link stringlengths 75 84 | letter stringclasses 5
values | answer float64 0 2,935,363,332B | problem stringlengths 14 5.33k | solution listlengths 1 13 |
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https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_15 | B | 18 | A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$ | [
"For each digit $a=1,2,\\ldots,9$ there are $10\\cdot10$ (ways of choosing $b$ and $c$ ) palindromes. So the $a$ s contribute $(1+2+\\cdots+9)(100)(10^4+1)$ to the sum.\nFor each digit $b=0,1,2,\\ldots,9$ there are $9\\cdot10$ (since $a \\neq 0$ ) palindromes. So the $b$ s contribute $(0+1+2+\\cdots+9)(90)(10^3+10)... |
https://artofproblemsolving.com/wiki/index.php/2005_AMC_8_Problems/Problem_16 | D | 13 | A five-legged Martian has a drawer full of socks, each of which is red, white or blue, and there are at least five socks of each color. The Martian pulls out one sock at a time without looking. How many socks must the Martian remove from the drawer to be certain there will be 5 socks of the same color?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 13\qquad\textbf{(E)}\ 15$ | [
"The Martian can pull out $12$ socks, $4$ of each color, without having $5$ of the same kind yet. However, the next one he pulls out must be the fifth of one of the colors so he must remove $\\boxed{13}$ socks."
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_10 | null | 2.4 | A flagpole is originally $5$ meters tall. A hurricane snaps the flagpole at a point $x$ meters above the ground so that the upper part, still attached to the stump, touches the ground $1$ meter away from the base. What is $x$
$\text{(A) } 2.0 \qquad \text{(B) } 2.1 \qquad \text{(C) } 2.2 \qquad \text{(D) } 2.3 \qquad \text{(E) } 2.4$ | [
"The broken flagpole forms a right triangle with legs $1$ and $x$ , and hypotenuse $5-x$ . The Pythagorean theorem now states that $1^2 + x^2 = (5-x)^2$ , hence $10x = 24$ , and $x=\\boxed{2.4}$",
"Let $AB$ represent the flagpole in the diagram above. After the flagpole breaks at point $D$ , its tip lies at point... |
https://artofproblemsolving.com/wiki/index.php/2020_AIME_I_Problems/Problem_6 | null | 173 | A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$ | [
" Set the common radius to $r$ . First, take the cross section of the sphere sitting in the hole of radius $1$ . If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_10 | E | 70 | A flower bouquet contains pink roses, red roses, pink carnations, and red carnations. One third of the pink flowers are roses, three fourths of the red flowers are carnations, and six tenths of the flowers are pink. What percent of the flowers are carnations?
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 40 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 70$ | [
"Let the total amount of flowers be $x$ . Thus, the number of pink flowers is $0.6x$ , and the number of red flowers is $0.4x$ . The number of pink carnations is $\\frac{2}{3}(0.6x) = 0.4x$ and the number of red carnations is $\\frac{3}{4}(0.4x) = 0.3x$ . Summing these, the total number of carnations is $0.4x+0.... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_3 | A | 10 | A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?
$\text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24$ | [
"If the Cougars won by a margin of 14 points, then the Panthers' score would be half of (34-14). That's 10 $\\Rightarrow \\boxed{10}$",
"Let the Panthers' score be $x$ . The Cougars then scored $x+14$ . Since the teams combined scored $34$ , we get $x+x+14=34 \\\\ \\rightarrow 2x+14=34 \\\\ \\rightarrow 2x=20 \\\... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10B_Problems/Problem_3 | A | 10 | A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of $34$ points, and the Cougars won by a margin of $14$ points. How many points did the Panthers score?
$\textbf{(A) } 10\qquad \textbf{(B) } 14\qquad \textbf{(C) } 17\qquad \textbf{(D) } 20\qquad \textbf{(E) } 24$ | [
"Let $x$ be the number of points scored by the Cougars, and $y$ be the number of points scored by the Panthers. The problem is asking for the value of $y$ \\begin{align*} x+y &= 34 \\\\ x-y &= 14 \\\\ 2x &= 48 \\\\ x &= 24 \\\\ y &= \\boxed{10}"
] |
https://artofproblemsolving.com/wiki/index.php/2022_AMC_10A_Problems/Problem_20 | E | 206 | A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are $57$ $60$ , and $91$ . What is the fourth term of this sequence?
$\textbf{(A) } 190 \qquad \textbf{(B) } 194 \qquad \textbf{(C) } 198 \qquad \textbf{(D) } 202 \qquad \textbf{(E) } 206$ | [
"Let the arithmetic sequence be $a,a+d,a+2d,a+3d$ and the geometric sequence be $b,br,br^2,br^3.$\nWe are given that \\begin{align*} a+b&=57, \\\\ a+d+br&=60, \\\\ a+2d+br^2&=91, \\end{align*} and we wish to find $a+3d+br^3.$\nSubtracting the first equation from the second and the second equation from the third, we... |
https://artofproblemsolving.com/wiki/index.php/2012_AIME_I_Problems/Problem_11 | null | 373 | A frog begins at $P_0 = (0,0)$ and makes a sequence of jumps according to the following rule: from $P_n = (x_n, y_n),$ the frog jumps to $P_{n+1},$ which may be any of the points $(x_n + 7, y_n + 2),$ $(x_n + 2, y_n + 7),$ $(x_n - 5, y_n - 10),$ or $(x_n - 10, y_n - 5).$ There are $M$ points $(x, y)$ with $|x| + |y| \le 100$ that can be reached by a sequence of such jumps. Find the remainder when $M$ is divided by $1000.$ | [
"First of all, it is easy to see by induction that for any $P(x,y)$ in the frog's jump sequence, $x+y$ will be a multiple of $3$ and $x-y$ will be a multiple of $5.$ The base case $(x,y) = (0,0)$ obviously satisfies the constraints and if $x+y = 3n$ and $x-y = 5m,$ any of the four transformations will sustain this ... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_I_Problems/Problem_6 | null | 169 | A frog is placed at the origin on the number line , and moves according to the following rule: in a given move, the frog advances to either the closest point with a greater integer coordinate that is a multiple of 3, or to the closest point with a greater integer coordinate that is a multiple of 13. A move sequence is a sequence of coordinates that correspond to valid moves, beginning with 0 and ending with 39. For example, $0,\ 3,\ 6,\ 13,\ 15,\ 26,\ 39$ is a move sequence. How many move sequences are possible for the frog? | [
"Another way would be to use a table representing the number of ways to reach a certain number\n$\\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} 0 & 3 & 6 & 9 & 12 & 13 & 15 & 18 & 21 & 24 & 26 & 27 & 30 & 33 & 36 \\\\ \\hline 1 & 1 & 1 & 1 & 1 & 5 & 6 & 6 & 6 & 6 & 29 & 35 & 35 & 35 & 35 \\\\ \\end{tabular}$\nHow... |
https://artofproblemsolving.com/wiki/index.php/2018_AIME_II_Problems/Problem_8 | null | 556 | A frog is positioned at the origin of the coordinate plane. From the point $(x, y)$ , the frog can jump to any of the points $(x + 1, y)$ $(x + 2, y)$ $(x, y + 1)$ , or $(x, y + 2)$ . Find the number of distinct sequences of jumps in which the frog begins at $(0, 0)$ and ends at $(4, 4)$ | [
"We solve this problem by working backwards. Notice, the only points the frog can be on to jump to $(4,4)$ in one move are $(2,4),(3,4),(4,2),$ and $(4,3)$ . This applies to any other point, thus we can work our way from $(0,0)$ to $(4,4)$ , recording down the number of ways to get to each point recursively.\n$(0,0... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12B_Problems/Problem_11 | null | 3 | A frog located at $(x,y)$ , with both $x$ and $y$ integers, makes successive jumps of length $5$ and always lands on points with integer coordinates. Suppose that the frog starts at $(0,0)$ and ends at $(1,0)$ . What is the smallest possible number of jumps the frog makes?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | [
"Since the frog always jumps in length $5$ and lands on a lattice point, the sum of its coordinates must change either by $5$ (by jumping parallel to the x- or y-axis), or by $3$ or $4$ (3-4-5 right triangle).\nBecause either $1$ $5$ , or $7$ is always the change of the sum of the coordinates, the sum of the coordi... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_13 | B | 58 | A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$ , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0,0), (0,4), (4,4),$ and $(4,0)$ . What is the probability that the sequence of jumps ends on a vertical side of the square?
$\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78$ | [
"Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is $\\frac{1}{4} \\cdot 1 = \\frac{1}{4}$ . If the frog goes to the right, it will be in the center of the square at $(2,2)$ , and by symmet... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_11 | B | 58 | A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$ , and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0,0), (0,4), (4,4),$ and $(4,0)$ . What is the probability that the sequence of jumps ends on a vertical side of the square?
$\textbf{(A)}\ \frac12\qquad\textbf{(B)}\ \frac 58\qquad\textbf{(C)}\ \frac 23\qquad\textbf{(D)}\ \frac34\qquad\textbf{(E)}\ \frac 78$ | [
"Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is $\\frac{1}{4} \\cdot 1 = \\frac{1}{4}$ . If the frog goes to the right, it will be in the center of the square at $(2,2)$ , and by symmet... |
https://artofproblemsolving.com/wiki/index.php/2012_AMC_12A_Problems/Problem_5 | D | 64 | A fruit salad consists of blueberries, raspberries, grapes, and cherries. The fruit salad has a total of $280$ pieces of fruit. There are twice as many raspberries as blueberries, three times as many grapes as cherries, and four times as many cherries as raspberries. How many cherries are there in the fruit salad?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$ | [
"So let the number of blueberries be $b,$ the number of raspberries be $r,$ the number of grapes be $g,$ and finally the number of cherries be $c.$\nObserve that since there are $280$ pieces of fruit, \\[b+r+g+c=280.\\]\nSince there are twice as many raspberries as blueberries, \\[2b=r.\\]\nThe fact that there are ... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_12 | B | 6 | A function $f$ from the integers to the integers is defined as follows:
\[f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ n/2 &\text{if n is even}\end{cases}\]
Suppose $k$ is odd and $f(f(f(k))) = 27$ . What is the sum of the digits of $k$
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$ | [
"Going out the final step, if you have $f(f(f(k))) = 27$ , you would have to have $f(f(k))) = 51$ or $f(f(k)) = 108$\nIf you doubled either of these, $k$ would not be odd. So you must subtract $3$\nIf you subtract $3$ from $51$ , you would compute $f(48)$ , which would halve it, and not add the $3$ back.\nIf you s... |
https://artofproblemsolving.com/wiki/index.php/1999_AIME_Problems/Problem_9 | null | 259 | A function $f$ is defined on the complex numbers by $f(z)=(a+bi)z,$ where $a_{}$ and $b_{}$ are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin . Given that $|a+bi|=8$ and that $b^2=m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers, find $m+n.$ | [
"Suppose we pick an arbitrary point on the complex plane , say $(1,1)$ . According to the definition of $f(z)$ \\[f(1+i) = (a+bi)(1+i) = (a-b) + (a+b)i,\\] this image must be equidistant to $(1,1)$ and $(0,0)$ . Thus the image must lie on the line with slope $-1$ and which passes through $\\left(\\frac 12, \\frac12... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_20 | B | 2,017 | A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | [
"For all integers $n \\geq 7,$ note that \\begin{align*} f(n)&=f(n-1)-f(n-2)+n \\\\ &=[f(n-2)-f(n-3)+n-1]-f(n-2)+n \\\\ &=-f(n-3)+2n-1 \\\\ &=-[f(n-4)-f(n-5)+n-3]+2n-1 \\\\ &=-f(n-4)+f(n-5)+n+2 \\\\ &=-[f(n-5)-f(n-6)+n-4]+f(n-5)+n+2 \\\\ &=f(n-6)+6. \\end{align*} It follows that \\begin{align*} f(2018)&=f(2012)+6 \... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_18 | B | 2,017 | A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$ | [
"For all integers $n \\geq 7,$ note that \\begin{align*} f(n)&=f(n-1)-f(n-2)+n \\\\ &=[f(n-2)-f(n-3)+n-1]-f(n-2)+n \\\\ &=-f(n-3)+2n-1 \\\\ &=-[f(n-4)-f(n-5)+n-3]+2n-1 \\\\ &=-f(n-4)+f(n-5)+n+2 \\\\ &=-[f(n-5)-f(n-6)+n-4]+f(n-5)+n+2 \\\\ &=f(n-6)+6. \\end{align*} It follows that \\begin{align*} f(2018)&=f(2012)+6 \... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_21 | A | 28 | A game board consists of $64$ squares that alternate in color between black and white. The figure below shows square $P$ in the bottom row and square $Q$ in the top row. A marker is placed at $P.$ A step consists of moving the marker onto one of the adjoining white squares in the row above. How many $7$ -step paths are there from $P$ to $Q?$ (The figure shows a sample path.)
[asy]//diagram by SirCalcsALot size(200); int[] x = {6, 5, 4, 5, 6, 5, 6}; int[] y = {1, 2, 3, 4, 5, 6, 7}; int N = 7; for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)); if ((i+j) % 2 == 0) { filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,black); } } } for (int i = 0; i < N; ++i) { draw(circle((x[i],y[i])+(0.5,0.5),0.35)); } label("$P$", (5.5, 0.5)); label("$Q$", (6.5, 7.5)); [/asy]
$\textbf{(A) }28 \qquad \textbf{(B) }30 \qquad \textbf{(C) }32 \qquad \textbf{(D) }33 \qquad \textbf{(E) }35$ | [
"Notice that, in order to step onto any particular white square, the marker must have come from one of the $1$ or $2$ white squares immediately beneath it (since the marker can only move on white squares). This means that the number of ways to move from $P$ to that square is the sum of the numbers of ways to move f... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_8 | B | 37 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$ | [
"We look at a set of three rounds, where the players begin with $x+1$ $x$ , and $x-1$ tokens.\nAfter three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$ $2$ , and $1$ tokens, re... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_7 | B | 37 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$ | [
"We look at a set of three rounds, where the players begin with $x+1$ $x$ , and $x-1$ tokens.\nAfter three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$ $2$ , and $1$ tokens, re... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_I_Problems/Problem_9 | null | 420 | A game show offers a contestant three prizes A, B and C, each of which is worth a whole number of dollars from $$ 1$ to $$ 9999$ inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A, B, C. As a hint, the digits of the three prices are given. On a particular day, the digits given were $1, 1, 1, 1, 3, 3, 3$ . Find the total number of possible guesses for all three prizes consistent with the hint. | [
"[Clarification: You are supposed to find the number of all possible tuples of prices, $(A, B, C)$ , that could have been on that day.]\nSince we have three numbers, consider the number of ways we can put these three numbers together in a string of 7 digits. For example, if $A=113, B=13, C=31$ , then the string is\... |
https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_11 | null | 106 | A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$ | [
"First notice that there is no difference between the maple trees and the oak trees; we have only two types, birch trees and \"non-birch\" trees. (If you don't believe this reasoning, think about it. You could also differentiate the tall oak trees from the short oak trees, and the maple trees with many branches as ... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_20 | E | 8 | A geometric sequence $(a_n)$ has $a_1=\sin x$ $a_2=\cos x$ , and $a_3= \tan x$ for some real number $x$ . For what value of $n$ does $a_n=1+\cos x$
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$ | [
"By the defintion of a geometric sequence, we have $\\cos^2x=\\sin x \\tan x$ . Since $\\tan x=\\frac{\\sin x}{\\cos x}$ , we can rewrite this as $\\cos^3x=\\sin^2x$\nThe common ratio of the sequence is $\\frac{\\cos x}{\\sin x}$ , so we can write\n\\[a_1= \\sin x\\] \\[a_2= \\cos x\\] \\[a_3= \\frac{\\cos^2x}{\\si... |
https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_13 | null | 931 | A given sequence $r_1, r_2, \dots, r_n$ of distinct real numbers can be put in ascending order by means of one or more "bubble passes". A bubble pass through a given sequence consists of comparing the second term with the first term, and exchanging them if and only if the second term is smaller, then comparing the third term with the second term and exchanging them if and only if the third term is smaller, and so on in order, through comparing the last term, $r_n$ , with its current predecessor and exchanging them if and only if the last term is smaller.
The example below shows how the sequence 1, 9, 8, 7 is transformed into the sequence 1, 8, 7, 9 by one bubble pass. The numbers compared at each step are underlined.
Suppose that $n = 40$ , and that the terms of the initial sequence $r_1, r_2, \dots, r_{40}$ are distinct from one another and are in random order. Let $p/q$ , in lowest terms, be the probability that the number that begins as $r_{20}$ will end up, after one bubble pass, in the $30^{\mbox{th}}$ place. Find $p + q$ | [
"If any of $r_1, \\ldots, r_{19}$ is larger than $r_{20}$ , one of these numbers will be compared with $r_{20}$ on the 19th step of the first bubble pass and $r_{20}$ will be moved back to the 19th position. Thus, $r_{20}$ must be the largest of the first 20 terms. In addition, $r_{20}$ must be larger than $r_{21... |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_23 | B | 26 | A gives $B$ as many cents as $B$ has and $C$ as many cents as $C$ has. Similarly, $B$ then gives $A$ and $C$ as many cents as each then has. $C$ , similarly, then gives $A$ and $B$ as many cents as each then has. If each finally has $16$ cents, with how many cents does $A$ start?
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 26\qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 32$ | [
"Let $a$ be number of cents $A$ originally had, $b$ be number of cents $B$ originally had, and $c$ be number of cents $C$ originally had.\nAfter $A$ gave his money away, $A$ has $a-b-c$ cents, $B$ has $2b$ cents, and $C$ has $2c$ cents.\nAfter $B$ gave his money away, $A$ has $2a-2b-2c$ cents, $B$ has $-a+3b-c$ cen... |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_10 | D | 10 | A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $100$ cans, how many rows does it contain?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }8\qquad\mathrm{(C)\ }9\qquad\mathrm{(D)\ }10\qquad\mathrm{(E)\ }11$ | [
"The sum of the first $n$ odd numbers is $n^2$ . As in our case $n^2=100$ , we have $n=\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_8 | D | 10 | A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains $100$ cans, how many rows does it contain?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }8\qquad\mathrm{(C)\ }9\qquad\mathrm{(D)\ }10\qquad\mathrm{(E)\ }11$ | [
"The sum of the first $n$ odd numbers is $n^2$ . As in our case $n^2=100$ , we have $n=\\boxed{10}$"
] |
https://artofproblemsolving.com/wiki/index.php/2004_AMC_10A_Problems/Problem_7 | C | 100 | A grocer stacks oranges in a pyramid-like stack whose rectangular base is $5$ oranges by $8$ oranges. Each orange above the first level rests in a pocket formed by four oranges below. The stack is completed by a single row of oranges. How many oranges are in the stack?
$\mathrm{(A) \ } 96 \qquad \mathrm{(B) \ } 98 \qquad \mathrm{(C) \ } 100 \qquad \mathrm{(D) \ } 101 \qquad \mathrm{(E) \ } 134$ | [
"There are $5\\times8=40$ oranges on the $1^{\\text{st}}$ layer of the stack. The $2^{\\text{nd}}$ layer that is added on top of the first will be a layer of $4\\times7=28$ oranges. When the third layer is added on top of the $2^{\\text{nd}}$ , it will be a layer of $3\\times6=18$ oranges, etc.\nTherefore, there a... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_10A_Problems/Problem_21 | D | 1,925 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 1296\qquad\textbf{(C)}\ 1728\qquad\textbf{(D)}\ 1925\qquad\textbf{(E)}\ 3850$ | [
"Let $x$ be the number of coins. After the $k^{\\text{th}}$ pirate takes his share, $\\frac{12-k}{12}$ of the original amount is left. Thus, we know that\n$x \\cdot \\frac{11}{12} \\cdot \\frac{10}{12} \\cdot \\frac{9}{12} \\cdot \\frac{8}{12} \\cdot \\frac{7}{12} \\cdot \\frac{6}{12} \\cdot \\frac{5}{12} \\cdot ... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_12A_Problems/Problem_17 | D | 1,925 | A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^\text{th}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
$\textbf{(A)} \ 720 \qquad \textbf{(B)} \ 1296 \qquad \textbf{(C)} \ 1728 \qquad \textbf{(D)} \ 1925 \qquad \textbf{(E)} \ 3850$ | [
"The first pirate takes $\\frac{1}{12}$ of the $x$ coins, leaving $\\frac{11}{12} x$\nThe second pirate takes $\\frac{2}{12}$ of the remaining coins, leaving $\\frac{10}{12}\\cdot \\frac{11}{12}*x$\nNote that\n$12^{11} = (2^2 \\cdot 3)^{11} = 2^{22} \\cdot 3^{11}$\n$11! = 11 \\cdot 10 \\cdot 9 \\cdot 8 \\cdot 7 \... |
https://artofproblemsolving.com/wiki/index.php/2009_AIME_II_Problems/Problem_4 | null | 89 | A group of children held a grape-eating contest. When the contest was over, the winner had eaten $n$ grapes, and the child in $k$ -th place had eaten $n+2-2k$ grapes. The total number of grapes eaten in the contest was $2009$ . Find the smallest possible value of $n$ | [
"The total number of grapes eaten can be computed as the sum of the arithmetic progression with initial term $n$ (the number of grapes eaten by the child in $1$ -st place), difference $d=-2$ , and number of terms $c$ . We can easily compute that this sum is equal to $c(n-c+1)$\nHence we have the equation $2009=c(n-... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_8_Problems/Problem_4 | C | 5 | A group of children riding on bicycles and tricycles rode past Billy Bob's house. Billy Bob counted $7$ children and $19$ wheels. How many tricycles were there?
$\mathrm{(A)}\ 2 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 5 \qquad\mathrm{(D)}\ 6 \qquad\mathrm{(E)}\ 7$ | [
"If all the children were riding bicycles, there would be $2 \\times 7=14$ wheels. Each tricycle adds an extra wheel and $19-14=5$ extra wheels are needed, so there are $\\boxed{5}$ tricycles.",
"Setting up an equation, we have $a+b=7$ children and $3a+2b=19$ . Solving for the variables, we get, $a=\\boxed{5}$ tr... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_7 | null | 945 | A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting. | [
"There are $x$ clerks at the beginning, and $t$ clerks are reassigned to another task at the end of each hour. So, $30x+30(x-t)+30(x-2t)+30\\cdot\\frac{10}{60} \\cdot (x-3t)=1775$ , and simplify that we get $19x-21t=355$ .\nNow the problem is to find a reasonable integer solution. Now we know $x= \\frac{355+21t}{19... |
https://artofproblemsolving.com/wiki/index.php/2024_AMC_8_Problems/Problem_21 | E | 24 | A group of frogs (called an army) is living in a tree. A frog turns green when in the shade and turns yellow
when in the sun. Initially, the ratio of green to yellow frogs was $3 : 1$ . Then $3$ green frogs moved to the
sunny side and $5$ yellow frogs moved to the shady side. Now the ratio is $4 : 1$ . What is the difference
between the number of green frogs and the number of yellow frogs now?
$\textbf{(A) } 10\qquad\textbf{(B) } 12\qquad\textbf{(C) } 16\qquad\textbf{(D) } 20\qquad\textbf{(E) } 24$ | [
"Since the original ratio is $3:1$ and the new ratio is $4:1$ , the number of frogs must be a multiple of $12$ , the only solutions left are $(B)$ and $(E)$\nLet's start with $12$ frogs:\nWe must have $9$ frogs in the shade and $3$ frogs in the sun. After the change, there would be $11$ frogs in the shade and $1$ f... |
https://artofproblemsolving.com/wiki/index.php/1994_AJHSME_Problems/Problem_21 | C | 10 | A gumball machine contains $9$ red, $7$ white, and $8$ blue gumballs. The least number of gumballs a person must buy to be sure of getting four gumballs of the same color is
$\text{(A)}\ 8 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18$ | [
"If a person gets three gumballs of each of the three colors, that is, $9$ gumballs, then the $10^{\\text{th}}$ gumball must be the fourth one for one of the colors. Therefore, the person must buy $\\boxed{10}$ gumballs."
] |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_4 | D | 30 | A haunted house has six windows. In how many ways can
Georgie the Ghost enter the house by one window and leave
by a different window?
$\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 18 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 36$ | [
"Georgie can enter the haunted house through any of the six windows. Then, he can leave through any of the remaining five windows.\nSo, Georgie has a total of $6 \\cdot 5$ ways he can enter the house by one window and leave\nby a different window.\nTherefore, we have $\\boxed{30}$ ways."
] |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | E | 409 | A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$ | [
"All angle measures are in degrees.\nLet the first trapezoid be $ABCD$ , where $AB=BC=CD=3$ . Then the second trapezoid is $AFED$ , where $AF=FE=ED=5$ . We look for $AD$\nSince $ABCD$ is an isosceles trapezoid, we know that $\\angle BAD=\\angle CDA$ and, since $AB=BC$ , if we drew $AC$ , we would see $\\angle BCA... |
https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | null | 409 | A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$ | [
"In hexagon $ABCDEF$ , let $AB=BC=CD=3$ and let $DE=EF=FA=5$ . Since arc $BAF$ is one third of the circumference of the circle, it follows that $\\angle BCF = \\angle BEF=60^{\\circ}$ . Similarly, $\\angle CBE =\\angle CFE=60^{\\circ}$ . Let $P$ be the intersection of $\\overline{BE}$ and $\\overline{CF}$ $Q$ that ... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_8 | null | 272 | A hexagon that is inscribed in a circle has side lengths $22$ $22$ $20$ $22$ $22$ , and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$ , where $p$ and $q$ are positive integers. Find $p+q$ | [
"Let us call the hexagon $ABCDEF$ , where $AB=CD=DE=AF=22$ , and $BC=EF=20$ . \nWe can just consider one half of the hexagon, $ABCD$ , to make matters simpler. \nDraw a line from the center of the circle, $O$ , to the midpoint of $BC$ $X$ . Now, draw a line from $O$ to the midpoint of $AB$ $Y$ . Clearly, $\\angle B... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_24 | E | 34 | A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$ | [
"Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know $r > 1$ because the sequence is said to be increasing. We also know that each of $a, ar, ar^2, ar^3$ is an integer. We start by showing that $r$ must also be an integer.\nSuppose not, and say $r = m/n$ where $m>n>1$ , and $\\gcd(m,n)=1$ . Th... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_19 | E | 34 | A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than $100$ points. What was the total number of points scored by the two teams in the first half?
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$ | [
"Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know $r > 1$ because the sequence is said to be increasing. We also know that each of $a, ar, ar^2, ar^3$ is an integer. We start by showing that $r$ must also be an integer.\nSuppose not, and say $r = m/n$ where $m>n>1$ , and $\\gcd(m,n)=1$ . Th... |
https://artofproblemsolving.com/wiki/index.php/2005_AIME_II_Problems/Problem_2 | null | 79 | A hotel packed breakfast for each of three guests. Each breakfast should have consisted of three types of rolls, one each of nut, cheese, and fruit rolls. The preparer wrapped each of the nine rolls and once wrapped, the rolls were indistinguishable from one another. She then randomly put three rolls in a bag for each of the guests. Given that the probability each guest got one roll of each type is $\frac mn,$ where $m$ and $n$ are relatively prime integers , find $m+n.$ | [
"Use construction . We need only calculate the probability the first and second person all get a roll of each type, since then the rolls for the third person are determined.\nOur answer is thus $\\frac{9}{28} \\cdot \\frac{2}{5} = \\frac{9}{70}$ , and $m + n = \\boxed{79}$",
"Call the three different types of rol... |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_14 | B | 1,000 | A house and store were sold for $\textdollar 12,000$ each. The house was sold at a loss of $20\%$ of the cost, and the store at a gain of $20\%$ of the cost. The entire transaction resulted in:
$\textbf{(A) \ }\text{no loss or gain} \qquad \textbf{(B) \ }\text{loss of }\textdollar 1000 \qquad \textbf{(C) \ }\text{gain of }\textdollar 1000 \qquad \textbf{(D) \ }\text{gain of }\textdollar 2000 \qquad \textbf{(E) \ }\text{none of these}$ | [
"Denote the original price of the house and the store as $h$ and $s$ , respectively. It is given that $\\frac{4h}{5}=\\textdollar 12,000$ , and that $\\frac{6s}{5}=\\textdollar 12,000$ . Thus, $h=\\textdollar 15,000$ $s=\\textdollar10,000$ , and $h+s=\\textdollar25,000$ . This value is $\\textdollar1000$ higher tha... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_8_Problems/Problem_15 | C | 42 | A jar contains $5$ different colors of gumdrops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and other $30$ gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?
$\textbf{(A)}\ 35\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$ | [
"We do $100-30-20-15-10$ to find the percent of gumdrops that are green. We find that $25\\%$ of the gumdrops are green. That means there are $120$ gumdrops. If we replace half of the blue gumdrops with brown gumdrops, then $15\\%$ of the jar's gumdrops are brown. $\\dfrac{35}{100} \\cdot 120=42 \\Rightarrow \\boxe... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_2 | null | 441 | A jar has $10$ red candies and $10$ blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is $m/n,$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$ | [
"The probability that Terry picks two red candies is $\\frac{10 \\cdot 9}{20 \\cdot 19} = \\frac{9}{38}$ , and the probability that Mary picks two red candies after Terry chooses two red candies is $\\frac{7\\cdot8}{18\\cdot17} = \\frac{28}{153}$ . So the probability that they both pick two red candies is $\\frac{... |
https://artofproblemsolving.com/wiki/index.php/2023_AMC_8_Problems/Problem_5 | B | 1,500 | A lake contains $250$ trout, along with a variety of other fish. When a marine biologist catches and releases a sample of $180$ fish from the lake, $30$ are identified as trout. Assume that the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake?
$\textbf{(A)}\ 1250 \qquad \textbf{(B)}\ 1500 \qquad \textbf{(C)}\ 1750 \qquad \textbf{(D)}\ 1800 \qquad \textbf{(E)}\ 2000$ | [
"Note that \\[\\frac{\\text{number of trout}}{\\text{total number of fish}} = \\frac{30}{180} = \\frac16.\\] So, the total number of fish is $6$ times the number of trout. Since the lake contains $250$ trout, there are $250\\cdot6=\\boxed{1500}$ fish in the lake."
] |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_4 | C | 342 | A large bag of coins contains pennies, dimes and quarters. There are twice as many dimes as pennies and three times as many quarters as dimes. An amount of money which could be in the bag is
$\mathrm{(A)\ } $306 \qquad \mathrm{(B) \ } $333 \qquad \mathrm{(C)\ } $342 \qquad \mathrm{(D) \ } $348 \qquad \mathrm{(E) \ } $360$ | [
"If there are $x$ pennies in the bag, then there are $2x$ dimes and $3(2x) = 6x$ quarters. Since pennies are $$0.01$ , dimes are $$0.10$ , and quarters are $$0.25$ , the total amount of money in the bag is \\[$ \\left(0.01x+(0.10)(2x)+(0.25)(6x)\\right) = $1.71x.\\] Therefore, the possible amounts of money are prec... |
https://artofproblemsolving.com/wiki/index.php/2013_AIME_II_Problems/Problem_3 | null | 350 | A large candle is $119$ centimeters tall. It is designed to burn down more quickly when it is first lit and more slowly as it approaches its bottom. Specifically, the candle takes $10$ seconds to burn down the first centimeter from the top, $20$ seconds to burn down the second centimeter, and $10k$ seconds to burn down the $k$ -th centimeter. Suppose it takes $T$ seconds for the candle to burn down completely. Then $\tfrac{T}{2}$ seconds after it is lit, the candle's height in centimeters will be $h$ . Find $10h$ | [
"We find that $T=10(1+2+\\cdots +119)$ . From Gauss's formula, we find that the value of $T$ is $10(7140)=71400$ . The value of $\\frac{T}{2}$ is therefore $35700$ . We find that $35700$ is $10(3570)=10\\cdot \\frac{k(k+1)}{2}$ , so $3570=\\frac{k(k+1)}{2}$ . As a result, $7140=k(k+1)$ , which leads to $0=k^2+k-714... |
https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_30 | D | 19 | A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$ | [
"Place the cube so that its space diagonal is perpendicular to the ground. The space diagonal has length of $3\\sqrt{3}$ , the altitude of the top vertex of the newly placed cube is $3\\sqrt{3}$ . The plane perpendicular and bisecting the space diagonal is now parallel to the ground and also bisecting the space dia... |
https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_30 | null | 19 | A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$ | [
"Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.\nNow consider the diagonal from $(0,0,0)$ to $(3,3,3)$ . The midpoint of this diagonal is at $\\left(\\frac 32,\\frac 32,\\frac 32\\right)$ . The plane that passes through this point and is orthogonal to... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_23 | C | 1,507,509 | A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have $3$ rows of small congruent equilateral triangles, with $5$ small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of $2003$ small equilateral triangles?
[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60); pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp; pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp; pair Jp=shift(Gp)*Hp; pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp}; draw(Ap--Dp--Jp--cycle); draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle); for(pair p : points) { fill(circle(p, 0.07),white); } pair[] Cn=new pair[5]; Cn[0]=centroid(Ap,Bp,Gp); Cn[1]=centroid(Gp,Bp,Fp); Cn[2]=centroid(Bp,Fp,Cp); Cn[3]=centroid(Cp,Fp,Ep); Cn[4]=centroid(Cp,Ep,Dp); label("$1$",Cn[0]); label("$2$",Cn[1]); label("$3$",Cn[2]); label("$4$",Cn[3]); label("$5$",Cn[4]); for (pair p : Cn) { draw(circle(p,0.1)); }[/asy]
$\mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018$ | [
"There are $1+3+5+...+2003=1002^{2}=1004004$ small equilateral triangles.\nEach small equilateral triangle needs $3$ toothpicks to make it.\nBut, each toothpick that isn't one of the $1002\\cdot3=3006$ toothpicks on the outside of the large equilateral triangle is a side for $2$ small equilateral triangles.\nSo, th... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10A_Problems/Problem_23 | null | 1,507,509 | A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have $3$ rows of small congruent equilateral triangles, with $5$ small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of $2003$ small equilateral triangles?
[asy] unitsize(15mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair Ap=(0,0), Bp=(1,0), Cp=(2,0), Dp=(3,0), Gp=dir(60); pair Fp=shift(Gp)*Bp, Ep=shift(Gp)*Cp; pair Hp=shift(Gp)*Gp, Ip=shift(Gp)*Fp; pair Jp=shift(Gp)*Hp; pair[] points={Ap,Bp,Cp,Dp,Ep,Fp,Gp,Hp,Ip,Jp}; draw(Ap--Dp--Jp--cycle); draw(Gp--Bp--Ip--Hp--Cp--Ep--cycle); for(pair p : points) { fill(circle(p, 0.07),white); } pair[] Cn=new pair[5]; Cn[0]=centroid(Ap,Bp,Gp); Cn[1]=centroid(Gp,Bp,Fp); Cn[2]=centroid(Bp,Fp,Cp); Cn[3]=centroid(Cp,Fp,Ep); Cn[4]=centroid(Cp,Ep,Dp); label("$1$",Cn[0]); label("$2$",Cn[1]); label("$3$",Cn[2]); label("$4$",Cn[3]); label("$5$",Cn[4]); for (pair p : Cn) { draw(circle(p,0.1)); }[/asy]
$\mathrm{(A) \ } 1,004,004 \qquad \mathrm{(B) \ } 1,005,006 \qquad \mathrm{(C) \ } 1,507,509 \qquad \mathrm{(D) \ } 3,015,018 \qquad \mathrm{(E) \ } 6,021,018$ | [
"Test out some fewer cases first.\nWhen there is just 1 equilateral triangle in the base, you need $3$ toothpicks.\nWhen there are 3 equilateral triangles in the base, you need $9$ toothpicks in all.\nWhen there are 5 equilateral triangles in the base, you need $18$ toothpicks in all.\nWhen there are 7 equilateral ... |
https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_2 | B | 15 | A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle? [asy] draw((0,0)--(10,0)--(10,7)--(0,7)--cycle); draw((0,5)--(10,5)); draw((3,0)--(3,7)); label("6", (1.5,6)); label("?", (1.5,2.5)); label("14", (6.5,6)); label("35", (6.5,2.5)); [/asy]
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20 \qquad\textbf{(D)}\ 21 \qquad\textbf{(E)}\ 25$ | [
"\nWe can easily see the dimensions of each small rectangle. So the area of the last rectangle is $3\\times 5=\\boxed{15}$"
] |
https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_10 | A | 49 | A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimter of each of the congruent rectangles is $14$ . What is the area of the large square?
$\mathrm{(A) \ }49 \qquad \mathrm{(B) \ }64 \qquad \mathrm{(C) \ }100 \qquad \mathrm{(D) \ }121 \qquad \mathrm{(E) \ }196$ | [
"Let the length of the longer side be $x$ , and the length of the shorter side be $y$ . We are given that $2x+2y=14\\implies x+y=7$ . However, note that $x+y$ is also the length of a side of the larger square. Thus the area of the larger square is $(x+y)^2=7^2=\\boxed{49}$",
"Expand the small square so it basi... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12A_Problems/Problem_1 | D | 50 | A large urn contains $100$ balls, of which $36 \%$ are red and the rest are blue. How many of the blue balls must be removed so that the percentage of red balls in the urn will be $72 \%$ ? (No red balls are to be removed.)
$\textbf{(A)}\ 28 \qquad\textbf{(B)}\ 32 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 64$ | [
"There are $36$ red balls; for these red balls to comprise $72 \\%$ of the urn, there must be only $14$ blue balls. Since there are currently $64$ blue balls, this means we must remove $\\boxed{50}$",
"There are $36$ red balls and $64$ blue balls. For the percentage of the red balls to double from $36 \\%$ to $72... |
https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_16 | B | 4 | A lattice point is a point in the plane with integer coordinates. How many lattice points are on the line segment whose endpoints are $(3,17)$ and $(48,281)$ ? (Include both endpoints of the segment in your count.)
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 46$ | [
"The difference in the $y$ -coordinates is $281 - 17 = 264$ , and the difference in the $x$ -coordinates is $48 - 3 = 45$ .\nThe gcd of 264 and 45 is 3, so the line segment joining $(3,17)$ and $(48,281)$ has slope \\[\\frac{88}{15}.\\] The points on the line have coordinates \\[\\left(3+t,17+\\frac{88}{15}t\\right... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_22 | C | 5 | A lemming sits at a corner of a square with side length $10$ meters. The lemming runs $6.2$ meters along a diagonal toward the opposite corner. It stops, makes a $90^{\circ}$ right turn and runs $2$ more meters. A scientist measures the shortest distance between the lemming and each side of the square. What is the average of these four distances in meters?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4.5 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6.2 \qquad \textbf{(E)}\ 7$ | [
"The shortest segments would be perpendicular to the square. The lemming went $x$ meters horizontally and $y$ meters vertically. No matter how much it went, the lemming would have been $x$ and $y$ meters from the sides and $10-x$ and $10-y$ meters from the remaining two. To find the average, add the lengths of the ... |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_17 | A | 6 | A line $x=k$ intersects the graph of $y=\log_5 x$ and the graph of $y=\log_5 (x + 4)$ . The distance between the points of intersection is $0.5$ . Given that $k = a + \sqrt{b}$ , where $a$ and $b$ are integers, what is $a+b$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | [
"Since the line $x=k$ is vertical, we are only concerned with vertical distance.\nIn other words, we want to find the value of $k$ for which the distance $|\\log_5 x - \\log_5 (x+4)| = \\frac{1}{2}$\nSince $\\log_5 x$ is a strictly increasing function, we have:\n$\\log_5 (x + 4) - \\log_5 x = \\frac{1}{2}$\n$\\log_... |
https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_30 | A | 2 | A line segment is divided so that the lesser part is to the greater part as the greater part is to the whole. If $R$ is the ratio of the lesser part to the greater part, then the value of
\[R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}\]
is
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }2R \qquad \mathrm{(C) \ } R^{-1} \qquad \mathrm{(D) \ } 2+R^{-1} \qquad \mathrm{(E) \ }2+R$ | [
"Let $w$ be the length of the shorter segment and $l$ be the length of the longer segment. We're given that $\\frac{w}{l}=\\frac{l}{w+l}$ . Cross-multiplying, we find that $w^2+wl=l^2\\implies w^2+wl-l^2=0$ . Now we divide both sides by $l^2$ to get $\\left(\\frac{w}{l}\\right)^2+\\left(\\frac{w}{l}\\right)-1=0$ . ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_3 | B | 10 | A line with slope $2$ intersects a line with slope $6$ at the point $(40,30)$ . What is the distance between the $x$ -intercepts of these two lines?
$\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50$ | [
"Using point slope form, we get the equations $y-30 = 6(x-40)$ and $y-30 = 2(x-40)$ . Simplifying, we get $6x-y=210$ and $2x-y=50$ . Letting $y=0$ in both equations and solving for $x$ gives the $x$ -intercepts: $x=35$ and $x=25$ , respectively. Thus the distance between them is $35-25=\\boxed{10}$",
"In order fo... |
https://artofproblemsolving.com/wiki/index.php/2003_AMC_10B_Problems/Problem_11 | A | 2 | A line with slope $3$ intersects a line with slope $5$ at point $(10,15)$ . What is the distance between the $x$ -intercepts of these two lines?
$\textbf{(A) } 2 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 20$ | [
"Using the point-slope form, the equation of each line is\n\\[y-15=3(x-10) \\longrightarrow y=3x-15\\] \\[y-15=5(x-10) \\longrightarrow y=5x-35\\]\nSubstitute in $y=0$ to find the $x$ -intercepts.\n\\[0=3x-15\\longrightarrow x=5\\] \\[0=5x-35\\longrightarrow x=7\\] The difference between them is $7-5=\\boxed{2}$",
... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_10B_Problems/Problem_18 | E | 35 | A list of $11$ positive integers has a mean of $10$ , a median of $9$ , and a unique mode of $8$ . What is the largest possible value of an integer in the list?
$\textbf {(A) } 24 \qquad \textbf {(B) } 30 \qquad \textbf {(C) } 31\qquad \textbf {(D) } 33 \qquad \textbf {(E) } 35$ | [
"We start off with the fact that the median is $9$ , so we must have $a, b, c, d, e, 9, f, g, h, i, j$ , listed in ascending order. Note that the integers do not have to be distinct.\nSince the mode is $8$ , we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list... |
https://artofproblemsolving.com/wiki/index.php/2014_AMC_12B_Problems/Problem_11 | E | 35 | A list of $11$ positive integers has a mean of $10$ , a median of $9$ , and a unique mode of $8$ . What is the largest possible value of an integer in the list?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 31\qquad\textbf{(D)}\ 33\qquad\textbf{(E)}\ 35$ | [
"We start off with the fact that the median is $9$ , so we must have $a, b, c, d, e, 9, f, g, h, i, j$ , listed in ascending order. Note that the integers do not have to be distinct.\nSince the mode is $8$ , we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_10B_Problems/Problem_14 | D | 225 | A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$ | [
"To minimize the number of distinct values, we want to maximize the number of times a number appears. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223 + 1 + 1 = \\boxed{225}.$",
"As in Solution 1, we want to maximize the number of ... |
https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_10 | D | 225 | A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$ | [
"To minimize the number of distinct values, we want to maximize the number of times a number appears. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223 + 1 + 1 = \\boxed{225}.$",
"As in Solution 1, we want to maximize the number of ... |
https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_25 | B | 6 | A list of five positive integers has mean $12$ and range $18$ . The mode and median are both $8$ . How many different values are possible for the second largest element of the list?
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$ | [
"Let $a$ be the smallest element, so $a+18$ is the largest element. Since the mode is $8$ , at least two of the five numbers must be $8$ . The last number we denote as $b$\nThen their average is $\\frac{a + (8) + (8) + b + (a+18)}5 = 12 \\Longrightarrow 2a + b = 26$ . Clearly $a \\le 8$ . Also we have $b \\le a + 1... |
https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_18 | E | 20 | A list of integers has mode $32$ and mean $22$ . The smallest number in the list is $10$ . The median $m$ of the list is a member of the list. If the list member $m$ were replaced by $m+10$ , the mean and median of the new list would be $24$ and $m+10$ , respectively. If were $m$ instead replaced by $m-8$ , the median of the new list would be $m-4$ . What is $m$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$ | [
"Let there be $n$ integers on the list. The list of $n$ integers has mean $22$ , so the sum of the integers is $22n$\nReplacing $m$ with $m+10$ will increase the sum of the list from $22n$ to $22n + 10$\nThe new mean of the list is $24$ , so the new sum of the list is also $24n$\nThus, we get $22n + 10 = 24n$ , le... |
https://artofproblemsolving.com/wiki/index.php/2024_AIME_II_Problems/Problem_2 | null | 236 | A list of positive integers has the following properties:
$\bullet$ The sum of the items in the list is $30$
$\bullet$ The unique mode of the list is $9$
$\bullet$ The median of the list is a positive integer that does not appear in the list itself.
Find the sum of the squares of all the items in the list. | [
"The third condition implies that the list's size must be an even number, as if it were an odd number, the median of hte list would surely appear in the list itself.\nTherefore, we can casework on what even numbers work.\nSay the size is 2. Clearly, this doesn't work as the only list would be $\\{9, 9\\}$ , which d... |
https://artofproblemsolving.com/wiki/index.php/1987_AHSME_Problems/Problem_13 | A | 36 | A long piece of paper $5$ cm wide is made into a roll for cash registers by wrapping it $600$ times around a cardboard tube of diameter $2$ cm,
forming a roll $10$ cm in diameter. Approximate the length of the paper in meters.
(Pretend the paper forms $600$ concentric circles with diameters evenly spaced from $2$ cm to $10$ cm.)
$\textbf{(A)}\ 36\pi \qquad \textbf{(B)}\ 45\pi \qquad \textbf{(C)}\ 60\pi \qquad \textbf{(D)}\ 72\pi \qquad \textbf{(E)}\ 90\pi$ | [
"Notice (by imagining unfolding the roll), that the length of the paper is equal to the sum of the circumferences of the concentric circles, which is $\\pi$ times the sum of the diameters. Now the, the diameters form an arithmetic series with first term $2$ , last term $10$ , and $600$ terms in total, so using the ... |
https://artofproblemsolving.com/wiki/index.php/2004_AIME_II_Problems/Problem_15 | null | 593 | A long thin strip of paper is $1024$ units in length, $1$ unit in width, and is divided into $1024$ unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a $512$ by $1$ strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a $256$ by $1$ strip of quadruple thickness. This process is repeated $8$ more times. After the last fold, the strip has become a stack of $1024$ unit squares. How many of these squares lie below the square that was originally the $942$ nd square counting from the left? | [
"Number the squares $0, 1, 2, 3, ... 2^{k} - 1$ . In this case $k = 10$ , but we will consider more generally to find an inductive solution. Call $s_{n, k}$ the number of squares below the $n$ square after the final fold in a strip of length $2^{k}$\nNow, consider the strip of length $1024$ . The problem asks fo... |
https://artofproblemsolving.com/wiki/index.php/1983_AIME_Problems/Problem_4 | null | 26 | A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy] size(150); defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy] | [
"Because we are given a right angle, we look for ways to apply the Pythagorean Theorem . Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$ . Let $OE=x$ and $OD=y$ . We're trying to find $x^2+y^2$\nApplying the Pythagorean Theorem, $OA^2... |
https://artofproblemsolving.com/wiki/index.php/2001_AIME_I_Problems/Problem_14 | null | 351 | A mail carrier delivers mail to the nineteen houses on the east side of Elm Street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day. How many different patterns of mail delivery are possible? | [
"Let $0$ represent a house that does not receive mail and $1$ represent a house that does receive mail. This problem is now asking for the number of $19$ -digit strings of $0$ 's and $1$ 's such that there are no two consecutive $1$ 's and no three consecutive $0$ 's.\nThe last two digits of any $n$ -digit string c... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_10 | B | 11 | A majority of the $30$ students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1$ . The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $$17.71$ . What was the cost of a pencil in cents?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 11 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 77$ | [
"The total cost of the pencils can be found by $(\\text{students}\\cdot\\text{pencils purchased by each}\\cdot\\text{price of each pencil})$\nSince $1771$ is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: $7, 11, 23$ . Since neither $(C)$ nor... |
https://artofproblemsolving.com/wiki/index.php/2011_AMC_12A_Problems/Problem_7 | B | 11 | A majority of the $30$ students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1$ . The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $$17.71$ . What was the cost of a pencil in cents?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 11 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 77$ | [
"The total cost of the pencils can be found by $(\\text{students}\\cdot\\text{pencils purchased by each}\\cdot\\text{price of each pencil})$\nSince $1771$ is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: $7, 11, 23$ . Since neither $(C)$ nor... |
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_44 | E | 1,806 | A man born in the first half of the nineteenth century was $x$ years old in the year $x^2$ . He was born in:
$\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806$ | [
"If a man born in the 19th century was $x$ years of in the year $x^2$ , it implies that the year the man was born was $x^2-x$ . So, if the man was born in the first half of the 19th century, it means that $x^2-x < 1850$ . Noticing that $40^2 - 40 = 1560$ and $50^2-50 = 2450$ , we see that $40 < x < 50$ . We can gue... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_23 | B | 83.33 | A man buys a house for $10,000 and rents it. He puts $12\frac{1}{2}\%$ of each month's rent aside for repairs and upkeep; pays $325 a year taxes and realizes $5\frac{1}{2}\%$ on his investment. The monthly rent (in dollars) is:
$\textbf{(A)} \ \ 64.82\qquad\textbf{(B)} \ \ 83.33\qquad\textbf{(C)} \ \ 72.08\qquad\textbf{(D)} \ \ 45.83\qquad\textbf{(E)} \ \ 177.08$ | [
"$12\\frac{1}{2}\\%$ is the same as $\\frac{1}{8}$ , so the man sets one eighth of each month's rent aside, so he only gains $\\frac{7}{8}$ of his rent. He also pays $325 each year, and he realizes $5.5\\%$ , or $550, on his investment. Therefore he must have collected a total of $325 +$550 = $875 in rent. This was... |
https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_17 | D | 16 | A man can commute either by train or by bus. If he goes to work on the train in the morning, he comes home on the bus in the afternoon; and if he comes home in the afternoon on the train, he took the bus in the morning. During a total of $x$ working days, the man took the bus to work in the morning $8$ times, came home by bus in the afternoon $15$ times, and commuted by train (either morning or afternoon) $9$ times. Find $x$
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 16 \qquad \\ \textbf{(E)}\ \text{ not enough information given to solve the problem}$ | [
"The man has three possible combinations of transportation: \\[\\text{Morning train, Afternoon bus (m.t., a.b.)}\\] \\[\\text{Morning bus, Afternoon train (m.b., a.t.)}\\] \\[\\text{Morning bus, Afternoon bus (m.b, a.b.)}\\]\nLet $y$ be the number of times the man takes the $\\text{a.t.}$ . Then, $9-y$ is the numbe... |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_4 | E | 6.4 | A man has $\textdollar{10,000 }$ to invest. He invests $\textdollar{4000}$ at 5% and $\textdollar{3500}$ at 4%.
In order to have a yearly income of $\textdollar{500}$ , he must invest the remainder at:
$\textbf{(A)}\ 6\%\qquad\textbf{(B)}\ 6.1\%\qquad\textbf{(C)}\ 6.2\%\qquad\textbf{(D)}\ 6.3\%\qquad\textbf{(E)}\ 6.4\%$ | [
"The man currently earns $4000 \\cdot \\frac{5}{1000} + 3500 \\cdot \\frac{4}{1000} = 340$ dollars. So, we need to find the value of $x$ such that \\[2500 \\cdot \\frac{x}{1000} = 160.\\] Solving, we get $x = \\boxed{6.4}$"
] |
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_17 | B | 4.8 | A man has part of $ $4500$ invested at $4$ % and the rest at $6$ %. If his annual return on each investment is the same, the average rate of interest which he realizes of the $4500 is:
$\textbf{(A)}\ 5\% \qquad \textbf{(B)}\ 4.8\% \qquad \textbf{(C)}\ 5.2\% \qquad \textbf{(D)}\ 4.6\% \qquad \textbf{(E)}\ \text{none of these}$ | [
"You are trying to find $\\frac{2(0.06x)}{4500}$ , where $x$ is the principle for one investment. To find $x$ , solve $0.04(4500-x) = 0.06x$ $X$ will come out to be $1800$ . Then, plug in x into the first equation, $\\frac{2(0.06)(1800)}{4500}$ , to get $0.048$ . Finally, convert that to a percentage and you get $\... |
https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_35 | B | 40 | A man on his way to dinner short after $6: 00$ p.m. observes that the hands of his watch form an angle of $110^{\circ}$ . Returning before $7: 00$ p.m. he notices that again the hands of his watch form an angle of $110^{\circ}$ . The number of minutes that he has been away is:
$\textbf{(A)}\ 36\frac{2}3\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 42.4\qquad\textbf{(E)}\ 45$ | [
"Let $n$ be the number of minutes after 6:00. Let $h(n)=180+\\frac{n}2$ be the angle, in degrees, of the hour hand (with $0^{\\circ}$ at the top and increasing in the clockwise direction); similarly, let $m(n)=6n$ be the angle of the minute hand. We want $|h(n)-m(n)|=110$ . This is equivalent to \\[180-\\frac{11n}2... |
https://artofproblemsolving.com/wiki/index.php/2007_AIME_II_Problems/Problem_1 | null | 372 | A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in 2007. No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in 2007. A set of plates in which each possible sequence appears exactly once contains N license plates. Find $\frac{N}{10}$ | [
"There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.\nThus, $N = 2520 + 1200 = 3720$ , and $\\frac{N}{10} = \\boxed{372}$"
] |
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_17 | C | 125 | A merchant bought some goods at a discount of $20\%$ of the list price. He wants to mark them at such a price that he can give a discount of $20\%$ of the marked price and still make a profit of $20\%$ of the selling price. The per cent of the list price at which he should mark them is:
$\textbf{(A) \ }20 \qquad \textbf{(B) \ }100 \qquad \textbf{(C) \ }125 \qquad \textbf{(D) \ }80 \qquad \textbf{(E) \ }120$ | [
"Let $C$ represent the cost of the goods, and let $L$ $S$ , and $M$ represent the list, selling, and marked prices of the goods, respectively. Hence, we have three equations, which we need to manipulate in order to relate $M$ and $L$\n$C=\\frac{4}{5}L$\n$S=C+\\frac{1}{5}S$\n$S=\\frac{4}{5}M$\nWe find that $M=\\frac... |
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_36 | null | 125 | A merchant buys goods at $25\%$ off the list price. He desires to mark the goods so that he can give a discount of $20\%$ on the marked price and still clear a profit of $25\%$ on the selling price. What percent of the list price must he mark the goods?
$\textbf{(A)}\ 125\% \qquad \textbf{(B)}\ 100\% \qquad \textbf{(C)}\ 120\% \qquad \textbf{(D)}\ 80\% \qquad \textbf{(E)}\ 75\%$ | [
"Without loss of generality, we can set the list price equal to $100$ . The merchant buys the goods for $100*.75=75$ . Let $x$ be the marked price.\nWe then use the equation $0.8x-75=25$ to solve for $x$ and get a marked price of $\\boxed{125}$"
] |
https://artofproblemsolving.com/wiki/index.php/2002_AMC_8_Problems/Problem_14 | B | 44 | A merchant offers a large group of items at $30\%$ off. Later, the merchant takes $20\%$ off these sale prices. The total discount is
$\text{(A)}\ 35\%\qquad\text{(B)}\ 44\%\qquad\text{(C)}\ 50\%\qquad\text{(D)}\ 56\%\qquad\text{(E)}\ 60\%$ | [
"Let's assume that each item is $100$ dollars. First we take off $30\\%$ off of $100$ dollars. $100\\cdot0.7=70$\nNext, we take off the extra $20\\%$ as asked by the problem. $70\\cdot0.80=56$\nSo the final price of an item is $56. We have to do $100-56$ because $56$ was the final price and we wanted the discount. ... |
https://artofproblemsolving.com/wiki/index.php/2007_AMC_8_Problems/Problem_17 | C | 40 | A mixture of $30$ liters of paint is $25\%$ red tint, $30\%$ yellow
tint and $45\%$ water. Five liters of yellow tint are added to
the original mixture. What is the percent of yellow tint
in the new mixture?
$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 35 \qquad \mathrm{(C)}\ 40 \qquad \mathrm{(D)}\ 45 \qquad \mathrm{(E)}\ 50$ | [
"Since $30\\%$ of the original $30$ liters of paint was yellow, and 5 liters of yellow paint were added to make the new mixture, there are $9+5=14$ liters of yellow tint in the new mixture. Since only 5 liters of paint were added to the original 30, there are a total of 35 liters of paint in the new mixture. This g... |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_10B_Problems/Problem_5 | B | 3 | A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | [
"$31 \\equiv 3 \\pmod {7}$ so the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays. Therefore, Monday, Thursday, and Friday are valid so the answer is $\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2010_AMC_12B_Problems/Problem_4 | B | 3 | A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | [
"$31 \\equiv 3 \\pmod {7}$ so the week cannot start with Saturday, Sunday, Tuesday or Wednesday as that would result in an unequal number of Mondays and Wednesdays. Therefore, Monday, Thursday, and Friday are valid so the answer is $\\boxed{3}$"
] |
https://artofproblemsolving.com/wiki/index.php/2019_AIME_I_Problems/Problem_5 | null | 252 | A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$ , it moves at random to one of the points $(a-1,b)$ $(a,b-1)$ , or $(a-1,b-1)$ , each with probability $\frac{1}{3}$ , independently of its previous moves. The probability that it will hit the coordinate axes at $(0,0)$ is $\frac{m}{3^n}$ , where $m$ and $n$ are positive integers such that $m$ is not divisible by $3$ . Find $m + n$ | [
"One could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as \\[P(x,y) = \\frac{1}{3} P(x-1,y) + \\frac{1}{3} P(x,y-1) + \\frac{1}{3} P(x-1,y-1)\\] for $x,y \\geq 1,$ and the base cases are $P(0,0) = 1, P(x,0) = P(y,0) = 0$ for any $x,y$ not equal to zero.\n... |
https://artofproblemsolving.com/wiki/index.php/1987_AJHSME_Problems/Problem_24 | D | 12 | A multiple choice examination consists of $20$ questions. The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question. John's score on the examination is $48$ . What is the maximum number of questions he could have answered correctly?
$\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 11 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 16$ | [
"Let $c$ be the number of questions correct, $w$ be the number of questions wrong, and $b$ be the number of questions left blank. We are given that \\begin{align} c+w+b &= 20 \\\\ 5c-2w &= 48 \\end{align}\nAdding equation $(2)$ to double equation $(1)$ , we get \\[7c+2b=88\\]\nSince we want to maximize the value ... |
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_5 | C | 6 | A nickel is placed on a table. The number of nickels which can be placed around it, each tangent to it and to two others is:
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$ | [
"Arranging the nickels in a hexagonal fashion, we see that only $\\boxed{6}$ nickels can be placed around the central nickel."
] |
https://artofproblemsolving.com/wiki/index.php/1985_AHSME_Problems/Problem_24 | C | 45,678 | A non-zero digit is chosen in such a way that the probability of choosing digit $d$ is $\log_{10}{(d+1)}-\log_{10}{d}$ . The probability that the digit $2$ is chosen is exactly $1/2$ the probability that the digit chosen is in the set
$\mathrm{(A)\ } \{2,3\} \qquad \mathrm{(B) \ }\{3,4\} \qquad \mathrm{(C) \ } \{4,5,6,7,8\} \qquad \mathrm{(D) \ } \{5,6,7,8,9\} \qquad \mathrm{(E) \ }\{4,5,6,7,8,9\}$ | [
"We have $\\log_{10}{(d+1)}-\\log_{10}{d} = \\log_{10}{\\left(\\frac{d+1}{d}\\right)}$ , so the probability of choosing $2$ is $\\log_{10}{\\left(\\frac{3}{2}\\right)}$ . The probability that the digit chosen is in the set must therefore be \\begin{align*}2\\log_{10}{\\left(\\frac{3}{2}\\right)} = &\\log_{10}{\\lef... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19 | B | 4 | A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$ | [
"A number is divisible by $15$ precisely if it is divisible by $3$ and $5$ . The latter means the last digit must be either $5$ or $0$ , and the former means the sum of the digits must be divisible by $3$ . If the last digit is $0$ , the first digit would be $0$ (because the digits alternate), which is not possible... |
https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_19 | null | 4 | A number is called flippy if its digits alternate between two distinct digits. For example, $2020$ and $37373$ are flippy, but $3883$ and $123123$ are not. How many five-digit flippy numbers are divisible by $15?$
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5 \qquad \textbf{(D) }6 \qquad \textbf{(E) }8$ | [
"assume the number is $ababa$ $10101a+1010b=0 (mod 15)\\newline$ $6a+5b=0 (mod 15)\\newline$ $a=0 (mod 5)\\newline$ $5b=0 (mod 15)\\newline$ $b=0 (mod 3)\\newline$ Solutions: $(5,0),(5,3),(5,6),(5,9)\\newline$ $\\boxed{4}$"
] |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_10A_Problems/Problem_14 | B | 173 | A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
[asy] size(7cm); pathpen = linewidth(0.7); D(CR((0,0),10)); D(CR((0,0),9.5)); D(CR((0,-18.5),9.5)); D(CR((0,-18.5),9)); MP("$\vdots$",(0,-31),(0,0)); D(CR((0,-39),3)); D(CR((0,-39),2.5)); D(CR((0,-43.5),2.5)); D(CR((0,-43.5),2)); D(CR((0,-47),2)); D(CR((0,-47),1.5)); D(CR((0,-49.5),1.5)); D(CR((0,-49.5),1.0)); D((12,-10)--(12,10)); MP('20',(12,0),E); D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);[/asy]
$\textbf{(A) } 171\qquad\textbf{(B) } 173\qquad\textbf{(C) } 182\qquad\textbf{(D) } 188\qquad\textbf{(E) } 210\qquad$ | [
"The inside diameters of the rings are the positive integers from $1$ to $18$ . The total distance needed is the sum of these values plus $2$ for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series , the answer is $\\frac{18 \\cdot 19}{2} + 2 = \\boxed{1... |
https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_12 | B | 173 | A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
[asy] size(7cm); pathpen = linewidth(0.7); D(CR((0,0),10)); D(CR((0,0),9.5)); D(CR((0,-18.5),9.5)); D(CR((0,-18.5),9)); MP("$\vdots$",(0,-31),(0,0)); D(CR((0,-39),3)); D(CR((0,-39),2.5)); D(CR((0,-43.5),2.5)); D(CR((0,-43.5),2)); D(CR((0,-47),2)); D(CR((0,-47),1.5)); D(CR((0,-49.5),1.5)); D(CR((0,-49.5),1.0)); D((12,-10)--(12,10)); MP('20',(12,0),E); D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);[/asy]
$\textbf{(A) } 171\qquad\textbf{(B) } 173\qquad\textbf{(C) } 182\qquad\textbf{(D) } 188\qquad\textbf{(E) } 210\qquad$ | [
"The inside diameters of the rings are the positive integers from $1$ to $18$ . The total distance needed is the sum of these values plus $2$ for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series , the answer is $\\frac{18 \\cdot 19}{2} + 2 = \\boxed{1... |
https://artofproblemsolving.com/wiki/index.php/2013_AMC_8_Problems/Problem_16 | E | 89 | A number of students from Fibonacci Middle School are taking part in a community service project. The ratio of $8^\text{th}$ -graders to $6^\text{th}$ -graders is $5:3$ , and the the ratio of $8^\text{th}$ -graders to $7^\text{th}$ -graders is $8:5$ . What is the smallest number of students that could be participating in the project?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 55 \qquad \textbf{(D)}\ 79 \qquad \textbf{(E)}\ 89$ | [
"We multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:\n$5:3 = 5(8):3(8) = 40:24$\n$8:5 = 8(5):5(5) = 40:25$\nTherefore, the ratio of 8th graders to 7th graders to 6th graders is $40:25:24$ . Since the rati... |
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_37 | D | 2,519 | A number which when divided by $10$ leaves a remainder of $9$ , when divided by $9$ leaves a remainder of $8$ , by $8$ leaves a remainder of $7$ , etc., down to where, when divided by $2$ , it leaves a remainder of $1$ , is:
$\textbf{(A)}\ 59\qquad\textbf{(B)}\ 419\qquad\textbf{(C)}\ 1259\qquad\textbf{(D)}\ 2519\qquad\textbf{(E)}\ \text{none of these answers}$ | [
"If we add $1$ to the number, it becomes divisible by $10, 9, 8, \\cdots, 2, 1$ . The LCM of $1$ through $10$ is $2520$ , therefore the number we want to find is $2520-1=\\boxed{2519}$"
] |
https://artofproblemsolving.com/wiki/index.php/2009_AMC_8_Problems/Problem_25 | E | 11 | A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $\frac{1}{2}$ foot from the top face. The second cut is $\frac{1}{3}$ foot below the first cut, and the third cut is $\frac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet? [asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label("A", (1,0,3/4), W); label("B", (1,0,1/3), W); label("C", (1,0,1/6-d/4), W); label("D", (1,0,d/2), W); label("1/2", (1,1,3/4), E); label("1/3", (1,1,1/3), E); label("1/17", (0,1,1/6-d/4), E);[/asy]
[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label("D", (1/2, 1, 0), SE); label("A", (1+1/2, 1, 0), SE); label("B", (2+1/2, 1, 0), SE); label("C", (3+1/2, 1, 0), SE);[/asy]
$\textbf{(A)}\:6\qquad\textbf{(B)}\:7\qquad\textbf{(C)}\:\frac{419}{51}\qquad\textbf{(D)}\:\frac{158}{17}\qquad\textbf{(E)}\:11$ | [
"The areas of the tops of $A$ $B$ $C$ , and $D$ in the figure formed has sum $1+1+1+1 = 4$ as do the bottoms. Thus, the total so far is $8$ . Now, one of the sides has an area of one, since it combines all of the heights of $A$ $B$ $C$ , and $D$ , which is $1$ . The other side is also the same. Thus the total area ... |
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