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Given that
\begin{eqnarray*}&(1)& x\text{ and }y\text{ are both integers between 100 and 999, inclusive;}\qquad \qquad \qquad \qquad \qquad \\ &(2)& y\text{ is the number formed by reversing the digits of }x\text{; and}\\ &(3)& z=|x-y|. \end{eqnarray*}
How many distinct values of $z$ are possible? | We express the numbers as $x=100a+10b+c$ and $y=100c+10b+a$. From this, we have
\begin{eqnarray*}z&=&|100a+10b+c-100c-10b-a|\\&=&|99a-99c|\\&=&99|a-c|\\ \end{eqnarray*}
Because $a$ and $c$ are digits, and $a$ and $c$ are both between 1 and 9 (from condition 1), there are $\boxed{009}$ possible values (since all digit... | 9 | Number Theory | math-word-problem | Yes | Yes | amc_aime | false |
Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$ | Using the properties of [logarithms](https://artofproblemsolving.com/wiki/index.php/Logarithm), we can simplify the first equation to $\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1$. Therefore, \[\sin x \cos x = \frac{1}{10}.\qquad (*)\]
Now, manipulate the second equation.
\begin{align*} \log_{10... | 12 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
Consider the polynomials $P(x) = x^{6} - x^{5} - x^{3} - x^{2} - x$ and $Q(x) = x^{4} - x^{3} - x^{2} - 1.$ Given that $z_{1},z_{2},z_{3},$ and $z_{4}$ are the roots of $Q(x) = 0,$ find $P(z_{1}) + P(z_{2}) + P(z_{3}) + P(z_{4}).$ | When we use long division to divide $P(x)$ by $Q(x)$, the remainder is $x^2-x+1$.
So, since $z_1$ is a root, $P(z_1)=(z_1)^2-z_1+1$.
Now this also follows for all roots of $Q(x)$
Now \[P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1\]
Now by [Vieta's](https://artofproblemsolving.com/wiki/ind... | 6 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
For each [positive integer](https://artofproblemsolving.com/wiki/index.php/Positive_integer) $k$, let $S_k$ denote the [ increasing](https://artofproblemsolving.com/wiki/index.php/Increasing_sequence) [arithmetic sequence](https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence) of [integers](https://artofpr... | Suppose that the $n$th term of the sequence $S_k$ is $2005$. Then $1+(n-1)k=2005$ so $k(n-1)=2004=2^2\cdot 3\cdot 167$. The [ordered pairs](https://artofproblemsolving.com/wiki/index.php/Ordered_pair) $(k,n-1)$ of positive integers that satisfy the last equation are $(1,2004)$,$(2,1002)$, $(3,668)$, $(4,501)$, $(6,334... | 12 | Number Theory | math-word-problem | Yes | Yes | amc_aime | false |
Given that $x, y,$ and $z$ are real numbers that satisfy:
\begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*}
and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are posit... | Let $\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$, and suppose this triangle is acute (so all [altitudes](https://artofproblemsolving.com/wiki/index.php/Altitude) are in the interior of the triangle).
Let the altitude to the side of length $x$ be of length $h_x$, and similarly for $y$ and $z$. Then... | 9 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
[Square](https://artofproblemsolving.com/wiki/index.php/Square) $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is [equilateral](https://artofproblemsolving.com/wiki/index.php/Equilateral). A square with vertex $B$ has sides that are [pa... | [asy] unitsize(32mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1); pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1); pair Ap = (0, (3 - sqrt(3))/6); pair Cp = ((3 - sqrt(3))/6, 0); pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6); pair[] dots = {A, B, C, D, ... | 12 | Geometry | math-word-problem | Yes | Yes | amc_aime | false |
Let $a$ and $b$ be positive real numbers with $a\ge b$. Let $\rho$ be the maximum possible value of $\frac {a}{b}$ for which the system of equations
\[a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2\]
has a solution in $(x,y)$ satisfying $0\le x < a$ and $0\le y < b$. Then $\rho^2$ can be expressed as a fraction $\frac {... | Solution 1
Notice that the given equation implies
$a^2 + y^2 = b^2 + x^2 = 2(ax + by)$
We have $2by \ge y^2$, so $2ax \le a^2 \implies x \le \frac {a}{2}$.
Then, notice $b^2 + x^2 = a^2 + y^2 \ge a^2$, so $b^2 \ge \frac {3}{4}a^2 \implies \rho^2 \le \frac {4}{3}$.
The solution $(a, b, x, y) = \left(1, \frac {\sqrt {3}... | 7 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
In right $\triangle ABC$ with hypotenuse $\overline{AB}$, $AC = 12$, $BC = 35$, and $\overline{CD}$ is the altitude to $\overline{AB}$. Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\o... | First, note that $AB=37$; let the tangents from $I$ to $\omega$ have length $x$. Then the perimeter of $\triangle ABI$ is equal to \[2(x+AD+DB)=2(x+37).\] It remains to compute $\dfrac{2(x+37)}{37}=2+\dfrac{2}{37}x$.
Observe $CD=\dfrac{12\cdot 35}{37}=\dfrac{420}{37}$, so the radius of $\omega$ is $\dfrac{210}{37}$. We... | 11 | Geometry | math-word-problem | Yes | Yes | amc_aime | false |
A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$, where $m$ and $n$ a... | The probability of three heads and five tails is $\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\binom {8}{3}p^5(1-p)^3$.
\begin{align*} 25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\ 25(1-p)^2&=p^2 \\ 25p^2-50p+25&=p^2 \\ 24p^2-50p+25&=0 \\ p&=\frac {5}{6}\end{align*}
Therefore,... | 11 | Combinatorics | math-word-problem | Yes | Yes | amc_aime | false |
Let $m$ be the number of solutions in positive integers to the equation $4x+3y+2z=2009$, and let $n$ be the number of solutions in positive integers to the equation $4x+3y+2z=2000$. Find the remainder when $m-n$ is divided by $1000$. | Solution 1
It is actually reasonably easy to compute $m$ and $n$ exactly.
First, note that if $4x+3y+2z=2009$, then $y$ must be odd. Let $y=2y'-1$. We get $4x + 6y' - 3 + 2z = 2009$, which simplifies to $2x + 3y' + z = 1006$. For any pair of positive integers $(x,y')$ such that $2x + 3y' < 1006$ we have exactly one $z$... | 0 | Combinatorics | math-word-problem | Yes | Yes | amc_aime | false |
[Triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) $ABC$ with [right angle](https://artofproblemsolving.com/wiki/index.php/Right_angle) at $C$, $\angle BAC < 45^\circ$ and $AB = 4$. Point $P$ on $\overline{AB}$ is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$. The ratio $\frac{AP}{BP}$ can b... | Let $O$ be the [circumcenter](https://artofproblemsolving.com/wiki/index.php/Circumcenter) of $ABC$ and let the intersection of $CP$ with the [circumcircle](https://artofproblemsolving.com/wiki/index.php/Circumcircle) be $D$. It now follows that $\angle{DOA} = 2\angle ACP = \angle{APC} = \angle{DPB}$. Hence $ODP$ is is... | 7 | Geometry | math-word-problem | Yes | Yes | amc_aime | false |
Let $ABCDEF$ be a [regular](https://artofproblemsolving.com/wiki/index.php/Regular_polygon) [hexagon](https://artofproblemsolving.com/wiki/index.php/Hexagon). Let $G$, $H$, $I$, $J$, $K$, and $L$ be the [midpoints](https://artofproblemsolving.com/wiki/index.php/Midpoint) of sides $AB$, $BC$, $CD$, $DE$, $EF$, and $AF$,... | [asy] defaultpen(0.8pt+fontsize(12pt)); pair A,B,C,D,E,F; pair G,H,I,J,K,L; A=dir(0); B=dir(60); C=dir(120); D=dir(180); E=dir(240); F=dir(300); draw(A--B--C--D--E--F--cycle,blue); G=(A+B)/2; H=(B+C)/2; I=(C+D)/2; J=(D+E)/2; K=(E+F)/2; L=(F+A)/2; int i; for (i=0; i<6; i+=1) { draw(rotate(60*i)*(A--H),dotted); } p... | 11 | Geometry | math-word-problem | Yes | Yes | amc_aime | false |
Let $R$ be the set of all possible remainders when a number of the form $2^n$, $n$ a nonnegative integer, is divided by 1000. Let $S$ be the sum of the elements in $R$. Find the remainder when $S$ is divided by 1000. | Note that $x \equiv y \pmod{1000} \Leftrightarrow x \equiv y \pmod{125}$ and $x \equiv y \pmod{8}$. So we must find the first two integers $i$ and $j$ such that $2^i \equiv 2^j \pmod{125}$ and $2^i \equiv 2^j \pmod{8}$ and $i \neq j$. Note that $i$ and $j$ will be greater than 2 since remainders of $1, 2, 4$ will not b... | 7 | Number Theory | math-word-problem | Yes | Yes | amc_aime | false |
Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$, where $a > 0$ and $a + b + c$ is an integer. The minimum possible value of $a$ can be written in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$. | If the vertex is at $\left(\frac{1}{4}, -\frac{9}{8}\right)$, the equation of the parabola can be expressed in the form \[y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}.\]
Expanding, we find that \[y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8},\] and \[y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}.\] From the pr... | 11 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
Ed has five identical green marbles, and a large supply of identical red marbles. He arranges the green marbles and some of the red ones in a row and finds that the number of marbles whose right hand neighbor is the same color as themselves is equal to the number of marbles whose right hand neighbor is the other color.... | We are limited by the number of marbles whose right hand neighbor is not the same color as the marble. By surrounding every green marble with red marbles - RGRGRGRGRGR. That's 10 "not the same colors" and 0 "same colors." Now, for every red marble we add, we will add one "same color" pair and keep all 10 "not the sa... | 3 | Combinatorics | math-word-problem | Yes | Yes | amc_aime | false |
Let $f_1(x) = \frac23 - \frac3{3x+1}$, and for $n \ge 2$, define $f_n(x) = f_1(f_{n-1}(x))$. The value of $x$ that satisfies $f_{1001}(x) = x-3$ can be expressed in the form $\frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | After evaluating the first few values of $f_k (x)$, we obtain $f_4(x) = f_1(x) = \frac{2}{3} - \frac{3}{3x+1} = \frac{6x-7}{9x+3}$. Since $1001 \equiv 2 \mod 3$, $f_{1001}(x) = f_2(x) = \frac{3x+7}{6-9x}$. We set this equal to $x-3$, i.e.
$\frac{3x+7}{6-9x} = x-3 \Rightarrow x = \frac{5}{3}$. The answer is thus $5+3 =... | 8 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the ... | We can set $AE=ED=m$. Set $BD=k$, therefore $CD=3k, AC=4k$. Thereafter, by Stewart's Theorem on $\triangle ACD$ and cevian $CE$, we get $2m^2+14=25k^2$. Also apply Stewart's Theorem on $\triangle CEB$ with cevian $DE$. After simplification, $2m^2=17-6k^2$. Therefore, $k=1, m=\frac{\sqrt{22}}{2}$. Finally, note that (us... | 10 | Geometry | math-word-problem | Yes | Yes | amc_aime | false |
Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$. Find $x_2(x_1+x_3)$. | Substituting $n$ for $2014$, we get \[\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2\] \[= x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0\] Noting that $nx^2 - 1$ factors as a difference of squares to \[(\sqrt{n}x - 1)(\sqrt{n}x+1)\] we can factor the left side as \[(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))\] This means... | 2 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$. At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a coun... | We divide this up into casework. The "directions" the bug can go are $\text{Clockwise}$, $\text{Counter-Clockwise}$, and $\text{Switching}$. Let an $I$ signal going clockwise (because it has to be in the inner circle), an $O$ signal going counter-clockwise, and an $S$ switching between inner and outer circles. An ex... | 4 | Combinatorics | math-word-problem | Yes | Yes | amc_aime | false |
Triangle $ABC$ has side lengths $AB=7, BC=8,$ and $CA=9.$ Circle $\omega_1$ passes through $B$ and is tangent to line $AC$ at $A.$ Circle $\omega_2$ passes through $C$ and is tangent to line $AB$ at $A.$ Let $K$ be the intersection of circles $\omega_1$ and $\omega_2$ not equal to $A.$ Then $AK=\tfrac mn,$ where $m$ an... | -Diagram by Brendanb4321
Note that from the tangency condition that the supplement of $\angle CAB$ with respects to lines $AB$ and $AC$ are equal to $\angle AKB$ and $\angle AKC$, respectively, so from tangent-chord, \[\angle AKC=\angle AKB=180^{\circ}-\angle BAC\] Also note that $\angle ABK=\angle KAC$$^{(*)}$, so $\... | 11 | Geometry | math-word-problem | Yes | Yes | amc_aime | false |
For a real number $x$ let $\lfloor x\rfloor$ be the greatest integer less than or equal to $x$, and define $\{x\} = x - \lfloor x \rfloor$ to be the fractional part of $x$. For example, $\{3\} = 0$ and $\{4.56\} = 0.56$. Define $f(x)=x\{x\}$, and let $N$ be the number of real-valued solutions to the equation $f(f(f(x))... | Note that the upper bound for our sum is $2019,$ and not $2020,$ because if it were $2020$ then the function composition cannot equal to $17.$ From there, it's not too hard to see that, by observing the function composition from right to left, $N$ is (note that the summation starts from the right to the left):
\[\sum_{... | 10 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
Find the remainder when\[\binom{\binom{3}{2}}{2} + \binom{\binom{4}{2}}{2} + \dots + \binom{\binom{40}{2}}{2}\]is divided by $1000$. | We first write the expression as a summation.
\begin{align*} \sum_{i=3}^{40} \binom{\binom{i}{2}}{2} & = \sum_{i=3}^{40} \binom{\frac{i \left( i - 1 \right)}{2}}{2} \\ & = \sum_{i=3}^{40} \frac{\frac{i \left( i - 1 \right)}{2} \left( \frac{i \left( i - 1 \right)}{2}- 1 \right)}{2} \\ & = \frac{1}{8} \sum_{i=3}^{40} i... | 4 | Combinatorics | math-word-problem | Yes | Yes | amc_aime | false |
The sum of all positive integers $m$ such that $\frac{13!}{m}$ is a perfect square can be written as $2^a3^b5^c7^d11^e13^f,$ where $a,b,c,d,e,$ and $f$ are positive integers. Find $a+b+c+d+e+f.$ | We first rewrite $13!$ as a prime factorization, which is $2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13.$
For the fraction to be a square, it needs each prime to be an even power. This means $m$ must contain $7\cdot11\cdot13$. Also, $m$ can contain any even power of $2$ up to $2^{10}$, any odd power of $3$ up to $3^{5}$,... | 12 | Number Theory | math-word-problem | Yes | Yes | amc_aime | false |
Two non-zero [real numbers](https://artofproblemsolving.com/wiki/index.php/Real_number), $a$ and $b,$ satisfy $ab = a - b$. Which of the following is a possible value of $\frac {a}{b} + \frac {b}{a} - ab$?
$\textbf{(A)} \ - 2 \qquad \textbf{(B)} \ \frac { -1 }{2} \qquad \textbf{(C)} \ \frac {1}{3} \qquad \textbf{(D)} \... | $\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = \frac{2(a-b)}{a-b} =2 \Rightarrow \boxed{\text{E}}$.
Another way is to solve the equation for $b,$ giving $b = \frac{a}{a+1};$ then substituting this into the expression and simplifyin... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The [Fibonacci sequence](https://artofproblemsolving.com/wiki/index.php/Fibonacci_sequence) $1,1,2,3,5,8,13,21,\ldots$ starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten [digits](https://artofproblemsolving.com/wiki/index.php/Digit) is the last to appear in the units p... | Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $\bmod{10}$:
$1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$
The last digit to appear in the units position of a number in the Fibonacci sequence is $6 \Longrightar... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The median of the list
$n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15$ is $10$. What is the mean?
$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }10\qquad\textbf{(E) }11$ | The median of the list is $10$, and there are $9$ numbers in the list, so the median must be the 5th number from the left, which is $n+6$.
We substitute the median for $10$ and the equation becomes $n+6=10$.
Subtract both sides by 6 and we get $n=4$.
$n+n+3+n+4+n+5+n+6+n+8+n+10+n+12+n+15=9n+63$.
The mean of those numbe... | 11 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes.
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)... | Drawing the problem out, we see we get a parallelogram with a height of $40$ and a base of $15$, giving an area of $600$.
If we look at it the other way, we see the distance between the stripes is the height and the base is $50$.
[asy] draw((0,0)--(5,0)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Let $P(n)$ and $S(n)$ denote the product and the sum, respectively, of the digits
of the integer $n$. For example, $P(23) = 6$ and $S(23) = 5$. Suppose $N$ is a
two-digit number such that $N = P(N)+S(N)$. What is the units digit of $N$?
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\... | Denote $a$ and $b$ as the tens and units digit of $N$, respectively. Then $N = 10a+b$. It follows that $10a+b=ab+a+b$, which implies that $9a=ab$. Since $a\neq0$, $b=9$. So the units digit of $N$ is $\boxed{(\textbf{E})\ 9}$. | 9 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A telephone number has the form $\text{ABC-DEF-GHIJ}$, where each letter represents
a different digit. The digits in each part of the number are in decreasing
order; that is, $A > B > C$, $D > E > F$, and $G > H > I > J$. Furthermore,
$D$, $E$, and $F$ are consecutive even digits; $G$, $H$, $I$, and $J$ are consecutive... | We start by noting that there are $10$ letters, meaning there are $10$ digits in total. Listing them all out, we have $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$. Clearly, the most restrictive condition is the consecutive odd digits, so we create casework based on that.
Case 1: $G$, $H$, $I$, and $J$ are $7$, $5$, $3$, and $1$ resp... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The ratio $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ is closest to which of the following numbers?
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$ | We factor $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ as $\frac{10^{2000}(1+100)}{10^{2001}(1+1)}=\frac{101}{20}$. As $\frac{101}{20}=5.05$, our answer is $\boxed{\textbf{(D)}\ 5 }$. | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 15$ | Solution 1
This is a Pythagorean triple (a $3-4-5$ actually) with legs $15$ and $20$. The area is then $\frac{(15)(20)}{2}=150$. Now, consider an altitude drawn to any side. Since the area remains constant, the altitude and side to which it is drawn are inversely proportional. To get the smallest altitude, it must be d... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$ | Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$. It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$. Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to [Vieta's Formulas]... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Points $A,B,C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$. Point $E$ is not on the line, and $BE = CE = 10$. The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$. Find $AB$.
$\text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)... | First, we draw an altitude to $BC$ from $E$. Let it intersect at $M$. As $\triangle BEC$ is isosceles, we immediately get $MB=MC=6$, so the altitude is $8$. Now, let $AB=CD=x$. Using the Pythagorean Theorem on $\triangle EMA$, we find $AE=\sqrt{x^2+12x+100}$. From symmetry, $DE=\sqrt{x^2+12x+100}$ as well. Now, we use ... | 9 | Geometry | MCQ | Yes | Yes | amc_aime | false |
There are 3 numbers A, B, and C, such that $1001C - 2002A = 4004$, and $1001B + 3003A = 5005$. What is the average of A, B, and C?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E) }\text{Not uniquely determined}$ | Notice that we don't need to find what $A, B,$ and $C$ actually are, just their average. In other words, if we can find $A+B+C$, we will be done.
Adding up the equations gives $1001(A+B+C)=9009=1001(9)$ so $A+B+C=9$ and the average is $\frac{9}{3}=3$. Our answer is $\boxed{\textbf{(B) }3}$. | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For which of the following values of $k$ does the equation $\frac{x-1}{x-2} = \frac{x-k}{x-6}$ have no solution for $x$?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$ | The domain over which we solve the equation is $\mathbb{R} \setminus \{2,6\}$.
We can now cross-multiply to get rid of the fractions, we get $(x-1)(x-6)=(x-k)(x-2)$.
Simplifying that, we get $7x-6 = (k+2)x - 2k$. Clearly for $k=5$ we get the equation $-6=-10$ which is never true. The answer is $\boxed{\mathrm{ (E)}\ 5}... | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For how many integers $n$ is $\dfrac n{20-n}$ the [square](https://artofproblemsolving.com/wiki/index.php/Perfect_square) of an integer?
$\mathrm{(A)}\ 1 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 3 \qquad\mathrm{(D)}\ 4 \qquad\mathrm{(E)}\ 10$ | Let $x^2 = \frac{n}{20-n}$, with $x \ge 0$ (note that the solutions $x < 0$ do not give any additional solutions for $n$). Then rewriting, $n = \frac{20x^2}{x^2 + 1}$. Since $\text{gcd}(x^2, x^2 + 1) = 1$, it follows that $x^2 + 1$ divides $20$. Listing the factors of $20$, we find that $x = 0, 1, 2 , 3$ are the only $... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Four distinct [circles](https://artofproblemsolving.com/wiki/index.php/Circle) are drawn in a [plane](https://artofproblemsolving.com/wiki/index.php/Plane). What is the maximum number of points where at least two of the circles intersect?
$\mathrm{(A)}\ 8 \qquad\mathrm{(B)}\ 9 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\... | For any given pair of circles, they can intersect at most $2$ times. Since there are ${4\choose 2} = 6$ pairs of circles, the maximum number of possible intersections is $6 \cdot 2 = 12$. We can construct such a situation as below, so the answer is $\boxed{\mathrm{(D)}\ 12}$.
[2002 12B AMC-14.png](https://artofproblems... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
For the nonzero numbers a, b, and c, define
$D(a,b,c)=\frac{abc}{a+b+c}$
Find $D(2,4,6)$.
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 24$ | $\frac{2\cdot 4\cdot 6}{2+4+6}=\frac{48}{12}=4\Longrightarrow\mathrm{ (C) \ }$ | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $a$, $b$, and $c$ be real numbers such that $a-7b+8c=4$ and $8a+4b-c=7$. Then $a^2-b^2+c^2$ is
$\mathrm{(A)\ }0\qquad\mathrm{(B)\ }1\qquad\mathrm{(C)\ }4\qquad\mathrm{(D)\ }7\qquad\mathrm{(E)\ }8$ | Solution 1
Rearranging, we get $a+8c=7b+4$ and $8a-c=7-4b$
Squaring both, $a^2+16ac+64c^2=49b^2+56b+16$ and $64a^2-16ac+c^2=16b^2-56b+49$ are obtained.
Adding the two equations and dividing by $65$ gives $a^2+c^2=b^2+1$, so $a^2-b^2+c^2=\boxed{(\text{B})1}$.
Solution 2
The easiest way is to assume a value for $a$ and ... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?
$\mathrm{(A) \ } 5\qquad \... | We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $\triangle OBC$ is a $30-60-90$ triangle. That means $\angle BOC = 60^\circ$ and the ferris wheel has made $\frac{60}{360} = \frac{1}{6}$ of a... | 10 | Geometry | MCQ | Yes | Yes | amc_aime | false |
When $15$ is appended to a list of integers, the mean is increased by $2$. When $1$ is appended to the enlarged list, the mean of the enlarged list is decreased by $1$. How many integers were in the original list?
$\mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(... | Let $x$ be the sum of the integers and $y$ be the number of elements in the list. Then we get the equations $\dfrac{x+15}{y+1}=\dfrac{x}{y}+2$ and $\dfrac{x+15+1}{y+1+1}=\dfrac{x+16}{y+2}=\frac{x}{y}+2-1=\frac{x}{y}+1$. With a lot of algebra, the solution is found to be $y= \boxed{\textbf{(A)}\ 4}$. | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The [arithmetic mean](https://artofproblemsolving.com/wiki/index.php/Arithmetic_mean) of the nine numbers in the set $\{9, 99, 999, 9999, \ldots, 999999999\}$ is a $9$-digit number $M$, all of whose digits are distinct. The number $M$ doesn't contain the digit
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ ... | We wish to find $\frac{9+99+\cdots +999999999}{9}$, or $\frac{9(1+11+111+\cdots +111111111)}{9}=123456789$. This doesn't have the digit 0, so the answer is $\boxed{\mathrm{(A)}\ 0}$ | 0 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the value of $(3x - 2)(4x + 1) - (3x - 2)4x + 1$ when $x=4$?
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 1 \qquad\mathrm{(C)}\ 10 \qquad\mathrm{(D)}\ 11 \qquad\mathrm{(E)}\ 12$ | By the distributive property,
\[(3x-2)[(4x+1)-4x] + 1 = 3x-2 + 1 = 3x-1 = 3(4) - 1 = \boxed{\mathrm{(D)}\ 11}\]. | 11 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The [polygon](https://artofproblemsolving.com/wiki/index.php/Polygon) enclosed by the solid lines in the figure consists of 4 [congruent](https://artofproblemsolving.com/wiki/index.php/Congruent) [ squares](https://artofproblemsolving.com/wiki/index.php/Square_(geometry)) joined [edge](https://artofproblemsolving.com/w... | Solution 1
[2003amc10a10solution.gif](https://artofproblemsolving.com/wiki/index.php/File:2003amc10a10solution.gif)
Let the squares be labeled $A$, $B$, $C$, and $D$.
When the polygon is folded, the "right" edge of square $A$ becomes adjacent to the "bottom edge" of square $C$, and the "bottom" edge of square $A$ becom... | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Let $n$ be the largest integer that is the product of exactly 3 distinct prime numbers $d$, $e$, and $10d+e$, where $d$ and $e$ are single digits. What is the sum of the digits of $n$?
$\mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24$ | Since we want $n$ to be as large as possible, we would like $d$ in $10d+e$ to be as large as possible. So, $d=7,$ the greatest single-digit prime. Then, $e$ cannot be $5$ because $10(7)+5 = 75,$ which is not prime. So $e = 3$. Therefore, $d \cdot e \cdot (10d+e) = 7 \cdot 3 \cdot 73 = 1533$.
So, the sum of the digits o... | 12 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the units digit of $13^{2003}$?
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9$ | $13^{2003}\equiv 3^{2003}\pmod{10}$
Since $3^4=81\equiv1\pmod{10}$:
$3^{2003}=(3^{4})^{500}\cdot3^{3}\equiv1^{500}\cdot27\equiv7\pmod{10}$
Therefore, the units digit is $7 \Rightarrow\boxed{\mathrm{(C)}\ 7}$ | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the sum of the reciprocals of the roots of the equation
$\frac{2003}{2004}x+1+\frac{1}{x}=0$?
$\mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003}$ | Multiplying both sides by $x$:
$\frac{2003}{2004}x^{2}+1x+1=0$
Let the roots be $a$ and $b$.
The problem is asking for $\frac{1}{a}+\frac{1}{b}= \frac{a+b}{ab}$
By [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas):
$a+b=(-1)^{1}\frac{1}{\frac{2003}{2004}}=-\frac{2004}{2003}$
$ab=(... | -1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Sally has five red cards numbered $1$ through $5$ and four blue cards numbered $3$ through $6$. She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?
$\mathrm{(A) ... | Let $R_i$ and $B_j$ designate the red card numbered $i$ and the blue card numbered $j$, respectively.
$B_5$ is the only blue card that $R_5$ evenly divides, so $R_5$ must be at one end of the stack and $B_5$ must be the card next to it.
$R_1$ is the only other red card that evenly divides $B_5$, so $R_1$ must be the ... | 12 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?
$\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qq... | Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km.
Since she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\frac{40}{60}=\frac{2}{3}$ hours.
Therefore her average speed in km/hr is $\frac{2}{\frac{2}{3}}=\boxed{\... | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $d$ and $e$ denote the solutions of $2x^{2}+3x-5=0$. What is the value of $(d-1)(e-1)$?
$\mathrm{(A) \ } -\frac{5}{2}\qquad \mathrm{(B) \ } 0\qquad \mathrm{(C) \ } 3\qquad \mathrm{(D) \ } 5\qquad \mathrm{(E) \ } 6$ | Solution 1
Using factoring:
$2x^{2}+3x-5=0$
$(2x+5)(x-1)=0$
$x = -\frac{5}{2}$ or $x=1$
So $d$ and $e$ are $-\frac{5}{2}$ and $1$.
Therefore the answer is $\left(-\frac{5}{2}-1\right)(1-1)=\left(-\frac{7}{2}\right)(0)=\boxed{\mathrm{(B)}\ 0}$
Solution 2
We can use the sum and product of a quadratic (a.k.a Vieta):
$(... | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
How many non-[congruent](https://artofproblemsolving.com/wiki/index.php/Congruent_(geometry)) [triangles](https://artofproblemsolving.com/wiki/index.php/Triangle) with [perimeter](https://artofproblemsolving.com/wiki/index.php/Perimeter) $7$ have [integer](https://artofproblemsolving.com/wiki/index.php/Integer) side le... | By the [triangle inequality](https://artofproblemsolving.com/wiki/index.php/Triangle_inequality), no side may have a length greater than the semiperimeter, which is $\frac{1}{2}\cdot7=3.5$.
Since all sides must be integers, the largest possible length of a side is $3$. Therefore, all such triangles must have all sides... | 2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
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
\[y-15=3(x-10) \longrightarrow y=3x-15\]
\[y-15=5(x-10) \longrightarrow y=5x-35\]
Substitute in $y=0$ to find the $x$-intercepts.
\[0=3x-15\longrightarrow x=5\]
\[0=5x-35\longrightarrow x=7\]
The difference between them is $7-5=\boxed{\textbf{(A) \ } 2}$. | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $\clubsuit(x)$ denote the sum of the digits of the positive integer $x$. For example, $\clubsuit(8)=8$ and $\clubsuit(123)=1+2+3=6$. For how many two-digit values of $x$ is $\clubsuit(\clubsuit(x))=3$?
$\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 9 \qquad\textbf{(E) } 10$ | Let $y=\clubsuit (x)$. Since $x \leq 99$, we have $y \leq 18$. Thus if $\clubsuit (y)=3$, then $y=3$ or $y=12$. The 3 values of $x$ for which $\clubsuit (x)=3$ are 12, 21, and 30, and the 7 values of $x$ for which $\clubsuit (x)=12$ are 39, 48, 57, 66, 75, 84, and 93. There are $\boxed{E=10}$ values in all. | 10 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$ | Solution 1
Let the total value, in cents, of the coins Paula has originally be $v$, and the number of coins she has be $n$. Then $\frac{v}{n}=20\Longrightarrow v=20n$ and $\frac{v+25}{n+1}=21$. Substituting yields: $20n+25=21(n+1),$ so $n=4$, $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if s... | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A [sequence](https://artofproblemsolving.com/wiki/index.php/Sequence) of three real numbers forms an [arithmetic progression](https://artofproblemsolving.com/wiki/index.php/Arithmetic_progression) with a first term of $9$. If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbe... | Let $d$ be the common difference. Then $9$, $9+d+2=11+d$, $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the [geometric mean](https://artofproblemsolving.com/wiki/index.php/Geometric_mean) of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In the overlapping [triangles](https://artofproblemsolving.com/wiki/index.php/Triangle) $\triangle{ABC}$ and $\triangle{ABE}$ sharing common [ side](https://artofproblemsolving.com/wiki/index.php/Edge) $AB$, $\angle{EAB}$ and $\angle{ABC}$ are [right angles](https://artofproblemsolving.com/wiki/index.php/Right_angle), ... | Solution 1
Since $AE \perp AB$ and $BC \perp AB$, $AE \parallel BC$. By alternate interior angles and $AA\sim$, we find that $\triangle ADE \sim \triangle CDB$, with side length ratio $\frac{4}{3}$. Their heights also have the same ratio, and since the two heights add up to $4$, we have that $h_{ADE} = 4 \cdot \frac{4}... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
In the United States, coins have the following thicknesses: penny, $1.55$ mm; nickel, $1.95$ mm; dime, $1.35$ mm; quarter, $1.75$ mm. If a stack of these coins is exactly $14$ mm high, how many coins are in the stack?
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 9 \qquad \mathrm{(D) \ } 10 \qquad ... | All numbers in this solution will be in hundredths of a millimeter.
The thinnest coin is the dime, with thickness $135$. A stack of $n$ dimes has height $135n$.
The other three coin types have thicknesses $135+20$, $135+40$, and $135+60$. By replacing some of the dimes in our stack by other, thicker coins, we can clear... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$?
$\mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 144\qquad \mathrm{(E) \ } 720$ | The product of all six numbers is $6!=720$. The products of numbers that can be visible are $720/1$, $720/2$, ..., $720/6$.
The answer to this problem is their greatest common divisor -- which is $720/L$, where $L$ is the least common multiple of $\{1,2,3,4,5,6\}$.
Clearly $L=60$ and the answer is $720/60=\boxed{\mathr... | 12 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In the expression $c\cdot a^b-d$, the values of $a$, $b$, $c$, and $d$ are $0$, $1$, $2$, and $3$, although not necessarily in that order. What is the maximum possible value of the result?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\ }6\qquad\mathrm{(C)\ }8\qquad\mathrm{(D)\ }9\qquad\mathrm{(E)\ }10$ | If $a=0$ or $c=0$, the expression evaluates to $-d<0$.
If $b=0$, the expression evaluates to $c-d\leq 2$.
Case $d=0$ remains.
In that case, we want to maximize $c\cdot a^b$ where $\{a,b,c\}=\{1,2,3\}$. Trying out the six possibilities we get that the greatest is $(a,b,c)=(3,2,1)$, where $c\cdot a^b=1\cdot 3^2=\boxed{... | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
On a trip from the United States to Canada, Isabella took $d$ U.S. dollars. At the border she exchanged them all, receiving $10$ Canadian dollars for every $7$ U.S. dollars. After spending $60$ Canadian dollars, she had $d$ Canadian dollars left. What is the sum of the digits of $d$?
$\mathrm{(A)\ }5\qquad\mathrm{(B)\... | Isabella had $60+d$ Canadian dollars. Setting up an equation we get $d=\frac{7}{10}\cdot(60+d)$, which solves to $d=140$, and the sum of digits of $d$ is $\boxed{\mathrm{(A)}\ 5}$. | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
While eating out, Mike and Joe each tipped their server $2$ dollars. Mike tipped $10\%$ of his bill and Joe tipped $20\%$ of his bill. What was the difference, in dollars between their bills?
$\textbf{(A) } 2\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 10\qquad \textbf{(E) } 20$ | Let $m$ be Mike's bill and $j$ be Joe's bill.
$\frac{10}{100}m=2$
$m=20$
$\frac{20}{100}j=2$
$j=10$
So the desired difference is $m-j=20-10=10 \Rightarrow \boxed{\textbf{(D) }10}$ | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$?
$\textbf{(A) } 3\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 6\qquad \textbf{(E) } 7$ | Since there are $n^2$ little [faces](https://artofproblemsolving.com/wiki/index.php/Face) on each face of the big wooden [ cube](https://artofproblemsolving.com/wiki/index.php/Cube_(geometry)), there are $6n^2$ little faces painted red.
Since each unit cube has $6$ faces, there are $6n^3$ little faces total.
Since on... | 4 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
How many positive cubes divide $3! \cdot 5! \cdot 7!$ ?
$\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 4\qquad \textbf{(D) } 5\qquad \textbf{(E) } 6$ | $3! \cdot 5! \cdot 7! = (3\cdot2\cdot1) \cdot (5\cdot4\cdot3\cdot2\cdot1) \cdot (7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1) = 2^{8}\cdot3^{4}\cdot5^{2}\cdot7^{1}$
Therefore, a [perfect cube](https://artofproblemsolving.com/wiki/index.php/Perfect_cube) that divides $3! \cdot 5! \cdot 7!$ must be in the form $2^{a}\cdot3^{b}... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In the five-sided star shown, the letters $A, B, C, D,$ and $E$ are replaced by the numbers $3, 5, 6, 7,$ and $9$, although not necessarily in this order. The sums of the numbers at the ends of the line segments $AB$, $BC$, $CD$, $DE$, and $EA$ form an arithmetic sequence, although not necessarily in that order. What i... | Each corner $(A,B,C,D,E)$ goes to two sides/numbers. ($A$ goes to $AE$ and $AB$, $D$ goes to $DC$ and $DE$). The sum of every term is equal to $2(3+5+6+7+9)=60$
Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is $\frac{60}{5}=\boxed{\textbf{(D) }12}$ | 12 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
For each pair of real numbers $a \neq b$, define the [operation](https://artofproblemsolving.com/wiki/index.php/Operation) $\star$ as
$(a \star b) = \frac{a+b}{a-b}$.
What is the value of $((1 \star 2) \star 3)$?
$\textbf{(A) } -\frac{2}{3}\qquad \textbf{(B) } -\frac{1}{5}\qquad \textbf{(C) } 0\qquad \textbf{(D) } \fra... | $((1 \star 2) \star 3) = \left(\left(\frac{1+2}{1-2}\right) \star 3\right) = (-3 \star 3) = \frac{-3+3}{-3-3} = 0 \Longrightarrow \boxed{\textbf{(C) }0}$ | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$?
$\textbf {(A)} -8 \qquad \textbf{(B)} -4 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 8$ | $2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = \boxed{\textbf{(B)}-4}$ | -4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In $\triangle ABC$, we have $AC=BC=7$ and $AB=2$. Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$. What is $BD$?
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | Solution 1
Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$. By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $CH=\sqrt{48}$. Since $CD=8$, $HD=\sqrt{8^2-48}=\sqrt{16}=4$, and $BD=HD-1$, so $BD=\boxed{\textbf{(A) }3}$.
Solution 2 (Trig)... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The [quadratic equation](https://artofproblemsolving.com/wiki/index.php/Quadratic_equation) $x^2+mx+n$ has roots twice those of $x^2+px+m$, and none of $m,n,$ and $p$ is zero. What is the value of $n/p$?
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \q... | Solution 1
Let $x^2 + px + m = 0$ have roots $a$ and $b$. Then
\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\]
so $p = -(a+b)$ and $m = ab$. Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$, so
\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\]
and $m = -2(a+b)$ and $n = 4ab$. Thus $\frac{n}{p} = \frac{4ab}{-(a... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
All of David's telephone numbers have the form $555-abc-defg$, where $a$, $b$, $c$, $d$, $e$, $f$, and $g$ are distinct digits and in increasing order, and none is either $0$ or $1$. How many different telephone numbers can David have?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } ... | The only digits available to use in the phone number are $2$, $3$, $4$, $5$, $6$, $7$, $8$, and $9$. There are only $7$ spots left among the $8$ numbers, so we need to find the number of ways to choose $7$ numbers from $8$. The answer is then $\dbinom{8}{7}=\dfrac{8!}{7!\,(8-7)!}=\boxed{\textbf{(D) } 8}$
Alternatively,... | 8 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
At the beginning of the school year, Lisa's goal was to earn an $A$ on at least $80\%$ of her $50$ quizzes for the year. She earned an $A$ on $22$ of the first $30$ quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an $A$?
$\textbf{(A) }\ 1 \qqua... | Lisa's goal was to get an $A$ on $80\% \cdot 50 = 40$ quizzes. She already has $A$'s on $22$ quizzes, so she needs to get $A$'s on $40-22=18$ more. There are $50-30=20$ quizzes left, so she can afford to get less than an $A$ on $20-18=\boxed{\textbf{(B) }2}$ of them.
Here, only the $A$'s matter... No complicated stuf... | 2 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
Doug and Dave shared a pizza with $8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was $8$ dollars, and there was an additional cost of $2$ dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. ... | Dave and Doug paid $8+2=10$ dollars in total. Doug paid for three slices of plain pizza, which cost $\frac{3}{8}\cdot 8=3$. Dave paid $10-3=7$ dollars. Dave paid $7-3=\boxed{\textbf{(D) }4}$ more dollars than Doug. | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The $8\times18$ [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle) $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$?
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}... | Since the two [hexagons](https://artofproblemsolving.com/wiki/index.php/Hexagon) are going to be repositioned to form a [ square](https://artofproblemsolving.com/wiki/index.php/Square_(geometry)) without overlap, the [area](https://artofproblemsolving.com/wiki/index.php/Area) will remain the same. The rectangle's area ... | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many [sets](https://artofproblemsolving.com/wiki/index.php/Set) of two or more consecutive positive integers have a sum of $15$?
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Notice that if the consecutive positive integers have a sum of $15$, then their average (which could be a fraction) must be a divisor of $15$. If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$, and $1$ is clearly not possible. The other two possibilities both work:
$1 + 2 + 3... | 3 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is $(-1)^{1} + (-1)^{2} + ... + (-1)^{2006}$ ?
$\textbf{(A)} -2006\qquad \textbf{(B)} -1\qquad \textbf{(C) } 0\qquad \textbf{(D) } 1\qquad \textbf{(E) } 2006$ | Since $-1$ raised to an odd integer is $-1$ and $-1$ raised to an even integer exponent is $1$:
$(-1)^{1} + (-1)^{2} + ... + (-1)^{2006} = (-1) + (1) + ... + (-1)+(1) = \boxed{\textbf{(C) }0}.$ | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $a_1 , a_2 , ...$ be a sequence for which $a_1=2$ , $a_2=3$, and $a_n=\frac{a_{n-1}}{a_{n-2}}$ for each positive integer $n \ge 3$. What is $a_{2006}$?
$\mathrm{(A) \ } \frac{1}{2}\qquad \mathrm{(B) \ } \frac{2}{3}\qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2\qquad \mathrm{(E) \ } 3$ | Looking at the first few terms of the sequence:
$a_1=2 , a_2=3 , a_3=\frac{3}{2}, a_4=\frac{1}{2} , a_5=\frac{1}{3} , a_6=\frac{2}{3} , a_7=2 , a_8=3 , ....$
Clearly, the sequence repeats every 6 terms.
Since $2006 \equiv 2\bmod{6}$,
$a_{2006} = a_2 = \boxed{\textbf{(E) }3}$ | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
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{\textbf{(A) }10} \\ \end{align*} | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Circles of diameter $1$ inch and $3$ inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?
[2006amc10b04.gif](https://artofproblemsolving.com/wiki/index... | The area painted red is equal to the area of the smaller circle and the area painted blue is equal to the area of the larger circle minus the area of the smaller circle.
So we have:
\begin{align*} A_{red}&=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}\\ A_{blue}&=\pi\left(\frac{3}{2}\right)^2-\pi\left(\frac{1}{2}\right)... | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The Dunbar family consists of a mother, a father, and some children. The [average](https://artofproblemsolving.com/wiki/index.php/Average) age of the members of the family is $20$, the father is $48$ years old, and the average age of the mother and children is $16$. How many children are in the family?
$\text{(A)}\ 2 \... | Let $n$ be the number of children. Then the total ages of the family is $48 + 16(n+1)$, and the total number of people in the family is $n+2$. So
\[20 = \frac{48 + 16(n+1)}{n+2} \Longrightarrow 20n + 40 = 16n + 64 \Longrightarrow n = 6\ \mathrm{(E)}.\] | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A [sphere](https://artofproblemsolving.com/wiki/index.php/Sphere) is inscribed in a [cube](https://artofproblemsolving.com/wiki/index.php/Cube) that has a [surface area](https://artofproblemsolving.com/wiki/index.php/Surface_area) of $24$ square meters. A second cube is then inscribed within the sphere. What is the sur... | Solution 1
[asy] import three; draw(((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0))^^((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1))^^((0,0,0)--(0,0,1))^^((0,1,0)--(0,1,1))^^((1,1,0)--(1,1,1))^^((1,0,0)--(1,0,1))); draw(shift((0.5,0.5,0.5))*scale3(1/sqrt(3))*shift((-0.5,-0.5,-0.5))*rotate(aTan(sqrt(2)),(0,0,0.5),(1,1,0.5))... | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many [ordered pairs](https://artofproblemsolving.com/wiki/index.php/Ordered_pair) $(m,n)$ of positive [integers](https://artofproblemsolving.com/wiki/index.php/Integer), with $m \ge n$, have the property that their squares differ by $96$?
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ ... | \[m^2 - n^2 = (m+n)(m-n) = 96 = 2^{5} \cdot 3\]
For every two factors $xy = 96$, we have $m+n=x, m-n=y \Longrightarrow m = \frac{x+y}{2}, n = \frac{x-y}{2}$. It follows that the number of ordered pairs $(m,n)$ is given by the number of ordered pairs $(x,y): xy=96, x > y > 0$. There are $(5+1)(1+1) = 12$ factors of $96$... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For each positive integer $n$, let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | For the sake of notation, let $T(n) = n + S(n) + S(S(n))$. Obviously $n 19$, and $n + S(n)$ clearly sum to $> 2007$.
Case 5: $200u$. So $S(n) = 2 + u$ and $S(S(n)) = 2 + u$ (recall that $n < 2007$), and $2000 + u + 2 + u + 2 + u = 2004 + 3u = 2007 \Longrightarrow u = 1$. Fourth solution.
In total we have $4 \mathrm{(D)... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
An aquarium has a [rectangular](https://artofproblemsolving.com/wiki/index.php/Rectangle) base that measures $100$ cm by $40$ cm and has a height of $50$ cm. It is filled with water to a height of $40$ cm. A brick with a rectangular base that measures $40$ cm by $20$ cm and a height of $10$ cm is placed in the aquarium... | The volume of the brick is $40 \times 20 \times 10 = 8000$. Thus the water volume rose $8000 = 100 \times 40 \times h \Longrightarrow h = 2\ \mathrm{(D)}$. | 2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Tom's age is $T$ years, which is also the sum of the ages of his three children. His age $N$ years ago was twice the sum of their ages then. What is $T/N$?
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 5 \qquad\textbf{(E) } 6$ | Tom's age $N$ years ago was $T-N$. The sum of the ages of his three children $N$ years ago was $T-3N,$ since there are three children. If his age $N$ years ago was twice the sum of the children's ages then,
\begin{align*}T-N&=2(T-3N)\\ T-N&=2T-6N\\ T&=5N\\ T/N&=\boxed{\mathrm{(D) \ } 5}\end{align*}
Note that actual val... | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Some boys and girls are having a car wash to raise money for a class trip to China. Initially $40\%$ of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then $30\%$ of the group are girls. How many girls were initially in the group?
$\textbf{(A) } 4 \qquad\textbf{(B) } 6 \qquad\textbf{(C... | If we let $p$ be the number of people initially in the group, then $0.4p$ is the number of girls. If two girls leave and two boys arrive, the number of people in the group is still $p$, but the number of girls is $0.4p-2$. Since only $30\%$ of the group are girls,
\begin{align*} \frac{0.4p-2}{p}&=\frac{3}{10}\\ 4p-20&=... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Each of the sides of a square $S_1$ with area $16$ is bisected, and a smaller square $S_2$ is constructed using the bisection points as vertices. The same process is carried out on $S_2$ to construct an even smaller square $S_3$. What is the area of $S_3$?
$\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C... | Since the area of the large square is $16$, the side length equals $4$. If all sides are bisected, the resulting square has side length $2\sqrt{2}$, thus making the area $8$. If we repeat this process again, we notice that the area is just half that of the previous square, so the area of $S_{3} = 4 \longrightarrow \fbo... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in more than 30 gallons of water. Steve starts rowing towards the shore at a constant rate of 4 miles per hour while LeRoy bails water out of the... | It will take $\frac{1}{4}$ of an hour or $15$ minutes to get to shore.
Since only $30$ gallons of water can enter the boat, only $\frac{30}{15}=2$ net gallons can enter the boat per minute.
Since $10$ gallons of water enter the boat each minute, LeRoy must bail $10-2=8$ gallons per minute $\Rightarrow\mathrm{(D)}$. | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $k={2008}^{2}+{2}^{2008}$. What is the units digit of $k^2+2^k$?
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8$ | $k \equiv 2008^2 + 2^{2008} \equiv 8^2 + 2^4 \equiv 4+6 \equiv 0 \pmod{10}$.
So, $k^2 \equiv 0 \pmod{10}$. Since $k = 2008^2+2^{2008}$ is a multiple of four and the units digit of powers of two repeat in cycles of four, $2^k \equiv 2^4 \equiv 6 \pmod{10}$.
Therefore, $k^2+2^k \equiv 0+6 \equiv 6 \pmod{10}$. So the ... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Suppose that $\tfrac{2}{3}$ of $10$ bananas are worth as much as $8$ oranges. How many oranges are worth as much as $\tfrac{1}{2}$ of $5$ bananas?
$\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ \frac{5}{2}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{7}{2}\qquad\mathrm{(E)}\ 4$ | If $\frac{2}{3}\cdot10\ \text{bananas}=8\ \text{oranges}$, then $\frac{1}{2}\cdot5\ \text{bananas}=\left(\frac{1}{2}\cdot 5\ \text{bananas}\right)\cdot\left(\frac{8\ \text{oranges}}{\frac{2}{3}\cdot10\ \text{bananas}}\right)=3\ \text{oranges}\Longrightarrow\mathrm{(C)}$. | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A triathlete competes in a triathlon in which the swimming, biking, and running segments are all of the same length. The triathlete swims at a rate of 3 kilometers per hour, bikes at a rate of 20 kilometers per hour, and runs at a rate of 10 kilometers per hour. Which of the following is closest to the triathlete's ave... | Let $d$ be the length of one segment of the race.
Average speed is total distance divided by total time. The total distance is $3d$, and the total time is $\frac{d}{3}+\frac{d}{20}+\frac{d}{10}=\frac{29d}{60}$.
Thus, the average speed is $3d\div\left(\frac{29d}{60}\right)=\frac{180}{29}$. This is closest to $6$, so the... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The fraction
\[\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}\]
simplifies to which of the following?
$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9$ | Simplifying, we get \[\frac{3^{4016}-3^{4012}}{3^{4014}-3^{4010}}.\] Factoring out $3^{4012}$ in the numerator and factoring out $3^{4010}$ in the denominator gives us \[\frac{(3^4-1)(3^{4012})}{(3^4-1)(3^{4010})}.\] Canceling out $3^4-1$ gives us $\frac{3^{4012}}{3^{4010}}=\frac{3^2}{3^0}=9\ \mathrm{(E)}.$ | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A basketball player made 5 baskets during a game. Each basket was worth either 2 or 3 points. How many different numbers could represent the total points scored by the player?
$\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 5\qquad\mathrm{(E)}\ 6$ | The number of points could have been 10, 11, 12, 13, 14, or 15. This is because the minimum is 2*5=10 and the maximum is 3*5=15. The numbers between 10 and 15 are possible as well. Thus, the answer is $\boxed{\mathrm{(E)}\ 6}$. | 6 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers and $b > a$. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies half the area of ... | Because the unpainted part of the floor covers half the area, then the painted rectangle covers half the area as well. Since the border width is 1 foot, the dimensions of the rectangle are $a-2$ by $b-2$. With this information we can make the equation:
\begin{eqnarray*} ab &=& 2\left((a-2)(b-2)\right) \\ ab &=& 2ab - 4... | 2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leav... | Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$.
Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds.
Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds.
Meetings occur whenever $D(t)=0$.
We have $D(0)=200$.
The tr... | 5 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
For [real numbers](https://artofproblemsolving.com/wiki/index.php/Real_number) $a$ and $b$, define $a * b=(a-b)^2$. What is $(x-y)^2*(y-x)^2$?
$\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ x^2+y^2\qquad\mathrm{(C)}\ 2x^2\qquad\mathrm{(D)}\ 2y^2\qquad\mathrm{(E)}\ 4xy$ | Since $(-a)^2 = a^2$, it follows that $(x-y)^2 = (y-x)^2$, and
\[(x-y)^2 * (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.\] | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets could be purchased for exactly 50 dollars?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 7 \qquad \mathrm{(C)}\ 9 \qquad ... | The cost of a rose is odd, hence we need an even number of roses. Let there be $2r$ roses for some $r\geq 0$. Then we have $50-3\cdot 2r = 50-6r$ dollars left. We can always reach the sum exactly $50$ by buying $(50-6r)/2 = 25-3r$ carnations. Of course, the number of roses must be such that the number of carnations is ... | 9 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
A quadratic equation $ax^2 - 2ax + b = 0$ has two real solutions. What is the average of these two solutions?
$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ \frac ba\qquad\mathrm{(D)}\ \frac{2b}a\qquad\mathrm{(E)}\ \sqrt{2b-a}$ | Dividing both sides by $a$, we get $x^2 - 2x + b/a = 0$. By Vieta's formulas, the sum of the roots is $2$, therefore their average is $1\Rightarrow \boxed{A}$. | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
One can holds $12$ ounces of soda, what is the minimum number of cans needed to provide a gallon ($128$ ounces) of soda?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$ | $10$ cans would hold $120$ ounces, but $128>120$, so $11$ cans are required. Thus, the answer is $\mathrm{\boxed{(E)}}$. | 11 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Circle $A$ has radius $100$. Circle $B$ has an integer radius $r<100$ and remains internally tangent to circle $A$ as it rolls once around the circumference of circle $A$. The two circles have the same points of tangency at the beginning and end of circle $B$'s trip. How many possible values can $r$ have?
$\mathrm{(A)}... | The circumference of circle $A$ is $200\pi$, and the circumference of circle $B$ with radius $r$ is $2r\pi$. Since circle $B$ makes a complete revolution and ends up on the same point, the circumference of $A$ must be a multiple of the circumference of $B$, therefore the quotient must be an integer.
Thus, $\frac{200\pi... | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$. Diagonals $AC$ and $BD$ intersect at $E$, $AC = 14$, and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE$?
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad ... | Let $[ABC]$ denote the area of triangle $ABC$. $[AED] = [BEC]$, so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$. Since triangles $ABD$ and $ABC$ share a base, they also have the same height and thus $\overline{AB}||\overline{CD}$ and $\triangle{AEB}\sim\triangle{CED}$ with a ratio of $3: 4$. $AE = \frac {3}{7}\times... | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
For $k > 0$, let $I_k = 10\ldots 064$, where there are $k$ zeros between the $1$ and the $6$. Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$. What is the maximum value of $N(k)$?
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$ | The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$.
For $k\in\{1,2,3\}$ we have $I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)$. The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\leq 5$.
For $k>4$ we have $I_k=2^6 \left( 5^{k+2}... | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
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