\n& =6^{2}-\\f... | [
"20"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,609 | Let $T=20$. The lengths of the sides of a rectangle are the zeroes of the polynomial $x^{2}-3 T x+T^{2}$. Compute the length of the rectangle's diagonal. | [
"Let $r$ and $s$ denote the zeros of the polynomial $x^{2}-3 T x+T^{2}$. The rectangle's diagonal has length $\\sqrt{r^{2}+s^{2}}=\\sqrt{(r+s)^{2}-2 r s}$. Recall that for a quadratic polynomial $a x^{2}+b x+c$, the sum of its zeros is $-b / a$, and the product of its zeros is $c / a$. In this particular instance, ... | [
"$20 \\sqrt{7}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,610 | Let $T=20 \sqrt{7}$. Let $w>0$ be a real number such that $T$ is the area of the region above the $x$-axis, below the graph of $y=\lceil x\rceil^{2}$, and between the lines $x=0$ and $x=w$. Compute $\lceil 2 w\rceil$. | [
"Write $w=k+\\alpha$, where $k$ is an integer, and $0 \\leq \\alpha<1$. Then\n\n$$\nT=1^{2}+2^{2}+\\cdots+k^{2}+(k+1)^{2} \\cdot \\alpha .\n$$\n\nComputing $\\lceil 2 w\\rceil$ requires computing $w$ to the nearest half-integer. First obtain the integer $k$. As $\\sqrt{7}>2$, with $T=20 \\sqrt{7}$, one obtains $T>4... | [
"10"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,611 | Compute the least positive integer $n$ such that $\operatorname{gcd}\left(n^{3}, n !\right) \geq 100$. | [
"Note that if $p$ is prime, then $\\operatorname{gcd}\\left(p^{3}, p !\\right)=p$. A good strategy is to look for values of $n$ with several (not necessarily distinct) prime factors so that $n^{3}$ and $n$ ! will have many factors in common. For example, if $n=6, n^{3}=216=2^{3} \\cdot 3^{3}$ and $n !=720=2^{4} \\c... | [
"8"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,612 | Let $T=8$. At a party, everyone shakes hands with everyone else exactly once, except Ed, who leaves early. A grand total of $20 T$ handshakes take place. Compute the number of people at the party who shook hands with Ed. | [
"If there were $n$ people at the party, including Ed, and if Ed had not left early, there would have been $\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)$ handshakes. Because Ed left early, the number of handshakes is strictly less than that, but greater than $\\left(\\begin{array}{c}n-1 \\\\ 2\\end{array}\\r... | [
"7"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,613 | Let $T=7$. Given the sequence $u_{n}$ such that $u_{3}=5, u_{6}=89$, and $u_{n+2}=3 u_{n+1}-u_{n}$ for integers $n \geq 1$, compute $u_{T}$. | [
"By the recursive definition, notice that $u_{6}=89=3 u_{5}-u_{4}$ and $u_{5}=3 u_{4}-u_{3}=3 u_{4}-5$. This is a linear system of equations. Write $3 u_{5}-u_{4}=89$ and $-3 u_{5}+9 u_{4}=15$ and add to obtain $u_{4}=13$. Now apply the recursive definition to obtain $u_{5}=34$ and $u_{7}=\\mathbf{2 3 3}$.",
"Not... | [
"233"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,614 | In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group.
It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make.
| ARMLton | |
| :--- | :--- |
| Resident | Dishes |
| Paul | pie, turkey |
| Arnold | pie, salad |
| Kelly | salad, broth |
| ARMLville | |
| :--- | :--- |
| Resident | Dishes |
| Sally | steak, calzones |
| Ross | calzones, pancakes |
| David | steak, pancakes |
The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}.
A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad.
Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$.
Compute $\operatorname{pop}\left(\mathcal{F}_{17}\right)$. | [
"There are $\\left(\\begin{array}{c}17 \\\\ 2\\end{array}\\right)=136$ possible pairs of dishes, so $\\mathcal{F}_{17}$ must have 136 people."
] | [
"136"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,615 | In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group.
It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make.
| ARMLton | |
| :--- | :--- |
| Resident | Dishes |
| Paul | pie, turkey |
| Arnold | pie, salad |
| Kelly | salad, broth |
| ARMLville | |
| :--- | :--- |
| Resident | Dishes |
| Sally | steak, calzones |
| Ross | calzones, pancakes |
| David | steak, pancakes |
The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}.
A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad.
Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$.
Let $n=\operatorname{pop}\left(\mathcal{F}_{d}\right)$. In terms of $n$, compute $d$. | [
"With $d$ dishes there are $\\left(\\begin{array}{l}d \\\\ 2\\end{array}\\right)=\\frac{d^{2}-d}{2}$ possible pairs, so $n=\\frac{d^{2}-d}{2}$. Then $2 n=d^{2}-d$, or $d^{2}-d-2 n=0$. Using the quadratic formula yields $d=\\frac{1+\\sqrt{1+8 n}}{2}$ (ignoring the negative value)."
] | [
"$d=\\frac{1+\\sqrt{1+8 n}}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,625 | In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group.
It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make.
| ARMLton | |
| :--- | :--- |
| Resident | Dishes |
| Paul | pie, turkey |
| Arnold | pie, salad |
| Kelly | salad, broth |
| ARMLville | |
| :--- | :--- |
| Resident | Dishes |
| Sally | steak, calzones |
| Ross | calzones, pancakes |
| David | steak, pancakes |
The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}.
A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad.
Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$.
In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$.
For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups.
For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$.
If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$.
A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles.
Compute the number of distinguishable resident cycles of length 6 in $\mathcal{F}_{8}$. | [
"Because the town is full, each pair of dishes is cooked by exactly one resident, so it is simplest to identify residents by the pairs of dishes they cook. Suppose the first resident cooks $\\left(d_{1}, d_{2}\\right)$, the second resident $\\left(d_{2}, d_{3}\\right)$, the third resident $\\left(d_{3}, d_{4}\\righ... | [
"1680"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,626 | In each town in ARMLandia, the residents have formed groups, which meet each week to share math problems and enjoy each others' company over a potluck-style dinner. Each town resident belongs to exactly one group. Every week, each resident is required to make one dish and to bring it to his/her group.
It so happens that each resident knows how to make precisely two dishes. Moreover, no two residents of a town know how to make the same pair of dishes. Shown below are two example towns. In the left column are the names of the town's residents. Adjacent to each name is the list of dishes that the corresponding resident knows how to make.
| ARMLton | |
| :--- | :--- |
| Resident | Dishes |
| Paul | pie, turkey |
| Arnold | pie, salad |
| Kelly | salad, broth |
| ARMLville | |
| :--- | :--- |
| Resident | Dishes |
| Sally | steak, calzones |
| Ross | calzones, pancakes |
| David | steak, pancakes |
The population of a town $T$, denoted $\operatorname{pop}(T)$, is the number of residents of $T$. Formally, the town itself is simply the set of its residents, denoted by $\left\{r_{1}, \ldots, r_{\mathrm{pop}(T)}\right\}$ unless otherwise specified. The set of dishes that the residents of $T$ collectively know how to make is denoted $\operatorname{dish}(T)$. For example, in the town of ARMLton described above, pop(ARMLton) $=3$, and dish(ARMLton) $=$ \{pie, turkey, salad, broth\}.
A town $T$ is called full if for every pair of dishes in $\operatorname{dish}(T)$, there is exactly one resident in $T$ who knows how to make those two dishes. In the examples above, ARMLville is a full town, but ARMLton is not, because (for example) nobody in ARMLton knows how to make both turkey and salad.
Denote by $\mathcal{F}_{d}$ a full town in which collectively the residents know how to make $d$ dishes. That is, $\left|\operatorname{dish}\left(\mathcal{F}_{d}\right)\right|=d$.
In order to avoid the embarrassing situation where two people bring the same dish to a group dinner, if two people know how to make a common dish, they are forbidden from participating in the same group meeting. Formally, a group assignment on $T$ is a function $f: T \rightarrow\{1,2, \ldots, k\}$, satisfying the condition that if $f\left(r_{i}\right)=f\left(r_{j}\right)$ for $i \neq j$, then $r_{i}$ and $r_{j}$ do not know any of the same recipes. The group number of a town $T$, denoted $\operatorname{gr}(T)$, is the least positive integer $k$ for which there exists a group assignment on $T$.
For example, consider once again the town of ARMLton. A valid group assignment would be $f($ Paul $)=f($ Kelly $)=1$ and $f($ Arnold $)=2$. The function which gives the value 1 to each resident of ARMLton is not a group assignment, because Paul and Arnold must be assigned to different groups.
For a dish $D$, a resident is called a $D$-chef if he or she knows how to make the dish $D$. Define $\operatorname{chef}_{T}(D)$ to be the set of residents in $T$ who are $D$-chefs. For example, in ARMLville, David is a steak-chef and a pancakes-chef. Further, $\operatorname{chef}_{\text {ARMLville }}($ steak $)=\{$ Sally, David $\}$.
If $\operatorname{gr}(T)=\left|\operatorname{chef}_{T}(D)\right|$ for some $D \in \operatorname{dish}(T)$, then $T$ is called homogeneous. If $\operatorname{gr}(T)>\left|\operatorname{chef}_{T}(D)\right|$ for each dish $D \in \operatorname{dish}(T)$, then $T$ is called heterogeneous. For example, ARMLton is homogeneous, because $\operatorname{gr}($ ARMLton $)=2$ and exactly two chefs make pie, but ARMLville is heterogeneous, because even though each dish is only cooked by two chefs, $\operatorname{gr}($ ARMLville $)=3$.
A resident cycle is a sequence of distinct residents $r_{1}, \ldots, r_{n}$ such that for each $1 \leq i \leq n-1$, the residents $r_{i}$ and $r_{i+1}$ know how to make a common dish, residents $r_{n}$ and $r_{1}$ know how to make a common dish, and no other pair of residents $r_{i}$ and $r_{j}, 1 \leq i, j \leq n$ know how to make a common dish. Two resident cycles are indistinguishable if they contain the same residents (in any order), and distinguishable otherwise. For example, if $r_{1}, r_{2}, r_{3}, r_{4}$ is a resident cycle, then $r_{2}, r_{1}, r_{4}, r_{3}$ and $r_{3}, r_{2}, r_{1}, r_{4}$ are indistinguishable resident cycles.
In terms of $k$ and $d$, find the number of distinguishable resident cycles of length $k$ in $\mathcal{F}_{d}$. | [
"First, we compute the number of distinguishable resident cycles of length 6 in $\\mathcal{F}_{8}$.\n\nBecause the town is full, each pair of dishes is cooked by exactly one resident, so it is simplest to identify residents by the pairs of dishes they cook. Suppose the first resident cooks $\\left(d_{1}, d_{2}\\rig... | [
"$\\frac{d !}{2 k(d-k) !}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English |
2,635 | A student computed the repeating decimal expansion of $\frac{1}{N}$ for some integer $N$, but inserted six extra digits into the repetend to get $.0 \overline{0231846597}$. Compute the value of $N$. | [
"Because the given repetend has ten digits, the original had four digits. If $\\frac{1}{N}=.0 \\underline{A} \\underline{B} \\underline{C} \\underline{D}=$ $\\frac{\\underline{A} \\underline{B} \\underline{C} \\underline{D}}{99990}$, then the numerator must divide $99990=10 \\cdot 99 \\cdot 101=2 \\cdot 3^{2} \\cdo... | [
"606"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,636 | Let $n$ be a four-digit number whose square root is three times the sum of the digits of $n$. Compute $n$. | [
"Because $\\sqrt{n}$ is a multiple of $3, n$ must be a multiple of 9 . Therefore the sum of the digits of $n$ is a multiple of 9 . Thus $\\sqrt{n}$ must be a multiple of 27 , which implies that $n$ is a multiple of $27^{2}$. The only candidates to consider are $54^{2}(=2916)$ and $81^{2}(=6561)$, and only 2916 sati... | [
"2916"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,637 | Compute the sum of the reciprocals of the positive integer divisors of 24. | [
"The map $n \\mapsto 24 / n$ establishes a one-to-one correspondence among the positive integer divisors of 24 . Thus\n\n$$\n\\begin{aligned}\n\\sum_{\\substack{n \\mid 24 \\\\\nn>0}} \\frac{1}{n} & =\\sum_{\\substack{n \\mid 24 \\\\\nn>0}} \\frac{1}{24 / n} \\\\\n& =\\frac{1}{24} \\sum_{\\substack{n \\mid 24 \\\\\... | [
"$\\frac{5}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,638 | There exists a digit $Y$ such that, for any digit $X$, the seven-digit number $\underline{1} \underline{2} \underline{3} \underline{X} \underline{5} \underline{Y} \underline{7}$ is not a multiple of 11. Compute $Y$. | [
"Consider the ordered pairs of digits $(X, Y)$ for which $\\underline{1} \\underline{2} \\underline{3} \\underline{X} \\underline{5} \\underline{Y} \\underline{7}$ is a multiple of 11 . Recall that a number is a multiple of 11 if and only if the alternating sum of the digits is a multiple of 11 . Because $1+3+5+7=1... | [
"4"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | English |
2,639 | A point is selected at random from the interior of a right triangle with legs of length $2 \sqrt{3}$ and 4 . Let $p$ be the probability that the distance between the point and the nearest vertex is less than 2. Then $p$ can be written in the form $a+\sqrt{b} \pi$, where $a$ and $b$ are rational numbers. Compute $(a, b)$. | [
"Label the triangle as $\\triangle A B C$, with $A B=2 \\sqrt{3}$ and $B C=4$. Let $D$ and $E$ lie on $\\overline{A B}$ such that $D B=A E=2$. Let $F$ be the midpoint of $\\overline{B C}$, so that $B F=F C=2$. Let $G$ and $H$ lie on $\\overline{A C}$, with $A G=H C=2$. Now draw the arcs of radius 2 between $E$ and ... | [
"$(\\frac{1}{4}, \\frac{1}{27})$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Geometry | Math | English |
2,640 | The square $A R M L$ is contained in the $x y$-plane with $A=(0,0)$ and $M=(1,1)$. Compute the length of the shortest path from the point $(2 / 7,3 / 7)$ to itself that touches three of the four sides of square $A R M L$. | [
"Consider repeatedly reflecting square $A R M L$ over its sides so that the entire plane is covered by copies of $A R M L$. A path starting at $(2 / 7,3 / 7)$ that touches one or more sides and returns to $(2 / 7,3 / 7)$ corresponds to a straight line starting at $(2 / 7,3 / 7)$ and ending at the image of $(2 / 7,3... | [
"$\\frac{2}{7} \\sqrt{53}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,641 | For each positive integer $k$, let $S_{k}$ denote the infinite arithmetic sequence of integers with first term $k$ and common difference $k^{2}$. For example, $S_{3}$ is the sequence $3,12,21, \ldots$ Compute the sum of all $k$ such that 306 is an element of $S_{k}$. | [
"If 306 is an element of $S_{k}$, then there exists an integer $m \\geq 0$ such that $306=k+m k^{2}$. Thus $k \\mid 306$ and $k^{2} \\mid 306-k$. The second relation can be rewritten as $k \\mid 306 / k-1$, which implies that $k \\leq \\sqrt{306}$ unless $k=306$. The prime factorization of 306 is $2 \\cdot 3^{2} \\... | [
"326"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,642 | Compute the sum of all values of $k$ for which there exist positive real numbers $x$ and $y$ satisfying the following system of equations.
$$
\left\{\begin{aligned}
\log _{x} y^{2}+\log _{y} x^{5} & =2 k-1 \\
\log _{x^{2}} y^{5}-\log _{y^{2}} x^{3} & =k-3
\end{aligned}\right.
$$ | [
"Let $\\log _{x} y=a$. Then the first equation is equivalent to $2 a+\\frac{5}{a}=2 k-1$, and the second equation is equivalent to $\\frac{5 a}{2}-\\frac{3}{2 a}=k-3$. Solving this system by eliminating $k$ yields the quadratic equation $3 a^{2}+5 a-8=0$, hence $a=1$ or $a=-\\frac{8}{3}$. Substituting each of these... | [
"$\\frac{43}{48}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,643 | Let $W=(0,0), A=(7,0), S=(7,1)$, and $H=(0,1)$. Compute the number of ways to tile rectangle $W A S H$ with triangles of area $1 / 2$ and vertices at lattice points on the boundary of WASH. | [
"Define a fault line to be a side of a tile other than its base. Any tiling of $W A S H$ can be represented as a sequence of tiles $t_{1}, t_{2}, \\ldots, t_{14}$, where $t_{1}$ has a fault line of $\\overline{W H}, t_{14}$ has a fault line of $\\overline{A S}$, and where $t_{k}$ and $t_{k+1}$ share a fault line fo... | [
"3432"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,644 | Compute $\sin ^{2} 4^{\circ}+\sin ^{2} 8^{\circ}+\sin ^{2} 12^{\circ}+\cdots+\sin ^{2} 176^{\circ}$. | [
"Because $\\cos 2 x=1-2 \\sin ^{2} x, \\sin ^{2} x=\\frac{1-\\cos 2 x}{2}$. Thus the desired sum can be rewritten as\n\n$$\n\\frac{1-\\cos 8^{\\circ}}{2}+\\frac{1-\\cos 16^{\\circ}}{2}+\\cdots+\\frac{1-\\cos 352^{\\circ}}{2}=\\frac{44}{2}-\\frac{1}{2}\\left(\\cos 8^{\\circ}+\\cos 16^{\\circ}+\\cdots+\\cos 352^{\\ci... | [
"$\\frac{45}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,645 | Compute the area of the region defined by $x^{2}+y^{2} \leq|x|+|y|$. | [
"Call the region $R$, and let $R_{q}$ be the portion of $R$ in the $q^{\\text {th }}$ quadrant. Noting that the point $(x, y)$ is in $R$ if and only if $( \\pm x, \\pm y)$ is in $R$, it follows that $\\left[R_{1}\\right]=\\left[R_{2}\\right]=\\left[R_{3}\\right]=\\left[R_{4}\\right]$, and so $[R]=4\\left[R_{1}\\rig... | [
"$2+\\pi$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English |
2,646 | The arithmetic sequences $a_{1}, a_{2}, a_{3}, \ldots, a_{20}$ and $b_{1}, b_{2}, b_{3}, \ldots, b_{20}$ consist of 40 distinct positive integers, and $a_{20}+b_{14}=1000$. Compute the least possible value for $b_{20}+a_{14}$. | [
"Write $a_{n}=a_{1}+r(n-1)$ and $b_{n}=b_{1}+s(n-1)$. Then $a_{20}+b_{14}=a_{1}+b_{1}+19 r+13 s$, while $b_{20}+a_{14}=a_{1}+b_{1}+13 r+19 s=a_{20}+b_{14}+6(s-r)$. Because both sequences consist only of integers, $r$ and $s$ must be integers, so $b_{20}+a_{14} \\equiv a_{20}+b_{14} \\bmod 6$. Thus the least possibl... | [
"10"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
2,647 | Compute the ordered triple $(x, y, z)$ representing the farthest lattice point from the origin that satisfies $x y-z^{2}=y^{2} z-x=14$. | [
"First, eliminate $x: y\\left(y^{2} z-x\\right)+\\left(x y-z^{2}\\right)=14(y+1) \\Rightarrow z^{2}-y^{3} z+14(y+1)=0$. Viewed as a quadratic in $z$, this equation implies $z=\\frac{y^{3} \\pm \\sqrt{y^{6}-56(y+1)}}{2}$. In order for $z$ to be an integer, the discriminant must be a perfect square. Because $y^{6}=\\... | [
"$(-266,-3,-28)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Tuple | null | Open-ended | Algebra | Math | English |
2,648 | The sequence $a_{1}, a_{2}, a_{3}, \ldots$ is a geometric sequence with $a_{20}=8$ and $a_{14}=2^{21}$. Compute $a_{21}$. | [
"Let $r$ be the common ratio of the sequence. Then $a_{20}=r^{20-14} \\cdot a_{14}$, hence $8=r^{6} \\cdot 2^{21} \\Rightarrow r^{6}=$ $\\frac{2^{3}}{2^{21}}=2^{-18}$, so $r=2^{-3}=\\frac{1}{8}$. Thus $a_{21}=r \\cdot a_{20}=\\frac{1}{8} \\cdot 8=\\mathbf{1}$."
] | [
"1"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Algebra | Math | English |
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