problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
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$\mathrm{P}_{1}, \mathrm{P}_{2}, \ldots, \mathrm{P}_{\mathrm{n}}$ are points in the plane and $\mathrm{r}_{1}, \mathrm{r}_{2}, \ldots, \mathrm{r}_{\mathrm{n}}$ are real numbers such that the distance between $P_{i}$ and $P_{j}$ is $r_{i}+r_{j}$ (for $i$ not equal to $j$ ). Find the largest $n$ for which this is possibl... | Answer: $\mathrm{n}=4$.
Draw a circle of radius $r_{i}$ at $P_{i}$. Then each pair of circles must touch. But that is possible if and only if $n \leq 4$.
 | 4 | Geometry | math-word-problem | Yes | Yes | olympiads_ref | false |
For any set $A=\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$ of five distinct positive integers denote by $S_{A}$ the sum of its elements, and denote by $T_{A}$ the number of triples $(i, j, k)$ with $1 \leqslant i<j<k \leqslant 5$ for which $x_{i}+x_{j}+x_{k}$ divides $S_{A}$.
Find the largest possible value of $... | We will prove that the maximum value that $T_{A}$ can attain is 4. Let $A=$ $\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$ be a set of five positive integers such that $x_{1} < x_{2} < x_{3} < x_{4} < x_{5}$. Analogously we can show that any triple of form $(x, y, 5)$ where $y > 2$ isn't good.
By above, the number... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads_ref | false |
Let $n$ three-digit numbers satisfy the following properties:
(1) No number contains the digit 0.
(2) The sum of the digits of each number is 9.
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers are different.
Fi... | Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0. We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of 6 A's (which means that we add 1 to the current digit) and $2 G$'s (which means go... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads_ref | false |
Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression
$$
A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}
$$
$19^{\text {th }}$ Junior B... | We can rewrite $A$ as follows:
$$
\begin{aligned}
& A=\frac{2-a^{3}}{a}+\frac{2-b^{3}}{b}+\frac{2-c^{3}}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^{2}-b^{2}-c^{2}= \\
& 2\left(\frac{a b+b c+c a}{a b c}\right)-\left(a^{2}+b^{2}+c^{2}\right)=2\left(\frac{a b+b c+c a}{a b c}\right)-\left((a+b+c)^{2}-2(a b+b ... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads_ref | false |
Let \( x, y \) be positive real numbers such that \( x^{3} + y^{3} \leq x^{2} + y^{2} \). Find the greatest possible value of the product \( x y \). | We have $(x+y)\left(x^{2}+y^{2}\right) \geq(x+y)\left(x^{3}+y^{3}\right) \geq\left(x^{2}+y^{2}\right)^{2}$, hence $x+y \geq x^{2}+y^{2}$. Now $2(x+y) \geq(1+1)\left(x^{2}+y^{2}\right) \geq(x+y)^{2}$, thus $2 \geq x+y$. Because $x+y \geq 2 \sqrt{x y}$, we will obtain $1 \geq x y$. Equality holds when $x=y=1$.
So the gr... | 1 | Inequalities | math-word-problem | Yes | Yes | olympiads_ref | false |
Let $a, b, c$ be positive real numbers with $a b c=1$. Prove the inequality:
$$
\frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1}+\frac{2 b^{2}+\frac{1}{b}}{c+\frac{1}{b}+1}+\frac{2 c^{2}+\frac{1}{c}}{a+\frac{1}{c}+1} \geq 3
$$ | By $A M-G M$ we have $2 x^{2}+\frac{1}{x}=x^{2}+x^{2}+\frac{1}{x} \geq 3 \sqrt[3]{\frac{x^{4}}{x}}=3 x$ for all $x>0$, so we have:
$\sum_{\text {cyc }} \frac{2 a^{2}+\frac{1}{a}}{b+\frac{1}{a}+1} \geq \sum_{c y c} \frac{3 a}{1+b+b c}=3\left(\sum_{c y c} \frac{a^{2}}{1+a+a b}\right) \geq \frac{3(a+b+c)^{2}}{3+a+b+c+a b... | 3 | Inequalities | proof | Yes | Yes | olympiads_ref | false |
Let $A B C$ be a scalene triangle with $B C=a, A C=b$ and $A B=c$, where $a, b, c$ are positive integers. Prove that
$$
\left|a b^{2}+b c^{2}+c a^{2}-a^{2} b-b^{2} c-c^{2} a\right| \geq 2
$$ | Denote \( E = a b^{2} + b c^{2} + c a^{2} - a^{2} b - b^{2} c - c^{2} a \). We have
\[
\begin{aligned}
E = & \left(a b c - c^{2} a\right) + \left(c a^{2} - a^{2} b\right) + \left(b c^{2} - b^{2} c\right) + \left(a b^{2} - a b c\right) = \\
& (b - c)\left(a c - a^{2} - b c + a b\right) = (b - c)\left(a (c - a) - b (c -... | 2 | Algebra | proof | Yes | Yes | olympiads_ref | false |
Prove that amongst any 29 natural numbers there are 15 such that sum of them is divisible by 15. | Amongst any 5 natural numbers there are 3 such that the sum of them is divisible by 3. Amongst any 29 natural numbers we can choose 9 groups with 3 numbers such that the sum of numbers in every group is divisible by 3. In that way we get 9 natural numbers such that all of them are divisible by 3. It is easy to see that... | 4 | Number Theory | proof | Yes | Yes | olympiads_ref | false |
The vertices $A$ and $B$ of an equilateral $\triangle A B C$ lie on a circle $k$ of radius 1, and the vertex $C$ is inside $k$. The point $D \neq B$ lies on $k, A D=A B$ and the line $D C$ intersects $k$ for the second time in point $E$. Find the length of the segment $C E$. | As $A D=A C, \triangle C D A$ is isosceles. If $\angle A D C=\angle A C D=\alpha$ and $\angle B C E=\beta$, then $\beta=120^{\circ}-\alpha$. The quadrilateral $A B E D$ is cyclic, so $\angle A B E=180^{\circ}-\alpha$. Then $\angle C B E=120^{\circ}-\alpha$ so $\angle C B E=\beta$. Thus $\triangle C B E$ is isosceles, s... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads_ref | false |
Let $s(a)$ denote the sum of digits of a given positive integer $a$. The sequence $a_{1}, a_{2}, \ldots a_{n}, \ldots$ of positive integers is such that $a_{n+1}=a_{n}+s\left(a_{n}\right)$ for each positive integer $n$. Find the greatest possible $n$ for which it is possible to have $a_{n}=2008$.
Let $s(a)$ denote the... | Since $a_{n-1} \equiv s\left(a_{n-1}\right)$ (all congruences are modulo 9), we have $2 a_{n-1} \equiv a_{n} \equiv 2008 \equiv 10$, so $a_{n-1} \equiv 5$. But $a_{n-1}<2008$, so $s\left(a_{n-1}\right) \leq 28$ and thus $s\left(a_{n-1}\right)$ can equal 5, 14, or 23. We check $s(2008-5)=s(2003)=5, s(2008-14)=s(1994)=23... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads_ref | false |
Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function, satisfying the following condition:
for every integer $n>1$, there exists a prime divisor $p$ of $n$ such that $f(n)=f\left(\frac{n}{p}\right)-f(p)$. If
$$
f\left(2^{2007}\right)+f\left(3^{2008}\right)+f\left(5^{2009}\right)=2006
$$
determine the value of
$$... | If $n=p$ is a prime number, we have
$$
f(p)=f\left(\frac{p}{p}\right)-f(p)=f(1)-f(p)
$$
i.e.
$$
f(p)=\frac{f(1)}{2}
$$
If $n=p q$, where $p$ and $q$ are prime numbers, then
$$
f(n)=f\left(\frac{n}{p}\right)-f(p)=f(q)-f(p)=\frac{f(1)}{2}-\frac{f(1)}{2}=0
$$
If $n$ is a product of three prime numbers, we have
$$
f... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads_ref | false |
Five players $(A, B, C, D, E)$ take part in a bridge tournament. Every two players must play (as partners) against every other two players. Any two given players can be partners not more than once per day. What is the least number of days needed for this tournament? | A given pair must play with three other pairs and these plays must be on different days, so at least three days are needed. Suppose that three days suffice. Let the pair $A B$ play against $C D$ on day $x$. Then $A B-D E$ and $C D-B E$ cannot play on day $x$. Then one of the other two plays of $D E$ (with $A C$ and $B ... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads_ref | false |
In a right trapezoid $A B C D(A B \| C D)$ the angle at vertex $B$ measures $75^{\circ}$. Point $H$ is the foot of the perpendicular from point $A$ to the line $B C$. If $B H=D C$ and $A D+A H=8$, find the area of $A B C D$. | Produce the legs of the trapezoid until they intersect at point \( E \). The triangles \( ABH \) and \( ECD \) are congruent (ASA). The area of \( ABCD \) is equal to the area of triangle \( EAH \) of hypotenuse
\[
AE = AD + DE = AD + AH = 8
\]
Let \( M \) be the midpoint of \( AE \). Then
\[
ME = MA = MH = 4
\]
an... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads_ref | false |
Let $A=1 \cdot 4 \cdot 7 \cdot \ldots \cdot 2014$ be the product of the numbers less or equal to 2014 that give remainder 1 when divided by 3. Find the last non-zero digit of $A$. | Grouping the elements of the product by ten we get:
$$
\begin{aligned}
& (30 k+1)(30 k+4)(30 k+7)(30 k+10)(30 k+13)(30 k+16) \\
& (30 k+19)(30 k+22)(30 k+25)(30 k+28)= \\
& =(30 k+1)(15 k+2)(30 k+7)(120 k+40)(30 k+13)(15 k+8) \\
& (30 k+19)(15 k+11)(120 k+100)(15 k+14)
\end{aligned}
$$
(We divide all even numbers not... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads_ref | false |
Find the largest positive integer $n$ for which the inequality
$$
\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c} \leq \frac{5}{2}
$$
holds for all $a, b, c \in[0,1]$. Here $\sqrt[1]{a b c}=a b c$. | Let $n_{\max }$ be the sought largest value of $n$, and let $E_{a, b, c}(n)=\frac{a+b+c}{a b c+1}+\sqrt[n]{a b c}$. Then $E_{a, b, c}(m)-E_{a, b, c}(n)=\sqrt[m]{a b c}-\sqrt[n]{a b c}$ and since $a b c \leq 1$ we clearly have $E_{a, b, c}(m) \geq E_{a, b, c}(n)$ for $m \geq n$. So if $E_{a, b, c}(n) \geq \frac{5}{2}$ f... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads_ref | false |
MNE
Let $a, b, c$ be positive real numbers. Prove that
$$
\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}>2
$$ | Starting from the double expression on the left-hand side of given inequality, and applying twice the Arithmetic-Geometric mean inequality, we find that
$$
\begin{aligned}
2 \frac{a}{b}+2 \sqrt{\frac{b}{c}}+2 \sqrt[3]{\frac{c}{a}} & =\frac{a}{b}+\left(\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt{\frac{b}{c}}\right)+2 \sqrt[3]... | 3 | Inequalities | proof | Yes | Yes | olympiads_ref | false |
GRE
Let $n \geq 1$ be a positive integer. A square of side length $n$ is divided by lines parallel to each side into $n^{2}$ squares of side length 1. Find the number of parallelograms which have vertices among the vertices of the $n^{2}$ squares of side length 1, with both sides smaller or equal to 2, and which have ... | We can divide all these parallelograms into 7 classes (types I-VII), according to Figure.
Type 1: There are $n$ ways to choose the strip for the horizontal (shorter) side of the parallelog... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads_ref | false |
Find the maximum positive integer $k$ such that for any positive integers $m, n$ such that $m^{3}+n^{3}>$ $(m+n)^{2}$, we have
$$
m^{3}+n^{3} \geq(m+n)^{2}+k
$$ | We see that for $m=3$ and $n=2$ we have $m^{3}+n^{3}>(m+n)^{2}$, thus
$$
3^{3}+2^{3} \geq(3+2)^{2}+k \Rightarrow k \leq 10
$$
We will show that $k=10$ is the desired maximum. In other words, we have to prove that
$$
m^{3}+n^{3} \geq(m+n)^{2}+10
$$
The last inequality is equivalent to
$$
(m+n)\left(m^{2}+n^{2}-m n-... | 10 | Inequalities | math-word-problem | Yes | Yes | olympiads_ref | false |
A set $T$ of $n$ three-digit numbers has the following five properties:
(1) No number contains the digit 0.
(2) The sum of the digits of each number is 9.
(3) The units digits of any two numbers are different.
(4) The tens digits of any two numbers are different.
(5) The hundreds digits of any two numbers are diff... | Let $S$ denote the set of three-digit numbers that have digit sum equal to 9 and no digit equal to 0. We will first find the cardinality of $S$. We start from the number 111 and each element of $S$ can be obtained from 111 by a string of $6 A$'s (which means that we add 1 to the current digit) and 2 G's (which means go... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads_ref | false |
Let \( a, b, c \) be positive real numbers such that \( a b c = 8 \). Prove that
\[
\frac{a b + 4}{a + 2} + \frac{b c + 4}{b + 2} + \frac{c a + 4}{c + 2} \geq 6
\] | We have $a b+4=\frac{8}{c}+4=\frac{4(c+2)}{c}$ and similarly $b c+4=\frac{4(a+2)}{a}$ and $c a+4=\frac{4(b+2)}{b}$. It follows that
$$
(a b+4)(b c+4)(c a+4)=\frac{64}{a b c}(a+2)(b+2)(c+2)=8(a+2)(b+2)(c+2)
$$
so that
$$
\frac{(a b+4)(b c+4)(c a+4)}{(a+2)(b+2)(c+2)}=8
$$
Applying AM-GM, we conclude:
$$
\frac{a b+4}... | 6 | Inequalities | proof | Yes | Yes | olympiads_ref | false |
Determine the number of pairs of integers $(m, n)$ such that
$$
\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}} \in \mathbb{Q}
$$ | Let $r=\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}}$. Then
$$
n+m+2 \sqrt{n+\sqrt{2016}} \cdot \sqrt{m-\sqrt{2016}}=r^{2}
$$
and
$$
(m-n) \sqrt{2106}=\frac{1}{4}\left(r^{2}-m-n\right)^{2}-m n+2016 \in \mathbb{Q}
$$
Since $\sqrt{2016} \notin \mathbb{Q}$, it follows that $m=n$. Then
$$
\sqrt{n^{2}-2016}=\frac{1}{2}\lef... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads_ref | false |
Let $S_{n}$ be the sum of reciprocal values of non-zero digits of all positive integers up to (and including) $n$. For instance, $S_{13}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{1}{1}+\fr... | We will first calculate \( S_{999} \), then \( S_{1999} - S_{999} \), and then \( S_{2016} - S_{1999} \).
Writing the integers from 1 to 999 as 001 to 999, adding eventually also 000 (since 0 digits actually do not matter), each digit appears exactly 100 times in each position (as unit, ten, or hundred). Hence
\[
S_{... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads_ref | false |
Find the maximum number of natural numbers $x_{1}, x_{2}, \ldots, x_{m}$ satisfying the conditions:
a) No $x_{i}-x_{j}, 1 \leq i<j \leq m$ is divisible by 11 ; and
b) The sum $x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2} \ldots x_{m-1}$ is divisible by 11. | The required maximum is 10.
According to a), the numbers $x_{i}, 1 \leq i \leq m$, are all different $(\bmod 11)$ (1)
Hence, the number of natural numbers satisfying the conditions is at most 11.
If $x_{j} \equiv 0(\bmod 11)$ for some $j$, then
$$
x_{2} x_{3} \ldots x_{m}+x_{1} x_{3} \ldots x_{m}+\cdots+x_{1} x_{2}... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads_ref | false |
For a positive integer $a$, define a sequence of integers $x_{1}, x_{2}, \ldots$ by letting $x_{1}=a$ and $x_{n+1}=2 x_{n}+1$ for $n \geq 1$. Let $y_{n}=2^{x_{n}}-1$. Determine the largest possible $k$ such that, for some positive integer $a$, the numbers $y_{1}, \ldots, y_{k}$ are all prime.
(Russia) Valery Senderov | The largest such is $k=2$. Notice first that if $y_{i}$ is prime, then $x_{i}$ is prime as well. Actually, if $x_{i}=1$ then $y_{i}=1$ which is not prime, and if $x_{i}=m n$ for integer $m, n>1$ then $2^{m}-1 \mid 2^{x_{i}}-1=y_{i}$, so $y_{i}$ is composite. In particular, if $y_{1}, y_{2}, \ldots, y_{k}$ are primes fo... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads_ref | false |
A cubic sequence is a sequence of integers given by $a_{n}=n^{3}+b n^{2}+c n+d$, where $b, c$ and $d$ are integer constants and $n$ ranges over all integers, including negative integers.
(a) Show that there exists a cubic sequence such that the only terms of the sequence which are squares of integers are $a_{2015}$ and... | The only possible value of $a_{2015} \cdot a_{2016}$ is 0. For simplicity, by performing a translation of the sequence (which may change the defining constants $b, c$ and $d$), we may instead concern ourselves with the values $a_{0}$ and $a_{1}$, rather than $a_{2015}$ and $a_{2016}$.
Suppose now that we have a cubic ... | 0 | Algebra | proof | Yes | Yes | olympiads_ref | false |
Determine all positive integers $n$ satisfying the following condition: for every monic polynomial $P$ of degree at most $n$ with integer coefficients, there exists a positive integer $k \leq n$, and $k+1$ distinct integers $x_{1}, x_{2}, \ldots, x_{k+1}$ such that
$$
P\left(x_{1}\right)+P\left(x_{2}\right)+\cdots+P\l... | There is only one such integer, namely, $n=2$. In this case, if $P$ is a constant polynomial, the required condition is clearly satisfied; if $P=X+c$, then $P(c-1)+P(c+1)=$ $P(3c)$; and if $P=X^{2}+qX+r$, then $P(X)=P(-X-q)$.
To rule out all other values of $n$, it is sufficient to exhibit a monic polynomial $P$ of de... | 2 | Algebra | proof | Yes | Yes | olympiads_ref | false |
Xenia and Sergey play the following game. Xenia thinks of a positive integer $N$ not exceeding 5000. Then she fixes 20 distinct positive integers $a_{1}, a_{2}, \ldots, a_{20}$ such that, for each $k=1,2, \ldots, 20$, the numbers $N$ and $a_{k}$ are congruent modulo $k$. By a move, Sergey tells Xenia a set $S$ of posit... | Sergey can determine Xenia's number in 2 but not fewer moves.
We first show that 2 moves are sufficient. Let Sergey provide the set $\{17,18\}$ on his first move, and the set $\{18,19\}$ on the second move. In Xenia's two responses, exactly one number occurs twice, namely, $a_{18}$. Thus, Sergey is able to identify $a_... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads_ref | false |
Two players, Blake and Ruby, play the following game on an infinite grid of unit squares, all initially colored white. The players take turns starting with Blake. On Blake's turn, Blake selects one white unit square and colors it blue. On Ruby's turn, Ruby selects two white unit squares and colors them red. The players... | The answer is 4 squares. Algorithm for Blake to obtain at least 4 squares. We simply let Blake start with any cell blue, then always draw adjacent to a previously drawn blue cell until this is no longer possible. Note that for $n \leq 3$, any connected region of $n$ blue cells has more than $2 n$ liberties (non-blue ce... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads_ref | false |
Suppose that the set $\{1,2, \ldots, 1998\}$ has been partitioned into disjoint pairs $\left\{a_{i}, b_{i}\right\}$ $(1 \leq i \leq 999)$ so that for all $i,\left|a_{i}-b_{i}\right|$ equals 1 or 6. Prove that the sum
$$ \left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\cdots+\left|a_{999}-b_{999}\right| $$
ends in th... | Let $S$ be the sum. Modulo 2,
$$ S=\sum\left|a_{i}-b_{i}\right| \equiv \sum\left(a_{i}+b_{i}\right)=1+2+\cdots+1998 \equiv 1 \quad(\bmod 2) $$
Modulo 5,
$$ S=\sum\left|a_{i}-b_{i}\right|=1 \cdot 999 \equiv 4 \quad(\bmod 5) $$
So $S \equiv 9(\bmod 10)$. | 9 | Number Theory | proof | Yes | Yes | olympiads_ref | false |
The Planar National Park is an undirected 3-regular planar graph (i.e., all vertices have degree 3). A visitor walks through the park as follows: she begins at a vertex and starts walking along an edge. When she reaches the other endpoint, she turns left. On the next vertex she turns right, and so on, alternating left ... | The answer is 3. We consider the trajectory of the visitor as an ordered sequence of turns. A turn is defined by specifying a vertex, the incoming edge, and the outgoing edge. Hence there are six possible turns for each vertex. Claim - Given one turn in the sequence, one can reconstruct the entire sequence of turns. Th... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads_ref | false |
A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is essentially increasing if $f(s) \leq f(t)$ holds whenever $s \leq t$ are real numbers such that $f(s) \neq 0$ and $f(t) \neq 0$. Find the smallest integer $k$ such that for any 2022 real numbers $x_{1}, x_{2}, \ldots, x_{2022}$, there exist $k$ essentially increasing... | The answer is 11 and, more generally, if 2022 is replaced by $N$ then the answer is $\left\lfloor\log _{2} N\right\rfloor+1$. 『 Bound. Suppose for contradiction that $2^{k}-1>N$ and choose $x_{n}=-n$ for each $n=1, \ldots, N$. Now for each index $1 \leq n \leq N$, define $$ S(n)=\left\{\text { indices } i \text { for... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads_ref | false |
Let $A B C$ be an equilateral triangle with side length 1. Points $A_{1}$ and $A_{2}$ are chosen on side $B C$, points $B_{1}$ and $B_{2}$ are chosen on side $C A$, and points $C_{1}$ and $C_{2}$ are chosen on side $A B$ such that $B A_{1}<B A_{2}, C B_{1}<C B_{2}$, and $A C_{1}<A C_{2}$. Suppose that the three line se... | The only possible value of the common perimeter, denoted \( p \), is 1. 【 Synthetic approach (from author). We prove the converse of the problem first: Claim (\( p=1 \) implies concurrence) - Suppose the six points are chosen so that triangles \( A B_{2} C_{1}, B C_{2} A_{1}, C A_{2} B_{1} \) all have perimeter 1. Then... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads_ref | false |
The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Use properties of exponents to move the squares outside the brackets use difference of squares.
\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]
Using the binomial theorem, we can see that the number of terms is $\boxed{\mathrm{(B)}\ 5}$. | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
When $x^{13}+1$ is divided by $x-1$, the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$ | Solution 1
Using synthetic division, we get that the remainder is $\boxed{\textbf{(D)}\ 2}$.
Solution 2
By the remainder theorem, the remainder is equal to the expression $x^{13}+1$ when $x=1.$ This gives the answer of $\boxed{(\mathrm{D})\ 2.}$
Solution 3
Note that $x^{13} - 1 = (x - 1)(x^{12} + x^{11} \cdots + 1)$,... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$ | Solution 1
$\log_{5}\frac{(125)(625)}{25}$ can be simplified to $\log_{5}\ (125)(25)$ since $25^2 = 625$. $125 = 5^3$ and $5^2 = 25$ so $\log_{5}\ 5^5$ would be the simplest form. In $\log_{5}\ 5^5$, $5^x = 5^5$. Therefore, $x = 5$ and the answer is $\boxed{\mathrm{(D)}\ 5}$.
Solution 2
$\log_{5}\frac{(125)(625)}{25}$... | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$. $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:
$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph... | Let the speed of boy $A$ be $a$, and the speed of boy $B$ be $b$. Notice that $A$ travels $4$ miles per hour slower than boy $B$, so we can replace $b$ with $a+4$.
Now let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=rt$ to ma... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The sum of the roots of the equation $4x^{2}+5-8x=0$ is equal to:
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ -5\qquad\textbf{(C)}\ -\frac{5}{4}\qquad\textbf{(D)}\ -2\qquad\textbf{(E)}\ \text{None of these}$ | We can divide by 4 to get:
$x^{2}-2x+\dfrac{5}{4}=0.$
The Vieta's formula states that in quadratic equation $ax^2+bx+c$, the sum of the roots of the equation is $-\frac{b}{a}$.
Using Vieta's formula, we find that the roots add to $2$ or $\boxed{\textbf{(E)}\ \text{None of these}}$. | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A $25$ foot ladder is placed against a vertical wall of a building. The foot of the ladder is $7$ feet from the base of the building. If the top of the ladder slips $4$ feet, then the foot of the ladder will slide:
$\textbf{(A)}\ 9\text{ ft} \qquad \textbf{(B)}\ 15\text{ ft} \qquad \textbf{(C)}\ 5\text{ ft} \qquad \tex... | By the Pythagorean triple $(7,24,25)$, the point where the ladder meets the wall is $24$ feet above the ground. When the ladder slides, it becomes $20$ feet above the ground. By the $(15,20,25)$ Pythagorean triple, The foot of the ladder is now $15$ feet from the building. Thus, it slides $15-7 = \boxed{\textbf{(D)}\ 8... | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
In triangle $ABC$, $AC=24$ inches, $BC=10$ inches, $AB=26$ inches. The radius of the inscribed circle is:
$\textbf{(A)}\ 26\text{ in} \qquad \textbf{(B)}\ 4\text{ in} \qquad \textbf{(C)}\ 13\text{ in} \qquad \textbf{(D)}\ 8\text{ in} \qquad \textbf{(E)}\ \text{None of these}$ | The inradius is equal to the area divided by semiperimeter. The area is $\frac{(10)(24)}{2} = 120$ because it's a right triangle, as it's side length satisfies the Pythagorean Theorem. The semiperimeter is $30$. Therefore the inradius is $\boxed{\textbf{(B)}\ 4}$. | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The limit of $\frac {x^2-1}{x-1}$ as $x$ approaches $1$ as a limit is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \text{Indeterminate} \qquad \textbf{(C)}\ x-1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 1$ | Both $x^2-1$ and $x-1$ approach 0 as $x$ approaches $1$, using the L'Hôpital's rule, we have $\lim \limits_{x\to 1}\frac{x^2-1}{x-1} = \lim \limits_{x\to 1}\frac{2x}{1} = 2$.
Thus, the answer is $\boxed{\textbf{(D)}\ 2}$.
~ MATH__is__FUN | 2 | Calculus | MCQ | Yes | Yes | amc_aime | false |
$A$ can do a piece of work in $9$ days. $B$ is $50\%$ more efficient than $A$. The number of days it takes $B$ to do the same piece of work is:
$\textbf{(A)}\ 13\frac {1}{2} \qquad\textbf{(B)}\ 4\frac {1}{2} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \text{none of these answers}$ | Because $B$ is $50\%$ more efficient, he can do $1.5$ pieces of work in $9$ days. This is equal to 1 piece of work in $\textbf{(C)}\ 6$ days | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$, is:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$ | Factoring the polynomial gives $(n+1)(n)(n-1)$ According to the factorization, one of those factors must be a multiple of two because there are more than 2 consecutive integers. In addition, because there are three consecutive integers, one of the integers must be a multiple of 3. Therefore $6$ must divide the given e... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A total of $28$ handshakes were exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was:
$\textbf{(A)}\ 14\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 56\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 7$ | The handshake equation is $h=\frac{n(n-1)}{2}$, where $n$ is the number of people and $h$ is the number of handshakes. There were $28$ handshakes, so $28=\frac{n(n-1)}{2}$
$56=n(n-1)$
The factors of $56$ are:
$1, 2, 4, 7, 8, 14, 28, 56$.
As we can see, only $7, 8$ fit the requirements $n(n-1)$ if $n$ was an integer.
Th... | 8 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
The value of $10^{\log_{10}7}$ is:
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ \log_{10}7\qquad\textbf{(E)}\ \log_{7}10$ | $\log_{10}7=x \Rightarrow 10^x=7$
Substitute $\log_{10}7=x$ in $10^{\log_{10}7} \Rightarrow 10^x=?$
It was already stated that $10^x=7$, so our answer is $\textbf{(A)}\ 7$ | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
$\left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2}$ equals:
$\textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2}$
$\textbf{(E)}\ [(x^{3}-1)^{2}]^{2}$ | First, note that we can pull the exponents out of every factor, since they are all squared. This results in
$\left(\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\right)^{4}\cdot\left(\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\right)^{4}$
Now, multiplying the numerators together gives
$\left(\frac{x^3+1}{x^3+1}\right)^{4}\cdot\left(\frac{x^3-1... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Tom, Dick and Harry started out on a $100$-mile journey. Tom and Harry went by automobile at the rate of $25$ mph, while Dick walked at the rate of $5$ mph. After a certain distance, Harry got off and walked on at $5$ mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. ... | Let $d_1$ be the distance (in miles) that Harry traveled on car, and let $d_2$ be the distance (in miles) that Tom backtracked to get Dick. Let $T$ be the time (in hours) that it took the three to complete the journey. We now examine Harry's journey, Tom's journey, and Dick's journey. These yield, respectively, the equ... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $\left(r+\frac1r\right)^2=3$, then $r^3+\frac1{r^3}$ equals
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 6$ | We know $r+\frac1r=\sqrt3$. Cubing this gives $r^3+3r+\frac3r+\frac1{r^3}=3\sqrt3$. But $3r+\frac3r=3\left(r+\frac1r\right)=3\sqrt3$, so subtracting this from the first equation gives
$r^3+\frac1{r^3}=\boxed{0\textbf{ (C)}}$.
(Actually, $r+\frac1r$ could have been equal to $-\sqrt3$ instead of $\sqrt3$, but this would... | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Increasing the radius of a cylinder by $6$ units increased the volume by $y$ cubic units. Increasing the height of the cylinder by $6$ units also increases the volume by $y$ cubic units. If the original height is $2$, then the original radius is:
$\text{(A) } 2 \qquad \text{(B) } 4 \qquad \text{(C) } 6 \qquad \text{(D... | We know that the volume of a cylinder is equal to $\pi r^2h$, where $r$ and $h$ are the radius and height, respectively. So we know that $2\pi (r+6)^2-2\pi r^2=y=\pi r^2(2+6)-2\pi r^2$. Expanding and rearranging, we get that $2\pi (12r+36)=6\pi r^2$. Divide both sides by $6\pi$ to get that $4r+12=r^2$, and rearrange to... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $f(a)=a-2$ and $F(a,b)=b^2+a$, then $F(3,f(4))$ is:
$\textbf{(A)}\ a^2-4a+7 \qquad \textbf{(B)}\ 28 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 11$ | We find $f(4)=(4)-2=2$, so $F(3,f(4))=F(3,2)=(2)^2+3=\boxed{\textbf{(C) }7}$. | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The product, $\log_a b \cdot \log_b a$ is equal to:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$ | \[a^x=b\]
\[b^y=a\]
\[{a^x}^y=a\]
\[xy=1\]
\[\log_a b\log_b a=1\]
As a result, the answer should be $\boxed{\textbf{(A) }1}$. | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ % alcohol to a lotion containing $30$ % alcohol is:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$ | Say we add $N$ ounces of water to the shaving lotion. Since half of a $9$ ounce bottle of shaving lotion is alcohol, we know that we have $\frac{9}{2}$ ounces of alcohol. We want $\frac{9}{2}=0.3(9+N)$ (because we want the amount of alcohol, $\frac{9}{2}$, to be $30\%$, or $0.3$, of the total amount of shaving lotion, ... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A right circular cone has for its base a circle having the same radius as a given sphere.
The volume of the cone is one-half that of the sphere. The ratio of the altitude of the cone to the radius of its base is:
$\textbf{(A)}\ \frac{1}{1}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}... | Because the circle has the same radius as the sphere, the cylinder and sphere have the same radius. Then from the volume of cylinder and volume of a sphere formulas, we have $\frac{1}{3} \pi r^2 h= \frac{2}{3} \pi r^3 \implies h=2r\implies \frac{h}{r}=2$ $\boxed{(\textbf{D})}$ | 2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
If the Highest Common Divisor of $6432$ and $132$ is diminished by $8$, it will equal:
$\textbf{(A)}\ -6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ -2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | $\gcd(6432, 132)$
$13\cdot 12=132$
$\frac{6432}{6}=1072\implies\frac{1072}{4}=268\implies\frac{268}{4}=67$, so $\gcd(2^5\cdot 3\cdot 67, 2^2\cdot 3\cdot 13)=2^2\cdot 3=12\implies 12-8=4, \fbox{E}$ | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
If $\left (a+\frac{1}{a} \right )^2=3$, then $a^3+\frac{1}{a^3}$ equals:
$\textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3}$ | $a+\frac{1}{a}=\sqrt{3}\implies (a+\frac{1}{a})^3=3\sqrt{3}\implies a^3+3\frac{a^2}{a}+\frac{3a}{a^2}+\frac{1}{a^3}\implies a^3+3(a+\frac{1}{a})+\frac{1}{a^3}=3\sqrt{3}\implies a^3+\frac{1}{a^3}+3\sqrt{3}=3\sqrt{3}\implies a^3+\frac{1}{a^3}=0$, $\fbox{C}$ | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The sum of all the roots of $4x^3-8x^2-63x-9=0$ is:
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ -8 \qquad \textbf{(D)}\ -2 \qquad \textbf{(E)}\ 0$ | By Vieta's Formulas, $\frac{--8}{4}=2$, $\fbox{B}$ | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A housewife saved $\textdollar{2.50}$ in buying a dress on sale. If she spent $\textdollar{25}$ for the dress, she saved about:
$\textbf{(A)}\ 8 \% \qquad \textbf{(B)}\ 9 \% \qquad \textbf{(C)}\ 10 \% \qquad \textbf{(D)}\ 11 \% \qquad \textbf{(E)}\ 12\%$ | Since she saved $\textdollar{2.50}$ and spent $\textdollar{25}$, the original price of the dress was $\textdollar{27.50}$. The percent saved can be modeled as $27.5x=2.5$, so the answer is $x=\frac{2.5}{27.5}$ or $\frac{1}{11}$, approximately $9\% \implies \textbf{(B)}$
[1954 AHSC](https://artofproblemsolving.com/w... | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:
$\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0$ | First, square both sides. This gives us
\[\sqrt{5x-1}^2+2\cdot\sqrt{5x-1}\cdot\sqrt{x-1}+\sqrt{x-1}^2=4 \Longrightarrow 5x-1+2\cdot\sqrt{(5x-1)\cdot(x-1)}+x-1=4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}+6x-2=4\]
Then, adding $-6x$ to both sides gives us
\[2\cdot\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \Longrightarrow 2\cdot\sqrt{5x... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The value of $x + x(x^x)$ when $x = 2$ is:
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 16 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 36 \qquad\textbf{(E)}\ 64$ | Simple substitution yields
\[2 + 2(2^2) = 2 + 2(4) = 10\]
Therefore, the answer is $\fbox{(A) 10}$ | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ \text{more than }6$ | Let the angle be $\theta$ and the sides around it be $a$ and $b$.
The area of the triangle can be written as \[A =\frac{a \cdot b \cdot \sin(\theta)}{2}\]
The doubled sides have length $2a$ and $2b$, while the angle is still $\theta$. Thus, the area is
\[\frac{2a \cdot 2b \cdot \sin(\theta)}{2}\]
\[\Rrightarrow \frac... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
If $n$ is any whole number, $n^2(n^2 - 1)$ is always divisible by
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ \text{any multiple of }12\qquad \textbf{(D)}\ 12-n\qquad \textbf{(E)}\ 12\text{ and }24$ | Suppose $n$ is even. So, we have $n^2(n+1)(n-1).$ Out of these three numbers, at least one of them is going to be a multiple of 3. $n^2$ is also a multiple of 4. Therefore, this expression is always divisible by $\boxed{\textbf{(A)} \quad 12}.$
-coolmath34 | 12 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The number of scalene triangles having all sides of integral lengths, and perimeter less than $13$ is:
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 18$ | We can write all possible triangles adding up to 12 or less
\[(2, 4, 5)=11\]
\[(3, 4, 5)=12\]
\[(2, 3, 4)=9\]
This leaves $\boxed{\textbf{(C)} \quad 3}$ scalene triangles.
-coolmath34
-rubslul
(If you see any cases I missed out, edit them in.) | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
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 $\boxed{\textbf{(C) }6}$ nickels can be placed around the central nickel. | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 14$ | Let there be $x$ cows and $y$ chickens. Then, there are $4x+2y$ legs and $x+y$ heads. Writing the equation:
\[4x+2y=14+2(x+y)\] \[4x+2y=14+2x+2y\] \[2x=14\] \[x=\boxed{\textbf{(B) }7}\] | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $8\cdot2^x = 5^{y + 8}$, then when $y = - 8,x =$
$\textbf{(A)}\ - 4 \qquad\textbf{(B)}\ - 3 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 8$ | Simple substitution yields
\[8 \cdot 2^{x} = 5^{0}\]
Reducing the equation gives
\[8 \cdot 2^{x} = 1\]
Dividing by 8 gives
\[2^{x}=\frac{1}{8}\]
This simply gives that $x=-3$.
Therefore, the answer is $\fbox{(B) -3}$. | -3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
At a dance party a group of boys and girls exchange dances as follows: The first boy dances with $5$ girls, a second boy dances with $6$ girls, and so on, the last boy dancing with all the girls. If $b$ represents the number of boys and $g$ the number of girls, then:
$\textbf{(A)}\ b = g\qquad \textbf{(B)}\ b = \frac{... | After inspection, we notice a general pattern: the $nth$ boy dances with $n + 4$ girls.
Since the last boy dances with all the girls, there must be four more girls than guys.
Therefore, the equation that relates them is $\fbox{(C) b = g - 4}$ | 4 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
The points $(2,-3)$, $(4,3)$, and $(5, k/2)$ are on the same straight line. The value(s) of $k$ is (are):
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ -12\qquad \textbf{(C)}\ \pm 12\qquad \textbf{(D)}\ {12}\text{ or }{6}\qquad \textbf{(E)}\ {6}\text{ or }{6\frac{2}{3}}$ | First find the slope. Then use the point-slope formula to find the equation of the line. Then substitute 5 for x to find y.
$\text{(A)}12$ | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The expression $2 + \sqrt{2} + \frac{1}{2 + \sqrt{2}} + \frac{1}{\sqrt{2} - 2}$ equals:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2 - \sqrt{2}\qquad \textbf{(C)}\ 2 + \sqrt{2}\qquad \textbf{(D)}\ 2\sqrt{2}\qquad \textbf{(E)}\ \frac{\sqrt{2}}{2}$ | To make this problem easier to solve, lets get the radicals out of the denominator. For $\frac{1}{2 + \sqrt2}$, we will multiply the numerator and denominator by $2- \sqrt2$ so,
$\frac{1}{2 + \sqrt2} \cdot \frac{2 - \sqrt2}{2 - \sqrt2} \Rightarrow \frac{2 - \sqrt2}{2}$.
Now, the other fraction we need to get the radica... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The expression$\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},$ when simplified is:
$\textbf{(A)}\ \frac{(x-1)(x-6)}{(x-3)(x-4)} \qquad\textbf{(B)}\ \frac{x+3}{x-3}\qquad\textbf{(C)}\ \frac{x+1}{x-1}\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$ | Factoring each of the binomials in the expression $\frac{x^2-3x+2}{x^2-5x+6}\div \frac{x^2-5x+4}{x^2-7x+12},$ will yield the result of \[\frac{(x-2)(x-1)}{(x-3)(x-2)}\div \frac{(x-4)(x-1)}{(x-3)(x-4)},\]
We can eliminate like terms to get $\frac {x-1}{x-3}\div \frac{x-1}{x-3}$, which, according to identity property, is... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $y=a+\frac{b}{x}$, where $a$ and $b$ are constants, and if $y=1$ when $x=-1$, and $y=5$ when $x=-5$, then $a+b$ equals:
$\textbf{(A)}\ -1 \qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | Plugging in the x and y values, we obtain the following system of equations:
\[\begin {cases} 1 = a - b \\ 5 = a - \frac{b}{5} \end {cases}\]
We can then subtract the equations to obtain the equation $4 = 0.8b$, which works out to $b = 5$.
Plugging that in to the original system of equations:
\[\begin {cases} 1 = a - 5... | 11 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let the roots of $x^2-3x+1=0$ be $r$ and $s$. Then the expression $r^2+s^2$ is: $\textbf{(A)}\ \text{a positive integer} \qquad\textbf{(B)}\ \text{a positive fraction greater than 1}\qquad\textbf{(C)}\ \text{a positive fraction less than 1}\qquad\textbf{(D)}\ \text{an irrational number}\qquad\textbf{(E)}\ \text{an imag... | You may recognize that $r^2+s^2$ can be written as $(r+s)^2-2rs$. Then, by [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas), \[r+s=-(-3)=3\]and \[rs=1.\]Therefore, plugging in the values for $r+s$ and $rs$, \[(r+s)^2-2rs=(3)^2-2(1)=9-2=7.\]Hence, we can say that the expression $r^2+... | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The value of $\left(256\right)^{.16}\left(256\right)^{.09}$ is:
$\textbf{(A)}\ 4 \qquad \\ \textbf{(B)}\ 16\qquad \\ \textbf{(C)}\ 64\qquad \\ \textbf{(D)}\ 256.25\qquad \\ \textbf{(E)}\ -16$ | When we multiply numbers with exponents, we add the exponents together and leave the bases unchanged. We can apply this concept to computate $256^{0.16} \cdot 256^{0.09}$:
\[256^{0.16} \cdot 256^{0.09} = 256^{0.16+0.09}=256^{0.25}.\]
Now we can convert the decimal exponent to a fraction: \[256^{0.25} = 256^{\frac{1}{4... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $2$ is a solution (root) of $x^3+hx+10=0$, then $h$ equals:
$\textbf{(A) }10\qquad \textbf{(B) }9 \qquad \textbf{(C) }2\qquad \textbf{(D) }-2\qquad \textbf{(E) }-9$ | Substitute $2$ for $x$. We are given that this equation is true. Solving for $h$ gives $h=-9$. The answer is $\boxed{\textbf{(E)}}$. | -9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
It takes $5$ seconds for a clock to strike $6$ o'clock beginning at $6:00$ o'clock precisely. If the strikings are uniformly spaced, how long, in seconds, does it take to strike $12$ o'clock?
$\textbf{(A)}9\frac{1}{5}\qquad \textbf{(B )}10\qquad \textbf{(C )}11\qquad \textbf{(D )}14\frac{2}{5}\qquad \textbf{(E )}\text{... | Between six strikes, there are five intervals of space. Since it takes five seconds total, each interval is one second long. Therefore, $12$ strikes will take $11$ seconds. $\boxed{\textbf{(C)}}$. | 11 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
If $\log_{2x}216 = x$, where $x$ is real, then $x$ is:
$\textbf{(A)}\ \text{A non-square, non-cube integer}\qquad$
$\textbf{(B)}\ \text{A non-square, non-cube, non-integral rational number}\qquad$
$\textbf{(C)}\ \text{An irrational number}\qquad$
$\textbf{(D)}\ \text{A perfect square}\qquad$
$\textbf{(E)}\ \text{A perf... | Rewrite the equation to $(2x)^x = 216$. Since $2x$ is the base of the logarithm, we only need to check x-values that are positive.
With trial and error, $3$ is a solution because $(2 \cdot 3)^3 = 6^3 = 216$. Since $(2x)^x$ gets larger as x gets larger, $3$ is the only solution, so the answer is $\boxed{\textbf{(A)}}$... | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The first three terms of a geometric progression are $\sqrt{2}, \sqrt[3]{2}, \sqrt[6]{2}$. Find the fourth term.
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ \sqrt[7]{2}\qquad \textbf{(C)}\ \sqrt[8]{2}\qquad \textbf{(D)}\ \sqrt[9]{2}\qquad \textbf{(E)}\ \sqrt[10]{2}$ | After rewriting the radicals as fractional exponents, the sequence is $2^{1/2}, 2^{1/3}, 2^{1/6}$.
The common ratio of the [geometric sequence](https://artofproblemsolving.com/wiki/index.php/Geometric_sequence) is $\frac{2^{1/3}}{2^{1/2}} = 2^{-1/6}$. Multiplying that by the third term results in $2^0$. It simplifies... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In the base ten number system the number $526$ means $5 \times 10^2+2 \times 10 + 6$.
In the Land of Mathesis, however, numbers are written in the base $r$.
Jones purchases an automobile there for $440$ monetary units (abbreviated m.u).
He gives the salesman a $1000$ m.u bill, and receives, in change, $340$ m.u. The... | If Jones received $340$ m.u. change after paying $1000$ m.u. for something that costs $440$ m.u., then
\[440_r + 340_r = 1000_r\]
This equation can be rewritten as
\[4r^2 + 4r + 3r^2 + 4r = r^3\]
Bring all of the terms to one side to get
\[r^3 - 7r^2 - 8r = 0\]
Factor to get
\[r(r-8)(r+1)=0\]
Since [base numbers](https... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The yearly changes in the population census of a town for four consecutive years are,
respectively, 25% increase, 25% increase, 25% decrease, 25% decrease.
The net change over the four years, to the nearest percent, is:
$\textbf{(A)}\ -12 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1\qquad \te... | A 25% increase means the new population is $\frac{5}{4}$ of the original population. A 25% decrease means the new population is $\frac{3}{4}$ of the original population.
Thus, after four years, the population is $1 \cdot \frac{5}{4} \cdot \frac{5}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{225}{256}$ times the ori... | -12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $2137^{753}$ is multiplied out, the units' digit in the final product is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$ | $7^1$ has a unit digit of $7$.
$7^2$ has a unit digit of $9$.
$7^3$ has a unit digit of $3$.
$7^4$ has a unit digit of $1$.
$7^5$ has a unit digit of $7$.
Notice that the unit digit eventually cycles to itself when the exponent is increased by $4$. It also does not matter what the other digits are in the base because ... | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
If the graphs of $2y+x+3=0$ and $3y+ax+2=0$ are to meet at right angles, the value of $a$ is:
$\textbf{(A)}\ \pm \frac{2}{3} \qquad \textbf{(B)}\ -\frac{2}{3}\qquad \textbf{(C)}\ -\frac{3}{2} \qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ -6$ | The slope of the first graph is -1/2. The slope of the second is 2, since it is perpendicular, and it is also -a/3 by rearranging. Thus $a=-6$. Answer is $\boxed{\textbf{(E)}}$. | -6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A regular polygon of $n$ sides is inscribed in a circle of radius $R$. The area of the polygon is $3R^2$. Then $n$ equals:
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 15\qquad \textbf{(E)}\ 18$ | Note that the distance from the center of the [circle](https://artofproblemsolving.com/wiki/index.php/Circle) to each of the vertices of the inscribed [regular polygon](https://artofproblemsolving.com/wiki/index.php/Regular_polygon) equals the radius $R$. Since each side of a regular polygon is the same length, all th... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
When simplified, $\log{8} \div \log{\frac{1}{8}}$ becomes:
$\textbf{(A)}\ 6\log{2} \qquad \textbf{(B)}\ \log{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 0\qquad \textbf{(E)}\ -1$ | First, note that $\frac{1}{8} = 8^{-1}$. That means the expression can be rewritten as
\[\log{8} \div \log{8^{-1}}\]
\[\log{8} \cdot \frac{1}{\log{8^{-1}}}\]
\[\log{8} \cdot \frac{1}{-1 \cdot \log{8}}\]
This simplifies to $-1$, which is answer choice $\boxed{\textbf{(E)}}$. | -1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A man drives $150$ miles to the seashore in $3$ hours and $20$ minutes. He returns from the shore to the starting point in $4$ hours and $10$ minutes. Let $r$ be the average rate for the entire trip. Then the average rate for the trip going exceeds $r$ in miles per hour, by:
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 4\frac{... | Since the equation for rate is $r=\frac{d}{t}$, you only need to find $d$ and $t$. The distance from the seashore to the starting point is $150$ miles, and since he makes a round trip, $d=300$. Also, you know the times it took him to go both directions, so when you add them up ($3$ hours and $20$ minutes $+$ $4$ hours ... | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The difference between the larger root and the smaller root of $x^2 - px + (p^2 - 1)/4 = 0$ is:
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ p\qquad\textbf{(E)}\ p+1$ | Call the two roots $r$ and $s$, with $r \ge s$.
By Vieta's formulas, $p=r+s$ and $(p^2-1)/4=rs.$
(Multiplying both sides of the second equation by 4 gives $p^2-1=4rs$.)
The value we need to find, then, is $r-s$.
Since $p=r+s$, $p^2=r^2+2rs+s^2$.
Subtracting $p^2-1=4rs$ from both sides gives $1=r^2-2rs+s^2$.
Taking squ... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
When $\left ( 1 - \frac{1}{a} \right ) ^6$ is expanded the sum of the last three coefficients is:
$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ -10\qquad\textbf{(E)}\ -11$ | This is equivalent to $\frac{(a-1)^6}{a^6}.$
Its expansion has 7 terms, whose coefficients are the same as those of $(a-1)^6$.
By the Binomial Theorem, the sum of the last three coefficients is
$\binom{6}{2}-\binom{6}{1}+\binom{6}{0}=15-6+1=\boxed{10 \textbf{ (C)}}$. | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If the parabola $y = ax^2 + bx + c$ passes through the points $( - 1, 12)$, $(0, 5)$, and $(2, - 3)$, the value of $a + b + c$ is:
$\textbf{(A)}\ -4\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$ | Substituting in the $(x, y)$ pairs gives the following system of equations:
\[a-b+c=12\]
\[c=5\]
\[4a+2b+c=-3\]
We know $c=5$, so plugging this in reduces the system to two variables:
\[a-b=7\]
\[4a+2b=-8\]
Dividing the second equation by 2 gives $2a+b=-4$, which can be added to the first equation to get $3a=3$, or $a=... | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Given square $ABCD$ with side $8$ feet. A circle is drawn through vertices $A$ and $D$ and tangent to side $BC$. The radius of the circle, in feet, is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4\sqrt{2}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5\sqrt{2}\qquad\textbf{(E)}\ 6$ | Let $O$ be the center of the circle and $E$ be the point of tangency of the circle and $BC$ and let $F$ be the point of intersection of lines $OE$ and $AD$ Because of the symmetry, $BE=EC=AF=FD=4$ feet. Let the length of $OF$ be $x$. The length of $OE$ is $EF-OF=-x+8$. By Pythagorean Theorem, $OA=OD=\sqrt{x^2+4^2}=\sqr... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$, in the order shown. The value of $x$ is:
$\textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined}$ | Let $y$ represent the common difference between the terms. We have $(x+1)-y=(x-1)\implies y=2$.
Substituting gives us $(2x+3)-2=(x+1)\implies 2x+1=x+1\implies x=0$.
Therefore, our answer is $\boxed{\textbf{(B)}\ 0}$; | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Consider the statements:
$\textbf{(1)}\ \text{p and q are both true}\qquad\textbf{(2)}\ \text{p is true and q is false}\qquad\textbf{(3)}\ \text{p is false and q is true}\qquad\textbf{(4)}\ \text{p is false and q is false.}$
How many of these imply the negative of the statement "p and q are both true?"
$\textbf{(A)}\... | By De Morgan's Law, the negation of $p$ and $q$ are both true is that at least one of them is false, with the exception of the statement 1, i.e.;
$\text{p and q are both true}.$
The other three statements state that at least one statement is false. So, 2, 3, and 4 work, yielding an answer of 3, or $\textbf{(D)}.$ | 3 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
The ratio of the interior angles of two regular polygons with sides of unit length is $3: 2$. How many such pairs are there?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{infinitely many}$ | The formula for the measure of the interior angle of a regular polygon with $n$-sides is $180 - \frac{360}{n}$. Letting our two polygons have side length $r$ and $k$, we have that the ratio of the interior angles is $\frac{180 - \frac{360}{r}}{180 - \frac{360}{k}} = \frac{(r-2) \cdot k}{(k-2) \cdot r} = \frac{3}{2}$. ... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
If both $x$ and $y$ are both integers, how many pairs of solutions are there of the equation $(x-8)(x-10) = 2^y$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \text{more than 3}$ | The equality implies $x-8$ and $x-10$ are both powers of two; since they differ by two, it must be the case that $(x-8,x-10) = (4,2)$ or $(x-8,x-10) = (-2,-4)$. (Note that $(1,-1)$ is not allowed because then the product is negative.) These yield $(x,y) = (12,3)$ or $(x,y) = (6,3)$, for a total of $\boxed{2\textbf{ (C... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$, the population was one more than a perfect square. Now, with an additional increase of $100$, the population is again a perfect square.
The original population is a multiple of:
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 7... | Let $a^2$ $=$ original population count, $b^2+1$ $=$ the second population count, and $c^2$ $=$ the third population count
We first see that $a^2 + 100 = b^2 + 1$ or $99$ $=$ $b^2-a^2$.
We then factor the right side getting $99$ $=$ $(b-a)(b+a)$.
Since we can only have an nonnegative integral population, clearly $b+a... | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Three vertices of parallelogram $PQRS$ are $P(-3,-2), Q(1,-5), R(9,1)$ with $P$ and $R$ diagonally opposite.
The sum of the coordinates of vertex $S$ is:
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 9$ | Graph the three points on the coordinate grid. Noting that the opposite sides of a [parallelogram](https://artofproblemsolving.com/wiki/index.php/Parallelogram) are congruent and parallel, count boxes to find out that point $S$ is on $(5,4)$. The sum of the x-coordinates and y-coordinates is $9$, so the answer is $\b... | 9 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Given the equations $x^2+kx+6=0$ and $x^2-kx+6=0$. If, when the roots of the equation are suitably listed,
each root of the second equation is $5$ more than the corresponding root of the first equation, then $k$ equals:
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ -5 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ -7 \qquad \tex... | Let the two roots of $x^2 + kx + 6$ be $r$ and $s$. By [Vieta's Formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas),
\[r+s=-k\]
Each root of $x^2 - kx + 6$ is five more than each root of the original, so using Vieta's Formula again yields
\[r+5+s+5 = k\]
\[r+s+10=k\]
Substitute $r+s$ to get
\[... | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Three numbers $a,b,c$, none zero, form an arithmetic progression. Increasing $a$ by $1$ or increasing $c$ by $2$ results
in a geometric progression. Then $b$ equals:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$ | Let $d$ be the common difference of the [arithmetic sequence](https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence), so $a = b-d$ and $c = b+d$.
Since increasing $a$ by $1$ or $c$ by $2$ results in a [geometric sequence](https://artofproblemsolving.com/wiki/index.php/Geometric_sequence),
\[\frac{b}{b-d+1}... | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Two men at points $R$ and $S$, $76$ miles apart, set out at the same time to walk towards each other.
The man at $R$ walks uniformly at the rate of $4\tfrac{1}{2}$ miles per hour; the man at $S$ walks at the constant
rate of $3\tfrac{1}{4}$ miles per hour for the first hour, at $3\tfrac{3}{4}$ miles per hour for the ... | First, find the number of hours it takes for the two to meet together. After $h$ hours, the person at $R$ walks $4.5h$ miles. In the same amount of time, the person at $S$ has been walking at $3.25+0.5(h-1)$ mph for the past hour, so the person walks $\frac{h(6.5+0.5(h-1))}{2}$ miles.
In order for both to meet, the s... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Point $F$ is taken in side $AD$ of square $ABCD$. At $C$ a perpendicular is drawn to $CF$, meeting $AB$ extended at $E$.
The area of $ABCD$ is $256$ square inches and the area of $\triangle CEF$ is $200$ square inches. Then the number of inches in $BE$ is:
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}... | Because $ABCD$ is a [square](https://artofproblemsolving.com/wiki/index.php/Square), $DC = CB = 16$, $DC \perp DA$, and $CB \perp BA$. Also, because $\angle DCF + \angle FCB = \angle FCB + \angle BCE = 90^\circ$, $\angle DCF = \angle BCE$. Thus, by ASA Congruency, $\triangle DCF \cong \triangle BCF$.
From the congrue... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
In $\triangle ABC$ lines $CE$ and $AD$ are drawn so that
$\dfrac{CD}{DB}=\dfrac{3}{1}$ and $\dfrac{AE}{EB}=\dfrac{3}{2}$. Let $r=\dfrac{CP}{PE}$
where $P$ is the intersection point of $CE$ and $AD$. Then $r$ equals:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \dfrac{3}{2}\qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qqua... | Draw line $PB$, and let $[PEB] = 2b$, $[PDB] = a$, and $[CAP] = c$, so $[CPD] = 3a$ and $[APE] = 3b$. Because $\triangle CAE$ and $\triangle CEB$ share an altitude,
\[c + 3b = \tfrac{3}{2} (3a+a+2b)\]
\[c + 3b = 6a + 3b\]
\[c = 6a\]
Because $\triangle ACD$ and $\triangle ABD$ share an altitude,
\[6a+3a = 3(a+2b+3b)\]... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
What is the value of $[\log_{10}(5\log_{10}100)]^2$?
$\textbf{(A)}\ \log_{10}50 \qquad \textbf{(B)}\ 25\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 1$ | $[\log_{10}(5\log_{10}100)]^2$
Since $10^2 = 100$, we have $\log_{10} 100 = 2$, so:
$[\log_{10}(5\cdot 2)]^2$
Multiply:
$[\log_{10}(10)]^2$
Since $10^1 = 10$, we have $\log_{10} 10 = 1$, so:
$[1]^2$
$1$
$\fbox{E}$. | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A farmer bought $749$ sheep. He sold $700$ of them for the price paid for the $749$ sheep.
The remaining $49$ sheep were sold at the same price per head as the other $700$.
Based on the cost, the percent gain on the entire transaction is:
$\textbf{(A)}\ 6.5 \qquad \textbf{(B)}\ 6.75 \qquad \textbf{(C)}\ 7 \qquad \tex... | Let us say each sheep cost $x$ dollars. The farmer paid $749x$ for the sheep. He sold $700$ of them for $749x$, so each sheep sold for $\frac{749}{700} = 1.07x$.
Since every sheep sold for the same price per head, and since every sheep cost $x$ and sold for $1.07x$, there is an increase of $\frac{1.07x - 1x}{1x} = 0.... | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $2x-3y-z=0$ and $x+3y-14z=0, z \neq 0$, the numerical value of $\frac{x^2+3xy}{y^2+z^2}$ is:
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ -20/17\qquad \textbf{(E)}\ -2$ | If the value of $\frac{x^2+3xy}{y^2+z^2}$ is constant, as the answers imply, we can pick a value of $z$, and then solve the two linear equations for the corresponding $(x, y)$. We can then plug in $(x,y, z)$ into the expression to get the answer.
If $z=1$, then $2x - 3y = 1$ and $x + 3y = 14$. We can solve each equat... | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
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