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Let $N$ be the second smallest positive integer that is divisible by every positive integer less than $7$. What is the sum of the digits of $N$?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$ | $N$ must be divisible by every positive integer less than $7$, or $1, 2, 3, 4, 5,$ and $6$. Each number that is divisible by each of these is a multiple of their least common multiple. $LCM(1,2,3,4,5,6)=60$, so each number divisible by these is a multiple of $60$. The smallest multiple of $60$ is $60$, so the second sm... | 3 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Consider the polynomial
\[P(x)=\prod_{k=0}^{10}(x^{2^k}+2^k)=(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)\]
The coefficient of $x^{2012}$ is equal to $2^a$. What is $a$?
\[\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 24\] | Every term in the expansion of the product is formed by taking one term from each factor and multiplying them all together. Therefore, we pick a power of $x$ or a power of $2$ from each factor.
Every number, including $2012$, has a unique representation by the sum of powers of two, and that representation can be found ... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6.5\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$ | The ratio $\frac{400 \text{ euros}}{500 \text{ dollars}}$ can be simplified using conversion factors:\[\frac{400 \text{ euros}}{500 \text{ dollars}} \cdot \frac{1.3 \text{ dollars}}{1 \text{ euro}} = \frac{520}{500} = 1.04\] which means the money is greater by $\boxed{ \textbf{(B)} \ 4 }$ percent. | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Two integers have a sum of $26$. when two more integers are added to the first two, the sum is $41$. Finally, when two more integers are added to the sum of the previous $4$ integers, the sum is $57$. What is the minimum number of even integers among the $6$ integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{... | Since, $x + y = 26$, $x$ can equal $15$, and $y$ can equal $11$, so no even integers are required to make 26. To get to $41$, we have to add $41 - 26 = 15$. If $a+b=15$, at least one of $a$ and $b$ must be even because two odd numbers sum to an even number. Therefore, one even integer is required when transitioning fro... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The sequence $S_1, S_2, S_3, \cdots, S_{10}$ has the property that every term beginning with the third is the sum of the previous two. That is, \[S_n = S_{n-2} + S_{n-1} \text{ for } n \ge 3.\] Suppose that $S_9 = 110$ and $S_7 = 42$. What is $S_4$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 10\qquad\te... | $S_9 = 110$, $S_7 = 42$
$S_8 = S_9 - S_ 7 = 110 - 42 = 68$
$S_6 = S_8 - S_7 = 68 - 42 = 26$
$S_5 = S_7 - S_6 = 42 - 26 = 16$
$S_4 = S_6 - S_5 = 26 - 16 = 10$
Therefore, the answer is $\boxed{\textbf{(C) }{10}}$ | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Given that $x$ and $y$ are distinct nonzero real numbers such that $x+\tfrac{2}{x} = y + \tfrac{2}{y}$, what is $xy$?
$\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{1}{2}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 4\qquad$ | $x+\tfrac{2}{x}= y+\tfrac{2}{y}$
Since $x\not=y$, we may assume that $x=\frac{2}{y}$ and/or, equivalently, $y=\frac{2}{x}$.
Cross multiply in either equation, giving us $xy=2$.
$\boxed{\textbf{(D) }{2}}$ | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $ABCDE$ be an equiangular convex pentagon of perimeter $1$. The pairwise intersections of the lines that extend the sides of the pentagon determine a five-pointed star polygon. Let $s$ be the perimeter of this star. What is the difference between the maximum and the minimum possible values of $s$?
$\textbf{(A)}\ 0 ... | The five pointed star can be thought of as five triangles sitting on the five sides of the pentagon. Because the pentagon is equiangular, each of its angles has measure $\frac{180^\circ (5-2)}{5}=108^\circ$, and so the base angles of the aforementioned triangles (i.e., the angles adjacent to the pentagon) have measure ... | 0 | Geometry | MCQ | Yes | Yes | amc_aime | false |
What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$?
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$ | Looking at the prime numbers under $12$, we see that there are $\left\lfloor\frac{12}{2}\right\rfloor+\left\lfloor\frac{12}{2^2}\right\rfloor+\left\lfloor\frac{12}{2^3}\right\rfloor=6+3+1=10$ factors of $2$, $\left\lfloor\frac{12}{3}\right\rfloor+\left\lfloor\frac{12}{3^2}\right\rfloor=4+1=5$ factors of $3$, and $\left... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Let $a<b<c$ be three integers such that $a,b,c$ is an arithmetic progression and $a,c,b$ is a geometric progression. What is the smallest possible value of $c$?
$\textbf{(A) }-2\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6\qquad$ | We have $b-a=c-b$, so $a=2b-c$. Since $a,c,b$ is geometric, $c^2=ab=(2b-c)b \Rightarrow 2b^2-bc-c^2=(2b+c)(b-c)=0$. Since $a<b<c$, we can't have $b=c$ and thus $c=-2b$. Then our arithmetic progression is $4b,b,-2b$. Since $4b < b < -2b$, $b < 0$. The smallest possible value of $c=-2b$ is $(-2)(-1)=2$, or $\boxed{\textb... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For every real number $x$, let $\lfloor x\rfloor$ denote the greatest integer not exceeding $x$, and let \[f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).\] The set of all numbers $x$ such that $1\leq x<2014$ and $f(x)\leq 1$ is a union of disjoint intervals. What is the sum of the lengths of those intervals?
$\t... | Let $\lfloor x\rfloor=k$ for some integer $1\leq k\leq 2013$. Then we can rewrite $f(x)$ as $k(2014^{x-k}-1)$. In order for this to be less than or equal to $1$, we need $2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)$. Combining this with the fact that $\lfloor x\rfloor =k$ gives that... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The first three terms of a geometric progression are $\sqrt 3$, $\sqrt[3]3$, and $\sqrt[6]3$. What is the fourth term?
$\textbf{(A) }1\qquad\textbf{(B) }\sqrt[7]3\qquad\textbf{(C) }\sqrt[8]3\qquad\textbf{(D) }\sqrt[9]3\qquad\textbf{(E) }\sqrt[10]3\qquad$ | The terms are $\sqrt 3$, $\sqrt[3]3$, and $\sqrt[6]3$, which are equivalent to $3^{\frac{3}{6}}$, $3^{\frac{2}{6}}$, and $3^{\frac{1}{6}}$. So the next term will be $3^{\frac{0}{6}}=1$, so the answer is $\boxed{\textbf{(A)}}$. | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A set S consists of triangles whose sides have integer lengths less than 5, and no two elements of S are congruent or similar. What is the largest number of elements that S can have?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$ | Define $T$ to be the set of all integral triples $(a, b, c)$ such that $a \ge b \ge c$, $b+c > a$, and $a, b, c < 5$. Now we enumerate the elements of $T$:
$(4, 4, 4)$
$(4, 4, 3)$
$(4, 4, 2)$
$(4, 4, 1)$
$(4, 3, 3)$
$(4, 3, 2)$
$(3, 3, 3)$
$(3, 3, 2)$
$(3, 3, 1)$
$(3, 2, 2)$
$(2, 2, 2)$
$(2, 2, 1)$
$(1, 1, 1)$
It shoul... | 9 | Geometry | MCQ | Yes | Yes | amc_aime | false |
For how many positive integers $n$ is $\frac{n}{30-n}$ also a positive integer?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Solution 1
We know that $n \le 30$ or else $30-n$ will be negative, resulting in a negative fraction. We also know that $n \ge 15$ or else the fraction's denominator will exceed its numerator making the fraction unable to equal a positive integer value. Substituting all values $n$ from $15$ to $30$ gives us integer v... | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In the addition shown below $A$, $B$, $C$, and $D$ are distinct digits. How many different values are possible for $D$?
\[\begin{tabular}{cccccc}&A&B&B&C&B\\ +&B&C&A&D&A\\ \hline &D&B&D&D&D\end{tabular}\]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | From the first column, we see $A+B < 10$ because it yields a single digit answer. From the fourth column, we see that $C+D$ equals $D$ and therefore $C = 0$. We know that $A+B = D$. Therefore, the number of values $D$ can take is equal to the number of possible sums less than $10$ that can be formed by adding two di... | 7 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
On a sheet of paper, Isabella draws a circle of radius $2$, a circle of radius $3$, and all possible lines simultaneously tangent to both circles. Isabella notices that she has drawn exactly $k \ge 0$ lines. How many different values of $k$ are possible?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qq... | Isabella can get $0$ lines if the circles are concentric, $1$ if internally tangent, $2$ if overlapping, $3$ if externally tangent, and $4$ if non-overlapping and not externally tangent. There are $\boxed{\textbf{(D)}\ 5}$ values of $k$. | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$, $5$, and $8$, while those of $T'$ have lengths $a$, $a$, and $b$. Which of the following numbers is closest to $b$?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{... | The area of $T$ is $\dfrac{1}{2} \cdot 8 \cdot 3 = 12$ and the perimeter is 18.
The area of $T'$ is $\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$ and the perimeter is $2a + b$.
Thus $2a + b = 18$, so $2a = 18 - b$.
Thus $12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}$, so $48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
For each positive integer $n$, let $S(n)$ be the number of sequences of length $n$ consisting solely of the letters $A$ and $B$, with no more than three $A$s in a row and no more than three $B$s in a row. What is the remainder when $S(2015)$ is divided by $12$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6... | One method of approach is to find a recurrence for $S(n)$.
Let us define $A(n)$ as the number of sequences of length $n$ ending with an $A$, and $B(n)$ as the number of sequences of length $n$ ending in $B$. Note that $A(n) = B(n)$ and $S(n) = A(n) + B(n)$, so $S(n) = 2A(n)$.
For a sequence of length $n$ ending in $A$... | 8 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?
$\textbf{(A)}\; 3 \qquad\textbf{(B)}\; 4 \qquad\textbf{(C)}\; 5 \qquad\textbf{(D)}\; 6 \qquad\textbf{(E)}\; 7$ | Since we want non-congruent triangles that are neither isosceles nor equilateral, we can just list side lengths $(a,b,c)$ with $a<b<c$. Furthermore, "positive area" tells us that $c < a + b$ and the perimeter constraints means $a+b+c < 15$.
There are no triangles when $a = 1$ because then $c$ must be less than $b+1$, i... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,$ and $BC=6$. What is $AC$?
$\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7$ | $\angle ADB$ and $\angle ACB$ are both subtended by segment $AB$, hence $\angle ACB = \angle ADB = 40^\circ$. By considering $\triangle ABC$, it follows that $\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ$. Hence $\triangle ABC$ is isosceles, and $AC = BC = \boxed{\textbf{(B)}\; 6}.$ | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
An unfair coin lands on heads with a probability of $\tfrac{1}{4}$. When tossed $n>1$ times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of $n$?
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; 8 \qquad\textbf{(C)}\; 10 \qquad\textbf{(D)}\; 11 \qquad\textbf{(E)}\;... | When tossed $n$ times, the probability of getting exactly 2 heads and the rest tails is
\[\dbinom{n}{2} {\left( \frac{1}{4} \right)}^2 {\left( \frac{3}{4} \right) }^{n-2}.\]
Similarly, the probability of getting exactly 3 heads is
\[\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.\]
Now ... | 11 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?
$\textbf{(A)}\; 8 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 14 \qquad\textbf{(D)}\; 15 \qquad\textbf{(E)}\; 18$ | Let $a$ be the number written two times, and $b$ the number written three times. Then $2a + 3b = 100$. Plugging in $a = 28$ doesn't yield an integer for $b$, so it must be that $b = 28$, and we get $2a + 84 = 100$. Solving for $a$, we obtain $a = \boxed{\textbf{(A)}\; 8}$. | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$-coordinates lie on two or more of the graphs?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ ... | Setting the first two equations equal to each other, $\log_3 x = \log_x 3$.
Solving this, we get $\left(3, 1\right)$ and $\left(\frac{1}{3}, -1\right)$.
Similarly with the last two equations, we get $\left(3, -1\right)$ and $\left(\frac{1}{3}, 1\right)$.
Now, by setting the first and third equations equal to each other... | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textb... | The acceleration must be zero at the $x$-intercept; this intercept must be an inflection point for the minimum $a$ value.
Derive $f(x)$ so that the acceleration $f''(x)=0$. Using the power rule,
\begin{align*} f(x) &= x^3-ax^2+bx-a \\ f’(x) &= 3x^2-2ax+b \\ f’’(x) &= 6x-2a \end{align*}
So $x=\frac{a}{3}$ for the infle... | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \frac{1}{2}$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$ | By: Dragonfly
We find that $a^{-1}$ is the same as $2$, since a number to the power of $-1$ is just the reciprocal of that number. We then get the equation to be
\[\frac{2\times2+\frac{2}{2}}{\frac{1}{2}}\]
We can then simplify the equation to get $\boxed{\textbf{(D)}\ 10}$ | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A quadrilateral has vertices $P(a,b)$, $Q(b,a)$, $R(-a, -b)$, and $S(-b, -a)$, where $a$ and $b$ are integers with $a>b>0$. The area of $PQRS$ is $16$. What is $a+b$?
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 13$ | Note that the slope of $PQ$ is $\frac{a-b}{b-a}=-1$ and the slope of $PS$ is $\frac{b+a}{a+b}=1$. Hence, $PQ\perp PS$ and we can similarly prove that the other angles are right angles. This means that $PQRS$ is a rectangle. By distance formula we have $(a-b)^2+(b-a)^2*2*(a+b)^2 = 256$. Simplifying we get $(a-b)(a+b) ... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of $1$ and $2016$ is closest to which integer?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 504 \qquad \textbf{(D)}\ 1008 \qquad \textbf{(E)}\ 2015$ | Since the harmonic mean is $2$ times their product divided by their sum, we get the equation
$\frac{2\times1\times2016}{1+2016}$
which is then
$\frac{4032}{2017}$
which is finally closest to $\boxed{\textbf{(A)}\ 2}$.
-dragonfly
You can also think of $\frac{2\times1\times2016}{1+2016}$ as $2\times\frac{2016}{2017}$ w... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$, with $A$ at the origin and $\overline{BC}$ parallel to the $x$-axis. The area of the triangle is $64$. What is the length of $BC$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$ | By: Albert471
Plotting points $B$ and $C$ on the graph shows that they are at $\left( -x,x^2\right)$ and $\left( x,x^2\right)$, which is isosceles. By setting up the triangle area formula you get: $64=\frac{1}{2}*2x*x^2 = 64=x^3$ Making x=4, and the length of $BC$ is $2x$, so the answer is $\boxed{\textbf{(C)}\ 8}$. | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
There are $24$ different complex numbers $z$ such that $z^{24}=1$. For how many of these is $z^6$ a real number?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$ | Note that these $z$ such that $z^{24}=1$ are $e^{\frac{ni\pi}{12}}$ for integer $0\leq n<24$. So
$z^6=e^{\frac{ni\pi}{2}}$
This is real if $\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n$ is even$)$. Thus, the answer is the number of even $0\leq n<24$ which is $\boxed{(D)=\ 12}$. | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$ | Let $x, y$ be our two numbers. Then $x+y = 4xy$. Thus,
$\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = 4$.
$\boxed{ \textbf{C}}$. | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The data set $[6, 19, 33, 33, 39, 41, 41, 43, 51, 57]$ has median $Q_2 = 40$, first quartile $Q_1 = 33$, and third quartile $Q_3 = 43$. An outlier in a data set is a value that is more than $1.5$ times the interquartile range below the first quartle ($Q_1$) or more than $1.5$ times the interquartile range above the thi... | The interquartile range is defined as $Q3 - Q1$, which is $43 - 33 = 10$. $1.5$ times this value is $15$, so all values more than $15$ below $Q1$ = $33 - 15 = 18$ is an outlier. The only one that fits this is $6$. All values more than $15$ above $Q3 = 43 + 15 = 58$ are also outliers, of which there are none so there is... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The circle having $(0,0)$ and $(8,6)$ as the endpoints of a diameter intersects the $x$-axis at a second point. What is the $x$-coordinate of this point?
$\textbf{(A) } 4\sqrt{2} \qquad\textbf{(B) } 6 \qquad\textbf{(C) } 5\sqrt{2} \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Because the two points are on a diameter, the center must be halfway between them at the point (4,3). The distance from (0,0) to (4,3) is 5 so the circle has radius 5. Thus, the equation of the circle is $(x-4)^2+(y-3)^2=25$.
To find the x-intercept, y must be 0, so $(x-4)^2+(0-3)^2=25$, so $(x-4)^2=16$, $x-4=4$, $x=8... | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
A circle has center $(-10, -4)$ and has radius $13$. Another circle has center $(3, 9)$ and radius $\sqrt{65}$. The line passing through the two points of intersection of the two circles has equation $x+y=c$. What is $c$?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qq... | The equations of the two circles are $(x+10)^2+(y+4)^2=169$ and $(x-3)^2+(y-9)^2=65$. Rearrange them to $(x+10)^2+(y+4)^2-169=0$ and $(x-3)^2+(y-9)^2-65=0$, respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation $(x+10)^2+(y+4)^2-169=(x-3)^2... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Triangle $ABC$ is an isosceles right triangle with $AB=AC=3$. Let $M$ be the midpoint of hypotenuse $\overline{BC}$. Points $I$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$, respectively, so that $AI>AE$ and $AIME$ is a cyclic quadrilateral. Given that triangle $EMI$ has area $2$, the length $CI$ can be wri... | Observe that $\triangle{EMI}$ is isosceles right ($M$ is the midpoint of diameter arc $EI$ since $m\angle MEI = m\angle MAI = 45^\circ$), so $MI=2,MC=\frac{3}{\sqrt{2}}$. With $\angle{MCI}=45^\circ$, we can use Law of Cosines to determine that $CI=\frac{3\pm\sqrt{7}}{2}$. The same calculations hold for $BE$ also, and s... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
What is the sum of all possible values of $k$ for which the polynomials $x^2 - 3x + 2$ and $x^2 - 5x + k$ have a root in common?
$\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10$ | We factor $x^2-3x+2$ into $(x-1)(x-2)$. Thus, either $1$ or $2$ is a root of $x^2-5x+k$. If $1$ is a root, then $1^2-5\cdot1+k=0$, so $k=4$. If $2$ is a root, then $2^2-5\cdot2+k=0$, so $k=6$. The sum of all possible values of $k$ is $\boxed{\textbf{(E) }10}$. | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the value of \[\log_37\cdot\log_59\cdot\log_711\cdot\log_913\cdots\log_{21}25\cdot\log_{23}27?\]
$\textbf{(A) } 3 \qquad \textbf{(B) } 3\log_{7}23 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 10$ | From the Change of Base Formula, we have \[\frac{\prod_{i=3}^{13} \log (2i+1)}{\prod_{i=1}^{11}\log (2i+1)} = \frac{\log 25 \cdot \log 27}{\log 3 \cdot \log 5} = \frac{(2\log 5)\cdot(3\log 3)}{\log 3 \cdot \log 5} = \boxed{\textbf{(C) } 6}.\] | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $s_k$ denote the sum of the $\textit{k}$th powers of the roots of the polynomial $x^3-5x^2+8x-13$. In particular, $s_0=3$, $s_1=5$, and $s_2=9$. Let $a$, $b$, and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$, $3$, $....$ What is $a+b+c$?
$\textbf{(A)} \; -6 \qquad \te... | Applying [Newton's Sums](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/Newton's_Sums), we have\[s_{k+1}+(-5)s_k+(8)s_{k-1}+(-13)s_{k-2}=0,\]so\[s_{k+1}=5s_k-8s_{k-1}+13s_{k-2},\]we get the answer as $5+(-8)+13=10$. | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$, $\cos B = \frac{7}{8}$, and $\cos C = -\frac{1}{4}$. What is the least possible perimeter for $\triangle ABC$?
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$ | Solution 1
Notice that by the Law of Sines, $a:b:c = \sin{A}:\sin{B}:\sin{C}$, so let's flip all the cosines using $\sin^{2}{x} + \cos^{2}{x} = 1$ ($\sin{x}$ is positive for $0^{\circ} < x < 180^{\circ}$, so we're good there).
$\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\s... | 9 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\]for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsui... | By definition, the recursion becomes $a_n = \left(n^{\frac1{\log_7(n-1)}}\right)^{\log_7(a_{n-1})}=n^{\frac{\log_7(a_{n-1})}{\log_7(n-1)}}$. By the change of base formula, this reduces to $a_n = n^{\log_{n-1}(a_{n-1})}$. Thus, we have $\log_n(a_n) = \log_{n-1}(a_{n-1})$. Thus, for each positive integer $m \geq 3$, t... | 11 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The figure below shows line $\ell$ with a regular, infinite, recurring pattern of squares and line segments.
size(300);
defaultpen(linewidth(0.8));
real r = 0.35;
path P = (0,0)--(0,1)--(1,1)--(1,0), Q = (1,1)--(1+r,1+r);
path Pp = (0,0)--(0,-1)--(1,-1)--(1,0), Qp = (-1,-1)--(-1-r,-1-r);
for(int i=0;i <= 4;i=i+1)
{
dr... | Statement $1$ is true. A $180^{\circ}$ rotation about the point half way between an up-facing square and a down-facing square will yield the same figure.
Statement $2$ is also true. A translation to the left or right will place the image onto itself when the figures above and below the line realign (the figure goes on... | 2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$, without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.)
How ma... | Note that of the $12$ cities, $6$ of them ($2$ on the top, $2$ on the bottom, and $1$ on each side) have $3$ edges coming into/out of them (i.e., in graph theory terms, they have degree $3$). Therefore, at least $1$ edge connecting to each of these cities cannot be used. Additionally, the same applies to the start and ... | 4 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
How many nonzero complex numbers $z$ have the property that $0, z,$ and $z^3,$ when represented by points in the complex plane, are the three distinct vertices of an equilateral triangle?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }\text{infinitely many}$ | Convert $z$ and $z^3$ into modulus-argument (polar) form, giving $z=r\text{cis}(\theta)$ for some $r$ and $\theta$. Thus, by De Moivre's Theorem, $z^3=r^3\text{cis}(3\theta)$. Since the distance from $0$ to $z$ is $r$, and the triangle is equilateral, the distance from $0$ to $z^3$ must also be $r$, so $r^3=r$, giving ... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Let $f(x) = x^{2}(1-x)^{2}$. What is the value of the sum
\[f \left(\frac{1}{2019} \right)-f \left(\frac{2}{2019} \right)+f \left(\frac{3}{2019} \right)-f \left(\frac{4}{2019} \right)+\cdots + f \left(\frac{2017}{2019} \right) - f \left(\frac{2018}{2019} \right)?\]
$\textbf{(A) }0\qquad\textbf{(B) }\frac{1}{2019^{4}}\... | First, note that $f(x) = f(1-x)$. We can see this since
\[f(x) = x^2(1-x)^2 = (1-x)^2x^2 = (1-x)^{2}\left(1-\left(1-x\right)\right)^{2} = f(1-x)\]
Using this result, we regroup the terms accordingly:
\[\left( f \left(\frac{1}{2019} \right) - f \left(\frac{2018}{2019} \right) \right) + \left( f \left(\frac{2}{2019} \r... | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
There are integers $a, b,$ and $c,$ each greater than $1,$ such that
\[\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]
for all $N \neq 1$. What is $b$?
$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$ | $\sqrt[a]{N\sqrt[b]{N\sqrt[c]{N}}}$ can be simplified to $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}.$
The equation is then $N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}$ which implies that $\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}.$
$a$ has to be $2$ since $\frac{25}{36}>\frac{7}{12}$. $\fr... | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The vertices of a quadrilateral lie on the graph of $y=\ln{x}$, and the $x$-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$. What is the $x$-coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \t... | Let the coordinates of the quadrilateral be $(n,\ln(n)),(n+1,\ln(n+1)),(n+2,\ln(n+2)),(n+3,\ln(n+3))$. We have by shoelace's theorem, that the area is
\begin{align*} &\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)(n) + \ln(n+2)(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2} \\ &=\frac{\ln \left(... | 12 | Calculus | MCQ | Yes | Yes | amc_aime | false |
How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We count the intersections of the graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right):$
The graph of $y=\tan(2x)$ has a period of $\frac{\pi}{2},$ asymptotes at $x=\frac{\pi}{4}+\frac{k\pi}{2},$ and zeros at $x=\frac{k\pi}{2}$ for some integer $k.$
On the interval $[0,2\pi],$ the graph has five branches: \[\bigg... | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the value in simplest form of the following expression?\[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}\]
$\textbf{(A) }5 \qquad \textbf{(B) }4 + \sqrt{7} + \sqrt{10} \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$ | We have \[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16}\ = 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}.\]
Note: This comes from the fact that the sum of the first $n$ odds is $n^2$. | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the value of the following expression?
\[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}\]
$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{9951}{9950} \qquad \textbf{(C) } \frac{4780}{4779} \qquad \textbf{(D) } \frac{108}{107} \qquad \textbf{(E) } \frac{81}{80}$ | Using difference of squares to factor the left term, we get
\[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)} = \frac{(100-7)(100+7)}{(70-11)(70+11)} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}.\]
Cancelling all the terms, we get $\boxed{\textbf{(A) } 1}$ as the answer. | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
How many integers $n \geq 2$ are there such that whenever $z_1, z_2, ..., z_n$ are complex numbers such that
\[|z_1| = |z_2| = ... = |z_n| = 1 \text{ and } z_1 + z_2 + ... + z_n = 0,\]
then the numbers $z_1, z_2, ..., z_n$ are equally spaced on the unit circle in the complex plane?
$\textbf{(A)}\ 1 \qquad\textbf{... | For $n=2$, we see that if $z_{1}+z_{2}=0$, then $z_{1}=-z_{2}$, so they are evenly spaced along the unit circle.
For $n=3$, WLOG, we can set $z_{1}=1$. Notice that now Re$(z_{2}+z_{3})=-1$ and Im$\{z_{2}\}$ = $-$Im$\{z_{3}\}$. This forces $z_{2}$ and $z_{3}$ to be equal to $e^{i\frac{2\pi}{3}}$ and $e^{-i\frac{2\pi}{3}... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The five solutions to the equation\[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$. The eccentricity of $\mathcal E$ ca... | The solutions to this equation are $z = 1$, $z = -1 \pm i\sqrt 3$, and $z = -2\pm i\sqrt 2$. Consider the five points $(1,0)$, $\left(-1,\pm\sqrt 3\right)$, and $\left(-2,\pm\sqrt 2\right)$; these are the five points which lie on $\mathcal E$. Note that since these five points are symmetric about the $x$-axis, so must ... | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $d(n)$ denote the number of positive integers that divide $n$, including $1$ and $n$. For example, $d(1)=1,d(2)=2,$ and $d(12)=6$. (This function is known as the divisor function.) Let\[f(n)=\frac{d(n)}{\sqrt [3]n}.\]There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N$. Wh... | We consider the prime factorization of $n:$ \[n=\prod_{i=1}^{k}p_i^{e_i}.\] By the Multiplication Principle, we have \[d(n)=\prod_{i=1}^{k}(e_i+1).\] Now, we rewrite $f(n)$ as \[f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{\prod_{i=1}^{k}(e_i+1)}{\prod_{i=1}^{k}p_i^{e_i/3}}=\prod_{i=1}^{k}\frac{e_i+1}{p_i^{{e_i}/3}}.\] As $f(n)>... | 9 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is $\frac13$. When $4$ black cards are added to the deck, the probability of choosing red becomes $\frac14$. How many cards were in the deck originally?
$\textbf{(A) }6 \qquad \textbf{(B) }9 \qquad \textbf{(C) }12 \q... | If the probability of choosing a red card is $\frac{1}{3}$, the red and black cards are in ratio $1:2$. This means at the beginning there are $x$ red cards and $2x$ black cards.
After $4$ black cards are added, there are $2x+4$ black cards. This time, the probability of choosing a red card is $\frac{1}{4}$ so the ratio... | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Two distinct numbers are selected from the set $\{1,2,3,4,\dots,36,37\}$ so that the sum of the remaining $35$ numbers is the product of these two numbers. What is the difference of these two numbers?
$\textbf{(A) }5 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8\qquad \textbf{(D) }9 \qquad \textbf{(E) }10$ | The sum of the first $n$ integers is given by $\frac{n(n+1)}{2}$, so $\frac{37(37+1)}{2}=703$.
Therefore, $703-x-y=xy$
Rearranging, $xy+x+y=703$. We can factor this equation by [SFFT](https://artofproblemsolving.com/wiki/index.php/SFFT) to get
$(x+1)(y+1)=704$
Looking at the possible divisors of $704 = 2^6\cdot11$, $22... | 10 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
How many values of $\theta$ in the interval $0<\theta\le 2\pi$ satisfy \[1-3\sin\theta+5\cos3\theta = 0?\]
$\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8$ | We rearrange to get \[5\cos3\theta = 3\sin\theta-1.\]
We can graph two functions in this case: $y=5\cos{3x}$ and $y=3\sin{x} -1$.
Using transformation of functions, we know that $5\cos{3x}$ is just a cosine function with amplitude $5$ and period $\frac{2\pi}{3}$. Similarly, $3\sin{x} -1$ is just a sine function with am... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the value of\[\frac{\log_2 80}{\log_{40}2}-\frac{\log_2 160}{\log_{20}2}?\]$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }\frac54 \qquad \textbf{(D) }2 \qquad \textbf{(E) }\log_2 5$ | \[\frac{\log_{2}{80}}{\log_{40}{2}}-\frac{\log_{2}{160}}{\log_{20}{2}}\]
Note that $\log_{40}{2}=\frac{1}{\log_{2}{40}}$, and similarly $\log_{20}{2}=\frac{1}{\log_{2}{20}}$
\[= \log_{2}{80}\cdot \log_{2}{40}-\log_{2}{160}\cdot \log_{2}{20}\]
\[=(\log_{2}{4}+\log_{2}{20})(\log_{2}{2}+\log_{2}{20})-(\log_{2}{8}+\log_{2}... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $m\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1\le a_i \le m$ for all $i$ such that $m$ divides $a_1+a_2+a_3+a_4$. There is a polynomial
\[q(x) = c_3x^3+c_2x^2+c_1x+c_0\]such that $D(m) = q(m)$ for all odd integers $m\ge 5$. What is $c_1... | For a fixed value of $m,$ there is a total of $m(m-1)(m-2)(m-3)$ possible ordered quadruples $(a_1, a_2, a_3, a_4).$
Let $S=a_1+a_2+a_3+a_4.$ We claim that exactly $\frac1m$ of these $m(m-1)(m-2)(m-3)$ ordered quadruples satisfy that $m$ divides $S:$
Since $\gcd(m,4)=1,$ we conclude that \[\{k+4(0),k+4(1),k+4(2),\ldots... | 11 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Let $c = \frac{2\pi}{11}.$ What is the value of
\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]
$\textbf{(A)}\ {-}1 \qquad\textbf{(B)}\ {-}\frac{\sqrt{11}}{5} \qquad\textbf{(C)}\ \frac{\sqrt{11}}{5} \qquad\textbf{(D)}\ \frac{1... | Plugging in $c$, we get
\[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}=\frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Suppose that $P(z), Q(z)$, and $R(z)$ are polynomials with real coefficients, having degrees $2$, $3$, and $6$, respectively, and constant terms $1$, $2$, and $3$, respectively. Let $N$ be the number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z)=R(z)$. What is the minimum possible value of ... | The answer cannot be $0,$ as every nonconstant polynomial has at least $1$ distinct complex root (Fundamental Theorem of Algebra). Since $P(z) \cdot Q(z)$ has degree $2 + 3 = 5,$ we conclude that $R(z) - P(z)\cdot Q(z)$ has degree $6$ and is thus nonconstant.
It now suffices to illustrate an example for which $N = 1$: ... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Set $u_0 = \frac{1}{4}$, and for $k \ge 0$ let $u_{k+1}$ be determined by the recurrence \[u_{k+1} = 2u_k - 2u_k^2.\]
This sequence tends to a limit; call it $L$. What is the least value of $k$ such that \[|u_k-L| \le \frac{1}{2^{1000}}?\]
$\textbf{(A)}\: 10\qquad\textbf{(B)}\: 87\qquad\textbf{(C)}\: 123\qquad\textbf{(... | Note that terms of the sequence $(u_k)$ lie in the interval $\left(0,\frac12\right),$ strictly increasing.
Since the sequence $(u_k)$ tends to the limit $L,$ we set $u_{k+1}=u_k=L>0.$
The given equation becomes \[L=2L-2L^2,\] from which $L=\frac12.$
The given inequality becomes \[\frac12-\frac{1}{2^{1000}} \leq u_k \le... | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For real numbers $x$, let
\[P(x)=1+\cos(x)+i\sin(x)-\cos(2x)-i\sin(2x)+\cos(3x)+i\sin(3x)\]
where $i = \sqrt{-1}$. For how many values of $x$ with $0\leq x<2\pi$ does
\[P(x)=0?\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | Let $a=\cos(x)+i\sin(x)$. Now $P(a)=1+a-a^2+a^3$. $P(-1)=-2$ and $P(0)=1$ so there is a real root $a_1$ between $-1$ and $0$. The other $a$'s must be complex conjugates since all coefficients of the polynomial are real. The magnitude of those complex $a$'s squared is $\frac{1}{a_1}$ which is greater than $1$. If $x$ is... | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For $n$ a positive integer, let $R(n)$ be the sum of the remainders when $n$ is divided by $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, and $10$. For example, $R(15) = 1+0+3+0+3+1+7+6+5=26$. How many two-digit positive integers $n$ satisfy $R(n) = R(n+1)\,?$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\text... | Note that we can add $9$ to $R(n)$ to get $R(n+1)$, but must subtract $k$ for all $k|n+1$. Hence, we see that there are four ways to do that because $9=7+2=6+3=5+4=4+3+2$. Note that only $7+2$ is a plausible option, since $4+3+2$ indicates $n+1$ is divisible by $6$, $5+4$ indicates that $n+1$ is divisible by $2$, $6+3$... | 2 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the value of \[(\log 5)^{3}+(\log 20)^{3}+(\log 8)(\log 0.25)\] where $\log$ denotes the base-ten logarithm?
$\textbf{(A)}~\frac{3}{2}\qquad\textbf{(B)}~\frac{7}{4}\qquad\textbf{(C)}~2\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~3$ | Let $\text{log } 2 = x$. The expression then becomes \[(1+x)^3+(1-x)^3+(3x)(-2x)=\boxed{2}.\]
-bluelinfish | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $h_n$ and $k_n$ be the unique relatively prime positive integers such that \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}=\frac{h_n}{k_n}.\] Let $L_n$ denote the least common multiple of the numbers $1, 2, 3, \ldots, n$. For how many integers with $1\le{n}\le{22}$ is $k_n<L_n$?
$\textbf{(A) }0 \qquad\text... | We are given that \[\sum_{i=1}^{n}\frac1i = \frac{1}{L_n}\sum_{i=1}^{n}\frac{L_n}{i} = \frac{h_n}{k_n}.\] Since $k_n 1.$
For all primes $p$ such that $p\leq n,$ let $v_p(L_n)=e\geq1$ be the largest power of $p$ that is a factor of $L_n.$
It is clear that $L_n\equiv0\pmod{p},$ so we test whether $\sum_{i=1}^{n}\frac{L_n... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Let $f(n) = \left( \frac{-1+i\sqrt{3}}{2} \right)^n + \left( \frac{-1-i\sqrt{3}}{2} \right)^n$, where $i = \sqrt{-1}$. What is $f(2022)$?
$\textbf{(A)}\ -2 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ 2$ | Converting both summands to exponential form,
\begin{align*} -1 + i\sqrt{3} &= 2e^{\frac{2\pi i}{3}}, \\ -1 - i\sqrt{3} &= 2e^{-\frac{2\pi i}{3}} = 2e^{\frac{4\pi i}{3}}. \end{align*}
Notice that both are scaled copies of the third roots of unity.
When we replace the summands with their exponential form, we get \[f(n)... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The sequence $a_0,a_1,a_2,\cdots$ is a strictly increasing arithmetic sequence of positive integers such that \[2^{a_7}=2^{27} \cdot a_7.\] What is the minimum possible value of $a_2$?
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 22$ | We can rewrite the given equation as $2^{a_7-27}=a_7$. Hence, $a_7$ must be a power of $2$ and larger than $27$. The first power of 2 that is larger than $27$, namely $32$, does satisfy the equation: $2^{32 - 27} = 2^5 = 32$. In fact, this is the only solution; $2^{a_7-27}$ is exponential whereas $a_7$ is linear, so th... | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A real-valued function $f$ has the property that for all real numbers $a$ and $b,$ \[f(a + b) + f(a - b) = 2f(a) f(b).\] Which one of the following cannot be the value of $f(1)?$
$\textbf{(A) } 0 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } -1 \qquad \textbf{(D) } 2 \qquad \textbf{(E) } -2$ | Substituting $a = b$ we get
\[f(2a) + f(0) = 2f(a)^2\]
Substituting $a= 0$ we find \[2f(0) = 2f(0)^2 \implies f(0) \in \{0, 1\}.\]
This gives \[f(2a) = 2f(a)^2 - f(0) \geq 0-1\]
Plugging in $a = \frac{1}{2}$ implies $f(1) \geq -1$, so answer choice $\boxed{\textbf{(E) -2}}$ is impossible.
~AtharvNaphade | -2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
[katex]\dfrac{3\times 5}{9\times 11}\times \dfrac{7\times 9\times 11}{3\times 5\times 7}=[/katex]
[katex]\text{(A)}\ 1 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 49 \qquad \text{(D)}\ \frac{1}{49} \qquad \text{(E)}\ 50[/katex] | By the [associative property](https://artofproblemsolving.com/wiki/index.php/Associative_property), we can rearrange the numbers in the numerator and the denominator. [katex display=true]\frac{3}{3}\cdot \frac{5}{5}\cdot\frac{7}{7}\cdot\frac{9}{9}\cdot\frac{11}{11}=1\cdot1\cdot1\cdot1\cdot1=\boxed{\text{(A)} 1}[/katex... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If you walk for $45$ minutes at a [rate](https://artofproblemsolving.com/wiki/index.php/Rate) of $4 \text{ mph}$ and then run for $30$ minutes at a rate of $10\text{ mph}$, how many miles will you have gone at the end of one hour and $15$ minutes?
$\text{(A)}\ 3.5\text{ miles} \qquad \text{(B)}\ 8\text{ miles} \qquad \... | $45$ minutes is $\frac{3}{4}$ of an hour, so the walking contributes $\frac{3\text{ hr}}{4}\times \frac{4 \text{ miles}}{1\text{ hr}}=3\text{ miles}$.
Similarly, $30$ minutes is $\frac{1}{2}$ of an hour, so the running adds $\frac{1\text{ hr}}{2}\times\frac{10\text{ miles}}{1\text{ hr}}=5\text{ miles}$.
Their total is ... | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The [ratio](https://artofproblemsolving.com/wiki/index.php/Ratio) of boys to girls in Mr. Brown's math class is $2:3$. If there are $30$ students in the class, how many more girls than boys are in the class?
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 10$ | Let the number of boys be $2x$. It follows that the number of girls is $3x$. These two values add up to $30$ students, so \[2x+3x=5x=30\Rightarrow x=6\]
The [difference](https://artofproblemsolving.com/wiki/index.php/Subtraction) between the number of girls and the number of boys is $3x-2x=x$, which is $6$, so the ans... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If the [length](https://artofproblemsolving.com/wiki/index.php/Length) and width of a [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle) are each increased by $10\%$, then the [perimeter](https://artofproblemsolving.com/wiki/index.php/Perimeter) of the rectangle is increased by
$\text{(A)}\ 1\% \qqua... | Let the width be $w$ and the length be $l$. Then, the original perimeter is $2(w+l)$.
After the increase, the new width and new length are $1.1w$ and $1.1l$, so the new perimeter is $2(1.1w+1.1l)=2.2(w+l)$.
Therefore, the percent change is
\begin{align*} \frac{2.2(w+l)-2(w+l)}{2(w+l)} &= \frac{.2(w+l)}{2(w+l)} \\ &= ... | 10 | Geometry | MCQ | Yes | Yes | amc_aime | false |
If $a = - 2$, the largest number in the set $\{ - 3a, 4a, \frac {24}{a}, a^2, 1\}$ is
$\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1$ | Since all the numbers are small, we can just evaluate the set to be \[\{ (-3)(-2), 4(-2), \frac{24}{-2}, (-2)^2, 1 \}= \{ 6, -8, -12, 4, 1 \}\]
The largest number is $6$, which corresponds to $-3a$.
$\boxed{\text{A}}$ | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A picture $3$ feet across is hung in the center of a wall that is $19$ feet wide. How many feet from the end of the wall is the nearest edge of the picture?
$\text{(A)}\ 1\frac{1}{2} \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9\frac{1}{2} \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 22$ | Let's say that the distance from the picture to the wall is $x$. Since that distance will be on both sides of the picture (it's in the exact middle), we can say that
\begin{align*} x + 3 + x = 19 &\Rightarrow 2x+3=19 \\ &\Rightarrow 2x=16 \\ &\Rightarrow x=8 \\ \end{align*}
$\boxed{\text{B}}$ | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
If $\text{A}*\text{B}$ means $\frac{\text{A}+\text{B}}{2}$, then $(3*5)*8$ is
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16\qquad \text{(E)}\ 30$ | We just plug in and evaluate:
\begin{align*} (3*5)*8 &= \left( \frac{3+5}{2}\right) *8 \\ &= 4*8 \\ &= \frac{4+8}{2} \\ &= 6 \\ \end{align*}
$\boxed{\text{A}}$ | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $200\leq a \leq 400$ and $600\leq b\leq 1200$, then the largest value of the quotient $\frac{b}{a}$ is
$\text{(A)}\ \frac{3}{2} \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 300 \qquad \text{(E)}\ 600$ | $\frac{b}{a}$ will be largest if $b$ is the largest it can be, and $a$ is the smallest it can be.
Since $b$ can be no larger than $1200$, $b = 1200$. Since $a$ can be no less than $200$, $a = 200$. $\frac{1200}{200} = 6$
$6$ is $\boxed{\text{C}}$ | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A rectangular grazing area is to be fenced off on three sides using part of a $100$ meter rock wall as the fourth side. Fence posts are to be placed every $12$ meters along the fence including the two posts where the fence meets the rock wall. What is the fewest number of posts required to fence an area $36$ m by $60... | Since we want to minimize the amount of fence that we use, we should have the longer side of the rectangle have one side as the wall. The grazing area is a $36$m by $60$m rectangle, so the $60$m side should be parallel to the wall. That means the two fences perpendicular to the wall are $36$m. We can start by counting ... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The value of the expression $\frac{(304)^5}{(29.7)(399)^4}$ is closest to
$\text{(A)}\ .003 \qquad \text{(B)}\ .03 \qquad \text{(C)}\ .3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 30$ | \[\frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}\]
Which is closest to $3\rightarrow\boxed{\text{D}}$.
(The original expression is approximately equal to $3.44921198$.) | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Suppose one of the eight lettered identical squares is included with the four squares in the T-shaped figure outlined. How many of the resulting figures can be folded into a topless cubical box?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$ | Solution 1
The four squares we already have assemble nicely into four sides of the cube. Let the central one be the bottom, and fold the other three upwards to get the front, right, and back side. Currently, our box is missing its left side and its top side. We have to count the possibilities that would fold to one of ... | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The large circle has diameter $\text{AC}$. The two small circles have their centers on $\text{AC}$ and just touch at $\text{O}$, the center of the large circle. If each small circle has radius $1$, what is the value of the ratio of the area of the shaded region to the area of one of the small circles?
$\text{(A)}\ \... | The small circle has radius $1$, thus its area is $\pi$.
The large circle has radius $2$, thus its area is $4\pi$.
The area of the semicircle above $AC$ is then $2\pi$.
The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is $\pi$. This means that the are... | 1 | Geometry | MCQ | Yes | Yes | amc_aime | false |
$\frac{2}{1-\frac{2}{3}}=$
$\text{(A)}\ -3 \qquad \text{(B)}\ -\frac{4}{3} \qquad \text{(C)}\ \frac{2}{3} \qquad \text{(D)}\ 2 \qquad \text{(E)}\ 6$ | Just simplify the bottom as $\frac{3}{3}-\frac{2}{3}=\frac{1}{3}$, getting us $\frac{2}{\frac{1}{3}}$, with which we multiply top and bottom by 3, we get $\frac{6}{1}$, or $6$
$\boxed{\text{E}}$ | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
How many whole numbers are between $\sqrt{8}$ and $\sqrt{80}$?
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$ | No... of course you're not supposed to know what the square root of 8 is, or the square root of 80. There aren't any formulas, either. Approximation seems like the best strategy.
Clearly it must be true that for any positive integers $a$, $b$, and $c$ with $a>b>c$, \[\sqrt{a}>\sqrt{b}>\sqrt{c}\]
If we let $a=9$, $b=8$,... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In the product shown, $\text{B}$ is a digit. The value of $\text{B}$ is
\[\begin{array}{rr} &\text{B}2 \\ \times& 7\text{B} \\ \hline &6396 \\ \end{array}\]
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$ | Note that in any multiplication problem, the only 2 digits that will influence the last digit of the number will be the last digits of each number being multiplied.
So, $2 \times \text{B}$ has a units digit of $6$, so $\text{B}$ is either $3$ or $8$. If $\text{B}=3$, then the product is $32\times 73$, which is clearly ... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Joyce made $12$ of her first $30$ shots in the first three games of this basketball game, so her seasonal shooting [average](https://artofproblemsolving.com/wiki/index.php/Average) was $40\%$. In her next game, she took $10$ shots and raised her seasonal shooting average to $50\%$. How many of these $10$ shots did sh... | After the fourth game, she took $40$ shots, $50\%$ of which she made, so she made $40\times .5=20$ shots. Twelve of them were made in the first three games, so in the last game she made $20-12=8$ shots.
$\boxed{\text{E}}$ | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A calculator has a squaring key $\boxed{x^2}$ which replaces the current number displayed with its [square](https://artofproblemsolving.com/wiki/index.php/Perfect_square). For example, if the display is $\boxed{000003}$ and the $\boxed{x^2}$ key is depressed, then the display becomes $\boxed{000009}$. If the display ... | We just brute force this:
\begin{align*} 2 &\rightarrow 4 \\ &\rightarrow 16 \\ &\rightarrow 256 \\ &\rightarrow 65536>500 \end{align*}
Clearly we need to press the button $4$ times, so $\boxed{\text{A}}$ | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
"If a whole number $n$ is not prime, then the whole number $n-2$ is not prime." A value of $n$ which shows this statement to be false is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 23$ | To show this statement to be false, we need a non-prime value of $n$ such that $n-2$ is prime. Since $13$ and $23$ are prime, they won't prove anything relating to the truth of the statement.
Now we just check the statement for $n=9,12,16$. If $n=12$ or $n=16$, then $n-2$ is $10$ or $14$, which aren't prime. However... | 9 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Suppose $n^{*}$ means $\frac{1}{n}$, the [reciprocal](https://artofproblemsolving.com/wiki/index.php/Reciprocal) of $n$. For example, $5^{*}=\frac{1}{5}$. How many of the following statements are true?
i) $3^*+6^*=9^*$
ii) $6^*-4^*=2^*$
iii) $2^*\cdot 6^*=12^*$
iv) $10^*\div 2^* =5^*$
$\text{(A)}\ 0 \qquad \text{(B... | We can just test all of these statements:
\begin{align*} 3^*+6^* &= \frac{1}{3}+\frac{1}{6} \\ &= \frac{1}{2} \neq 9^* \\ 6^*-4^* &= \frac{1}{6}-\frac{1}{4} \\ &= \frac{-1}{12} \neq 2^* \\ 2^*\cdot 6^* &= \frac{1}{2}\cdot \frac{1}{6} \\ &= \frac{1}{12} = 12^* \\ 10^* \div 2^* &= \frac{1}{10}\div \frac{1}{2} \\ &= \frac... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A multiple choice examination consists of $20$ questions. The scoring is $+5$ for each correct answer, $-2$ for each incorrect answer, and $0$ for each unanswered question. John's score on the examination is $48$. What is the maximum number of questions he could have answered correctly?
$\text{(A)}\ 9 \qquad \text{(... | Solution 1
Let $c$ be the number of questions correct, $w$ be the number of questions wrong, and $b$ be the number of questions left blank. We are given that
\begin{align} c+w+b &= 20 \\ 5c-2w &= 48 \end{align}
Adding equation $(2)$ to double equation $(1)$, we get \[7c+2b=88\]
Since we want to maximize the value of... | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $\text{A}$ and $\text{B}$ are nonzero digits, then the number of digits (not necessarily different) in the sum of the three whole numbers is
\[\begin{tabular}[t]{cccc} 9 & 8 & 7 & 6 \\ & A & 3 & 2 \\ & B & 1 \\ \hline \end{tabular}\]
$\text{(A)}\ 4 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qq... | The minimum possible value of this sum is when $A=B=1$, which is \[9876+132+11=10019\]
The largest possible value of the sum is when $A=B=9$, making the sum \[9876+932+91=10899\]
Since all the possible sums are between $10019$ and $10899$, they must have $5$ digits.
$\boxed{\text{B}}$ | 5 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$ | If, for a moment, we disregard the white squares, we notice that the number of black squares in each row increases by 1 continuously as we go down the pyramid.
Thus, the number of black squares is $1 + 2 + \cdots + 8$.
Same goes for the white squares, except it starts a row later, making it $1 + 2 + \cdots + 7$.
Subtra... | 8 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
The letters $\text{A}$, $\text{J}$, $\text{H}$, $\text{S}$, $\text{M}$, $\text{E}$ and the digits $1$, $9$, $8$, $9$ are "cycled" separately as follows and put together in a numbered list:
\[\begin{tabular}[t]{lccc} & & AJHSME & 1989 \\ & & & \\ 1. & & JHSMEA & 9891 \\ 2. & & HSMEAJ & 8919 \\ 3. & & SMEAJH & 9198 \\ ... | Every $4\text{th}$ line has $1989$ as part of it and every $6\text{th}$ line has $\text{AJHSME}$ as part of it. In order for both to be part of line $n$, $n$ must be a multiple of $4$ and $6$, the least of which is $\text{lcm}(4,6)=12\rightarrow \boxed{\text{C}}$. | 12 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
$-15+9\times (6\div 3) =$
$\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12$ | We use the [order of operations](https://artofproblemsolving.com/wiki/index.php/Order_of_operations) here to get
\begin{align*} -15+9\times (6\div 3) &= -15+9\times 2 \\ &= -15+18 \\ &= 3 \rightarrow \boxed{\text{D}} \end{align*} | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If the markings on the [number line](https://artofproblemsolving.com/wiki/index.php/Number_line) are equally spaced, what is the number $\text{y}$?
$\text{(A)}\ 3 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 16$ | Five steps are taken to get from $0$ to $20$. Each step is of equal size, so each step is $4$. Three steps are taken from $0$ to $y$, so $y=3\times 4=12\rightarrow \boxed{\text{C}}$.
There are five steps, and y is step number three. We can get our answer by multiplying since all the steps are the same. \[20\cdot \f... | 12 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A straight concrete sidewalk is to be $3$ feet wide, $60$ feet long, and $3$ inches thick. How many cubic yards of concrete must a contractor order for the sidewalk if concrete must be ordered in a whole number of cubic yards?
$\text{(A)}\ 2 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 20 \qquad \text... | This is a $1$ yard by $20$ yard by $1/12$ yard sidewalk, so its volume in yards is
\[1\times 20\times \frac{1}{12} = 1.\overline{6}.\]
Since concrete must be ordered in a whole number of cubic yards, we need $2\rightarrow \boxed{\text{A}}$. | 2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Three $\Delta$'s and a $\diamondsuit$ will balance nine $\bullet$'s. One $\Delta$ will balance a $\diamondsuit$ and a $\bullet$.
How many $\bullet$'s will balance the two $\diamondsuit$'s in this balance?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$ | For simplicity, suppose $\Delta = a$, $\diamondsuit = b$ and $\bullet = c$. Then,
\[3a+b=9c\]
\[a=b+c\]
and we want to know what $2b$ is in terms of $c$. Substituting the second equation into the first, we have
\[4b=6c\Rightarrow 2b=3c\]
Thus, we need $3$ $\bullet$'s $\rightarrow \boxed{\text{C}}$. | 3 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
The area in square units of the region enclosed by parallelogram $ABCD$ is
$\text{(A)}\ 6 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18$ | The base is $\overline{BC}=4$. The height has a length of the difference of the y-coordinates of A and B, which is 2. Therefore the area is $4\cdot 2=8\Rightarrow \boxed{\mathrm{B}}$. | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The $16$ squares on a piece of paper are numbered as shown in the diagram. While lying on a table, the paper is folded in half four times in the following sequence:
(1) fold the top half over the bottom half
(2) fold the bottom half over the top half
(3) fold the right half over the left half
(4) fold the left half ov... | Suppose we undo each of the four folds, considering just the top square until we completely unfold the paper. $x$ will be marked in the square if the face that shows after all the folds is face up, $y$ if that face is facing down.
Step 0:
Step 1:
Step 2:
Step 3:
Step 4:
The marked square is in the same spot as th... | 9 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
$\frac{16+8}{4-2}=$
$\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$ | \begin{align*} \frac{16+8}{4-2} &= \frac{24}{2} \\ &= 12\rightarrow \boxed{\text{C}}. \end{align*} | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In the addition problem, each digit has been replaced by a letter. If different letters represent different digits then what is the value of $C$?
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9$ | From this we have
\[111A+11B+C=300.\]
Clearly, $A 201 \Rightarrow A\geq 2.\]
Thus, $A=2$ and $11B+C=78$. From here it becomes clear that $B=7$ and $C=1\rightarrow \boxed{\text{A}}$. | 1 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
For every $3^\circ$ rise in temperature, the volume of a certain gas expands by $4$ cubic centimeters. If the volume of the gas is $24$ cubic centimeters when the temperature is $32^\circ$, what was the volume of the gas in cubic centimeters when the temperature was $20^\circ$?
$\text{(A)}\ 8 \qquad \text{(B)}\ 12 \qq... | We know that
\[T=32-3k\Rightarrow V=24-4k.\]
Setting $k=4$, we get the volume when the temperature is $20^\circ$, which is $24-4(4)=8\rightarrow \boxed{\text{A}}$. | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The Pythagoras High School band has $100$ female and $80$ male members. The Pythagoras High School orchestra has $80$ female and $100$ male members. There are $60$ females who are members in both band and orchestra. Altogether, there are $230$ students who are in either band or orchestra or both. The number of male... | There are $100+80-60=120$ females in either band or orchestra, so there are $230-120=110$ males in either band or orchestra. Suppose $x$ males are in both band and orchestra.
\[80+100-x=110\Rightarrow x=70.\]
Thus, the number of males in band but not orchestra is $80-70=10\rightarrow \boxed{\text{A}}$.
PIE | 10 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Which number in the array below is both the largest in its column and the smallest in its row? (Columns go up and down, rows go right and left.)
\[\begin{tabular}[t]{ccccc} 10 & 6 & 4 & 3 & 2 \\ 11 & 7 & 14 & 10 & 8 \\ 8 & 3 & 4 & 5 & 9 \\ 13 & 4 & 15 & 12 & 1 \\ 8 & 2 & 5 & 9 & 3 \end{tabular}\]
$\text{(A)}\ 1 \qquad... | The largest numbers in the first, second, third, fourth and fifth columns are $13,7,15,12,9$ respectively. Of these, only $7$ is the smallest in its row $\rightarrow \boxed{\text{C}}$. | 7 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
What is the largest quotient that can be formed using two numbers chosen from the set $\{ -24, -3, -2, 1, 2, 8 \}$?
$\text{(A)}\ -24 \qquad \text{(B)}\ -3 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 24$ | Let the two chosen numbers be $a$ and $b$. To maximize the quotient, we first have either $a,b>0$ or $a,b0$, we have $a=8$ and $b=1$, which gives us $8/1=8$.
Since $12>8$, our answer is $\boxed{\text{D}}$. | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
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