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How many $7$-digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$, $2$, $3$, $3$, $5$, $5$, $5$?
$\text{(A) } 6 \qquad \text{(B) } 12 \qquad \text{(C) } 24 \qquad \text{(D) } 36 \qquad \text{(E) } 48$ | A seven-digit palindrome is a number of the form $\overline{abcdcba}$. Clearly, $d$ must be $5$, as we have an odd number of fives. We are then left with $\{a,b,c\} = \{2,3,5\}$. There are $3!$ permutations of these three numbers, since each is reflected over the midpoint we only have to count the first there. Each of ... | 6 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Distinct points $A$, $B$, $C$, and $D$ lie on a line, with $AB=BC=CD=1$. Points $E$ and $F$ lie on a second line, parallel to the first, with $EF=1$. A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?
$\text{(A) } 3 \qquad \text{(B... | Consider the classical formula for triangle area: $\frac 12 \cdot b \cdot h$.
Each of the triangles that we can make has exactly one side lying on one of the two parallel lines. If we pick this side to be the base, the height will always be the same - it will be the distance between the two lines.
Hence each area is u... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?
$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$ | Solution 1
The sum of any four consecutive powers of 3 is divisible by $3^0 + 3^1 + 3^2 +3^3 = 40$ and hence is divisible by 8. Therefore
$(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{2009})$
is divisible by 8. So the required remainder is $3^0 + 3^1 = \boxed {4}$. The answer is $\mathrm{(D... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
By inserting parentheses, it is possible to give the expression
\[2\times3 + 4\times5\]
several values. How many different values can be obtained?
$\text{(A) } 2 \qquad \text{(B) } 3 \qquad \text{(C) } 4 \qquad \text{(D) } 5 \qquad \text{(E) } 6$ | The three operations can be performed on any of $3! = 6$ orders. However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values of the four expressions
\begin{align*} (2\times3)... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Mary's top book shelf holds five books with the following widths, in centimeters: $6$, $\dfrac{1}{2}$, $1$, $2.5$, and $10$. What is the average book width, in centimeters?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | To find the average, we add up the widths $6$, $\dfrac{1}{2}$, $1$, $2.5$, and $10$, to get a total sum of $20$. Since there are $5$ books, the average book width is $\frac{20}{5}=4$ The answer is $\boxed{D}$. | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "Le... | Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
As Mike is a frog, his statement is false, hence there is at most one toad.
As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the... | 3 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \q... | Solution 1
Imagine making the cuts one at a time. The first cut removes a box $2\times 2\times 3$. The second cut removes two boxes, each of dimensions $2\times 2\times 0.5$, and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$.
Therefore the ... | 7 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The number obtained from the last two nonzero digits of $90!$ is equal to $n$. What is $n$?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 32 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 68$ | We will use the fact that for any integer $n$,
\begin{align*}(5n+1)(5n+2)(5n+3)(5n+4)&=[(5n+4)(5n+1)][(5n+2)(5n+3)]\\ &=(25n^2+25n+4)(25n^2+25n+6)\equiv 4\cdot 6\\ &=24\pmod{25}\equiv -1\pmod{25}.\end{align*}
First, we find that the number of factors of $10$ in $90!$ is equal to $\left\lfloor \frac{90}5\right\rfloor+\l... | 12 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Jim starts with a positive integer $n$ and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with $n = 55$, then his sequence contains $5$ numbers:
\[\begin{arra... | We can find the answer by working backwards. We begin with $1-1^2=0$ on the bottom row, then the $1$ goes to the right of the equal's sign in the row above. We find the smallest value $x$ for which $x-1^2=1$ and $x>1^2$, which is $x=2$.
We repeat the same procedure except with $x-1^2=1$ for the next row and $x-1^2=2$ f... | 3 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A circle with center $O$ has area $156\pi$. Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3}$, and point $O$ is outside $\triangle ABC$. What is the side length of $\triangle ABC$?
$\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\... | The formula for the area of a circle is $\pi r^2$ so the radius of this circle is $\sqrt{156}.$
Because $OA=4\sqrt{3} < \sqrt{156}, A$ must be in the interior of circle $O.$
Let $s$ be the unknown value, the sidelength of the triangle, and let $X$ be the point on $BC$ where $OX \perp BC.$ Since $\triangle ABC$ is equ... | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
A drawer contains red, green, blue, and white socks with at least 2 of each color. What is
the minimum number of socks that must be pulled from the drawer to guarantee a matching
pair?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$ | After you draw $4$ socks, you can have one of each color, so (according to the [pigeonhole principle](https://artofproblemsolving.com/wiki/index.php/Pigeonhole_principle)), if you pull $\boxed{\textbf{(C)}\ 5}$ then you will be guaranteed a matching pair. | 5 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
For a real number $x$, define $\heartsuit(x)$ to be the average of $x$ and $x^2$. What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 20$ | The average of two numbers, $a$ and $b$, is defined as $\frac{a+b}{2}$. Thus the average of $x$ and $x^2$ would be $\frac{x(x+1)}{2}$. With that said, we need to find the sum when we plug, $1$, $2$ and $3$ into that equation. So:
\[\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \... | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Lucky Larry's teacher asked him to substitute numbers for $a$, $b$, $c$, $d$, and $e$ in the expression $a-(b-(c-(d+e)))$ and evaluate the result. Larry ignored the parenthese but added and subtracted correctly and obtained the correct result by coincidence. The number Larry substituted for $a$, $b$, $c$, and $d$ were ... | We simply plug in the numbers
\[1 - 2 - 3 - 4 + e = 1 - (2 - (3 - (4 + e)))\]
\[-8 + e = -2 - e\]
\[2e = 6\]
\[e = 3 \;\;(D)\] | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A majority of the $30$ students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1$. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils ... | The total cost of the pencils can be found by $(\text{students}\cdot\text{pencils purchased by each}\cdot\text{price of each pencil})$.
Since $1771$ is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: $7, 11, 23$. Since neither $(C)$ nor $(E)$ are ... | 11 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
How many even integers are there between $200$ and $700$ whose digits are all different and come from the set $\left\{1,2,5,7,8,9\right\}$?
$\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$ | We split up into cases of the hundreds digits being $2$ or $5$. If the hundred digits is $2$, then the units digits must be $8$ in order for the number to be even and then there are $4$ remaining choices ($1,5,7,9$) for the tens digit, giving $1 \times 4 \times 1=4$ possibilities. Similarly, there are $1 \times 2 \time... | 12 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Circles $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad \text... | The requested area is the area of $C$ minus the area shared between circles $A$, $B$ and $C$.
Let $M$ be the midpoint of $\overline{AB}$ and $D$ be the other intersection of circles $C$ and $B$.
The area shared between $C$, $A$ and $B$ is $4$ of the regions between arc $\widehat {MD}$ and line $\overline{MD}$, which is... | 2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$ | The requested ratio is \[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\] Using the formula for a geometric series, we have \[10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},\] which is very close to $\dfrac{10^{10}}{9},$ so the ratio is very close to $\boxed{\mathrm{(B) \ } 9}.$ | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
There are $52$ people in a room. what is the largest value of $n$ such that the statement "At least $n$ people in this room have birthdays falling in the same month" is always true?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 12$ | Pretend you have $52$ people you want to place in $12$ boxes, because there are $12$ months in a year. By the [Pigeonhole Principle](https://artofproblemsolving.com/wiki/index.php/Pigeonhole_Principle), one box must have at least $\left\lceil \frac{52}{12} \right\rceil$ people $\longrightarrow \boxed{\textbf{(D)} 5}$ | 5 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
What is the hundreds digit of $2011^{2011}?$
$\textbf{(A) } 1 \qquad \textbf{(B) } 4 \qquad \textbf{(C) }5 \qquad \textbf{(D) } 6 \qquad \textbf{(E) } 9$ | Since $2011 \equiv 11 \pmod{1000},$ we know that $2011^{2011} \equiv 11^{2011} \pmod{1000}.$
To compute this, we use a clever application of the [binomial theorem](https://artofproblemsolving.com/wiki/index.php/Binomial_theorem).
$\begin{aligned} 11^{2011} &= (1+10)^{2011} \\ &= 1 + \dbinom{2011}{1} \cdot 10 + \dbino... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Mary divides a circle into 12 sectors. The central angles of these sectors, measured in degrees, are all integers and they form an arithmetic sequence. What is the degree measure of the smallest possible sector angle?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12... | Let $a_1$ be the first term of the arithmetic progression and $a_{12}$ be the last term of the arithmetic progression. From the formula of the sum of an arithmetic progression (or arithmetic series), we have $12*\frac{a_1+a_{12}}{2}=360$, which leads us to $a_1 + a_{12} = 60$. $a_{12}$, the largest term of the progress... | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Externally tangent circles with centers at points $A$ and $B$ have radii of lengths $5$ and $3$, respectively. A line externally tangent to both circles intersects ray $AB$ at point $C$. What is $BC$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4$ | Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$. $AB=8$. Also, let $BC=x$. As $\angle ADC$ and $\angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $\angle ACD$, $\triangle ADC \sim \triangle BEC$. From this we can ... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$. What is $a-b$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Since $a$ and $b$ are relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $\frac{73}{3}$, the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$. Looking at the answer choices, the only multiple of $3$ is $\boxed{\textbf{(C)}\ 3... | 3 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Real numbers $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$. The probability that no two of $x$, $y$, and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$. What is the smallest possible value of $n$?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ ... | Since $x,y,z$ are all reals located in $[0, n]$, the number of choices for each one is continuous so we use geometric probability.
WLOG([Without loss of generality](https://artofproblemsolving.com/wiki/index.php/Without_loss_of_generality)), assume that $n\geq x \geq y \geq z \geq 0$. Then the set of points $(x,y,z)$ i... | 10 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Let $\angle ABC = 24^\circ$ and $\angle ABD = 20^\circ$. What is the smallest possible degree measure for $\angle CBD$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 12$ | $\angle ABD$ and $\angle ABC$ share ray $AB$. In order to minimize the value of $\angle CBD$, $D$ should be located between $A$ and $C$.
$\angle ABC = \angle ABD + \angle CBD$, so $\angle CBD = 4$. The answer is $\boxed{\textbf{(C)}\ 4}$ | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | Cross-multiplying gives $MN=36.$ We write $36$ as a product of two positive integers:
\begin{align*} 36 &= 1\cdot36 \\ &= 2\cdot18 \\ &= 3\cdot12 \\ &= 4\cdot9 \\ &= 6\cdot6. \end{align*}
The products $1\cdot36, 2\cdot18, 3\cdot12,$ and $4\cdot9$ each produce $2$ ordered pairs $(M,N),$ as we can switch the order of the... | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end of... | The total number of games (and wins) in the tournament is $\frac{6 \times 5}{2}= 15$. A six-way tie is impossible as this would imply each team has 2.5 wins, so the maximum number of tied teams is five. Here's a chart of 15 games where five teams each have 3 wins:
| 1 2 3 4 5 6 |
|1 X W L W L W |
|2 L X W... | 5 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last ... | Solution 1
The last number that Bernardo says has to be between 950 and 999. Note that $1\rightarrow 2\rightarrow 52\rightarrow 104\rightarrow 154\rightarrow 308\rightarrow 358\rightarrow 716\rightarrow 766$ contains 4 doubling actions. Thus, we have $x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x... | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the sum of all integer solutions to $1<(x-2)^2<25$?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25$ | $(x-2)^2$ = perfect square.
1 < perfect square < 25
Perfect square can equal: 4, 9, or 16
Solve for $x$:
$(x-2)^2=4$
$x=4,0$
and
$(x-2)^2=9$
$x=5,-1$
and
$(x-2)^2=16$
$x=6,-2$
The sum of all integer solutions is
$4+5+6+0+(-1)+(-2)=\boxed{\textbf{(B)} 12}$ | 12 | Inequalities | MCQ | Yes | Yes | amc_aime | false |
Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C... | Out of the first two integers, it's possible for both to be even: for example, $10 + 16 = 26.$ But the next two integers, when added, increase the sum by $15,$ which is odd, so one of them must be odd and the other must be even: for example, $3 + 12 = 15.$ Finally, the next two integers increase the sum by $16,$ which ... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person plannin... | Let the number of students on the council be $x$. To select a two-person committee, we can select a "first person" and a "second person." There are $x$ choices to select a first person; subsequently, there are $x-1$ choices for the second person. This gives a preliminary count of $x(x-1)$ ways to choose a two-person c... | 10 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Two sides of a triangle have lengths $10$ and $15$. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$ | The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,
therefore the length of the side perpendicular to that altitude will be between $10$ and $15$. The only answer choice that meets this requirement is $\boxed{\te... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Alice is making a batch of cookies and needs $2\frac{1}{2}$ cups of sugar. Unfortunately, her measuring cup holds only $\frac{1}{4}$ cup of sugar. How many times must she fill that cup to get the correct amount of sugar?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 12 \qquad\textbf{(D)}\ 16 \qquad\tex... | To get how many cups we need, we realize that we simply need to divide the number of cups needed by the number of cups collected in her measuring cup each time. Thus, we need to evaluate the fraction $\frac{2\frac{1}{2}}{\frac{1}{4}}$. Simplifying, this is equal to $\frac{5}{2}(4) = \boxed{\textbf{(B) }10}$ | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\bigtriangleup ABE$ is $40$. What is $BE$?
$\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad$ | We are given that the area of $\triangle ABE$ is $40$, and that $AB = 10$.
The area of a triangle is:
$A = \frac{bh}{2}$
Using $AB$ as the height of $\triangle ABE$,
$40 = \frac{10b}{2}$
and solving for $b$,
$b = 8$, which is $\boxed{\textbf{(E)}}$ | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Joey and his five brothers are ages $3$, $5$, $7$, $9$, $11$, and $13$. One afternoon two of his brothers whose ages sum to $16$ went to the movies, two brothers younger than $10$ went to play baseball, and Joey and the $5$-year-old stayed home. How old is Joey?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 7 \qquad\textbf{... | Because the $5$-year-old stayed home, we know that the $11$-year-old did not go to the movies, as the $5$-year-old did not and $11+5=16$. Also, the $11$-year-old could not have gone to play baseball, as he is older than $10$. Thus, the $11$-year-old must have stayed home, so Joey is $\boxed{\textbf{(D) }11}$ | 11 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\... | Solution 1
Let us split this up into two cases.
Case $1$: The student chooses both algebra and geometry.
This means that $3$ courses have already been chosen. We have $3$ more options for the last course, so there are $3$ possibilities here.
Case $2$: The student chooses one or the other.
Here, we simply count how... | 9 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Real numbers $x$ and $y$ satisfy the equation $x^2+y^2=10x-6y-34$. What is $x+y$?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$ | If we move every term dependent on $x$ or $y$ to the LHS, we get $x^2 - 10x + y^2 + 6y = -34$. Adding $34$ to both sides, we have $x^2 - 10x + y^2 + 6y + 34 = 0$. We can split the $34$ into $25$ and $9$ to get $(x - 5)^2 + (y + 3)^2 = 0$. Notice this is a circle with radius $0$, which only contains one point. So, the o... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying , so Blair follows by saying . Jo then says , and so on. What is the $53^{\text{rd}}$ number said?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf... | We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", $1+2+3+4+5+6+7+8+9=45$ numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because $53-... | 8 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
The number $2013$ is expressed in the form $2013 = \frac {a_1!a_2!...a_m!}{b_1!b_2!...b_n!}$,where $a_1 \ge a_2 \ge \cdots \ge a_m$ and $b_1 \ge b_2 \ge \cdots \ge b_n$ are positive integers and $a_1 + b_1$ is as small as possible. What is $|a_1 - b_1|$?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 ... | The prime factorization of $2013$ is $61\cdot11\cdot3$. To have a factor of $61$ in the numerator and to minimize $a_1,$ $a_1$ must equal $61$. Now we notice that there can be no prime $p$ which is not a factor of $2013$ such that $b_1<p<61,$ because this prime will not be canceled out in the denominator, and will lead... | 2 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$-axis at the same point. What is the sum of all possible $x$-coordinates of these points of intersection?
$\textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-... | Note that when $y=0$, the $x$ values of the equations should be equal by the problem statement. We have that
\[0 = ax + 5 \implies x = -\dfrac{5}{a}\] \[0 = 3x+b \implies x= -\dfrac{b}{3}\]
Which means that \[-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15\]
The only possible pairs $(a,b)$ then are $(a,b) = (1,15), (3,5... | -8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?
$\textbf{(A)}\ 2\qqu... | Let's use casework on the yellow house. The yellow house $(\text{Y})$ is either the $3^\text{rd}$ house or the last house.
Case 1: $\text{Y}$ is the $3^\text{rd}$ house.
The only possible arrangement is $\text{B}-\text{O}-\text{Y}-\text{R}$
Case 2: $\text{Y}$ is the last house.
There are two possible arrangements:
$\te... | 3 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
On an algebra quiz, $10\%$ of the students scored $70$ points, $35\%$ scored $80$ points, $30\%$ scored $90$ points, and the rest scored $100$ points. What is the difference between the mean and median score of the students' scores on this quiz?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D... | Without loss of generality, let there be $20$ students (the least whole number possible) who took the test. We have $2$ students score $70$ points, $7$ students score $80$ points, $6$ students score $90$ points and $5$ students score $100$ points. Therefore, the mean
is $87$ and the median is $90$.
Thus, the solution i... | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$. How long is the third altitude of the triangle?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We find that the area of the triangle is $\frac{6\times 2\sqrt{3}}{2} =6\sqrt{3}$. By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$. Dropping an altitude from the right angle to the hypoten... | 3 | Geometry | 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 $D$?
\[\begin{array}[t]{r} ABBCB \\ + \ BCADA \\ \hline DBDDD \end{array}\]
$\textbf {(A) } 2 \qquad \textbf {(B) } 4 \qquad \textbf {(C) } 7 \qquad \textbf {(D) } 8 \qquad \textbf {(E) } 9$ | Note from the addition of the last digits that $A+B=D\text{ or }A+B=D+10$.
From the addition of the frontmost digits, $A+B$ cannot have a carry, since the answer is still a five-digit number. Also $A + B$ cant have a carry since then for the second column, $C + 1 + D$ cant equal $D$.
Therefore $A+B=D$.
Using the sec... | 7 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$. What is $|a-b|$?
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$ | Since points on the graph make the equation true, substitute $\sqrt{\pi}$ in to the equation and then solve to find $a$ and $b$.
$y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1$
$y^2 + \pi^2 = 2\pi y + 1$
$y^2 - 2\pi y + \pi^2 = 1$
$(y-\pi)^2 = 1$
$y-\pi = \pm 1$
$y = \pi + 1$
$y = \pi - 1$
There are only two solutions to th... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ : $1$ ?
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$ | This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.
The first statement can be written as $p-2=3(c-2)$. The second statement can be written as $p-4=4(c-4)$
To solve the system of equations:
$p=3c-4$
$p=4c-12$
$3c-4=4c-12$
$c=8$
$p=20.$
Let $x$ be the ... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For how many integers $x$ is the point $(x, -x)$ inside or on the circle of radius $10$ centered at $(5, 5)$?
$\textbf{(A) }11\qquad \textbf{(B) }12\qquad \textbf{(C) }13\qquad \textbf{(D) }14\qquad \textbf{(E) }15$ | [https://youtu.be/RZDFs3qrw7Y](https://artofproblemsolving.comhttps://youtu.be/RZDFs3qrw7Y)
~Education, the Study of Everything=
Solution
The equation of the circle is $(x-5)^2+(y-5)^2=100$. Plugging in the given conditions we have $(x-5)^2+(-x-5)^2 \leq 100$. Expanding gives: $x^2-10x+25+x^2+10x+25\leq 100$, which si... | 11 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. How many paths are there meeting these conditions?
$\textbf{(A) }\text{6}\qquad\textbf{(B) }\text{9}\q... | We label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled $A$.
If we define a "move" as each time Erin crawls along a single edge f... | 6 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$. What is the sum of the digits of $s$?
$\textbf{(A) }7\qquad\textbf{(B) }8\qquad\textb... | A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have:
\[\begin{array}{c|c|c|c|c|c|c} \mathrm{Factorial}&0!-4!&5!-9!&10!-14!&15!-19!&20!-24!&25!-29!\\\hlin... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A rectangular box measures $a \times b \times c$, where $a$, $b$, and $c$ are integers and $1\leq a \leq b \leq c$. The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
$\textbf{(A)}\; 4 \qquad\textbf{(B)}\; 10 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 21... | We need \[abc = 2(ab+bc+ac) \quad \text{ or } \quad (a-2)bc = 2a(b+c).\]
Since $a\le b, ac \le bc$, from the first equation we get $abc \le 6bc$. Thus $a\le 6$. From the second equation we see that $a > 2$. Thus $a\in \{3, 4, 5, 6\}$.
If $a=3$ we need $bc = 6(b+c) \Rightarrow (b-6)(c-6)=36$. We get five roots $\{(3, ... | 10 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Kaashish 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 the first number be $x$ and the second be $y$. We have $2x+3y=100$. We are given one of the numbers is $28$. If $x$ were to be $28$, $y$ would not be an integer, thus $y=28$. $2x+3(28)=100$, which gives $x=\boxed{\textbf{(A) }8}$. | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the length in feet of the inner rectangle?
$\textbf{(A) } 1 \qquad ... | Let the length of the inner rectangle be $x$.
Then the area of that rectangle is $x\cdot1 = x$.
The second largest rectangle has dimensions of $x+2$ and $3$, making its area $3x+6$. The area of the second shaded area, therefore, is $3x+6-x = 2x+6$.
The largest rectangle has dimensions of $x+4$ and $5$, making its area ... | 2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$:
\[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\]
Since the bases are equal, we can set the exponents equal, giving us $5x=15$. Solving the equation gives us $x = \boxed{\text... | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$th row. What is the sum of the digits of $N$?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | We are trying to find the value of $N$ such that \[1+2+3\cdots+(N-1)+N=\frac{N(N+1)}{2}=2016.\] Noticing that $\frac{63\cdot 64}{2}=2016,$ we have $N=63,$ so our answer is $\boxed{\textbf{(D) } 9}.$
Notice that we were attempting to solve $\frac{N(N+1)}{2} = 2016 \Rightarrow N(N+1) = 2016\cdot2 = 4032$. Approximating $... | 9 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the value of $\frac{2a^{-1}+\frac{a^{-1}}{2}}{a}$ when $a= \tfrac{1}{2}$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \frac{5}{2}\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 20$ | Factorizing the numerator, $\frac{\frac{1}{a}\cdot(2+\frac{1}{2})}{a}$ then becomes $\frac{\frac{5}{2}}{a^{2}}$ which is equal to $\frac{5}{2}\cdot 2^2$ which is $\boxed{\textbf{(D) }10}$. | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$. What is the number in the center?
$\textbf{(A)}\ 5\qquad\tex... | Consecutive numbers share an edge. That means that it is possible to walk from $1$ to $9$ by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity:[asy]size(4cm); for(int i=0;i<3;++i)for(int j=0;j<3;++j)filldraw(box((i,j),(i+1,j+1)),gray((i+j)%2*.... | 7 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | Factor $345=3\cdot 5\cdot 23$.
Suppose we take an odd number $k$ of consecutive integers, with the median as $m$. Then $mk=345$ with $\tfrac12k<m$.
Looking at the factors of $345$, the possible values of $k$ are $3,5,15,23$ with medians as $115,69,23,15$ respectively.
Suppose instead we take an even number $2k$ of cons... | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number $S$. What is the smallest possible value for the sum of the digits of $S$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$ | Let the two three-digit numbers she added be $a$ and $b$ with $a+b=S$ and $a<b$. The hundreds digits of these numbers must be at least $1$ and $2$, so $a\ge 100$ and $b\ge 200$.
Say $a=100+p$ and $b=200+q$; then we just need $p+q=100$ with $p$ and $q$ having different digits which aren't $1$ or $2$.There are many solut... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the tens digit of $2015^{2016}-2017?$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 8$ | Notice that, for $n\ge 2$, $2015^n\equiv 15^n$ is congruent to $25\pmod{100}$ when $n$ is even and $75\pmod{100}$ when $n$ is odd. (Check for yourself). Since $2016$ is even, $2015^{2016} \equiv 25\pmod{100}$ and $2015^{2016}-2017 \equiv 25 - 17 \equiv \underline{0}8\pmod{100}$.
So the answer is $\textbf{(A)}\ \boxed{... | 0 | Number Theory | 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 $BC$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$ | The area of the triangle is $\frac{(2r)(r^2)}{2} = r^3$, so $r^3=64\implies r=4$, giving a total distance across the top of $8$, which is answer $\textbf{(C)}$. | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf... | A pattern starts to emerge as the function is continued. The repeating pattern is $0,1,1,2,0,2,2,1\ldots$ The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating pattern, we just need to find the sum of the numbers in the sequence, which is $\boxed{\text... | 9 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Distinct points $P$, $Q$, $R$, $S$ lie on the circle $x^{2}+y^{2}=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$?
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 3\sqrt{5} \qquad \textbf{(D) } 7 \qquad ... | Because $P$, $Q$, $R$, and $S$ are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are $(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),$ and $(\pm 5,0).$ We want to maximize $PQ$ and minimize $RS.$ They also have to be non perfect squares, because they are both irrational.... | 7 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Amelia has a coin that lands heads with probability $\frac{1}{3}\,$, and Blaine has a coin that lands on heads with probability $\frac{2}{5}$. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability t... | Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P$, since if she gets to her turn again, she is back where she started with probability of winning $P$. The chance she wins on her first turn is $\frac{1}{3}$. The chance she m... | 4 | Combinatorics | 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 the two real numbers be $x,y$. We are given that $x+y=4xy,$ and dividing both sides by $xy$, $\frac{x}{xy}+\frac{y}{xy}=4.$
\[\frac{1}{y}+\frac{1}{x}=\boxed{\textbf{(C) } 4}.\]
Note: we can easily verify that this is the correct answer; for example, $\left(\frac{1}{2}, \frac{1}{2}\right)$ works, and the sum of thei... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$. Then she switched the digits of the result, obtaining a number between $71$ and $75$, inclusive. What was Mary's number?
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$ | Let her $2$-digit number be $x$. Multiplying by $3$ makes it a multiple of $3$, meaning that the sum of its digits is divisible by $3$. Adding on $11$ increases the sum of the digits by $1+1 = 2,$ (we can ignore numbers such as $39+11=50$) and reversing the digits keeps the sum of the digits the same; this means that t... | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at l... | By [PIE](https://artofproblemsolving.com/wiki/index.php/PIE) (Property of Inclusion/Exclusion), we have
$|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|.$
Number of people in at least two sets is $\sum |A_i \cap A_j| - 2|A_1 \cap A_2 \cap A_3| = 9.$
So, $20 = (10 + 13 + 9) - (9 + 2x... | 3 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Sofia ran $5$ laps around the $400$-meter track at her school. For each lap, she ran the first $100$ meters at an average speed of $4$ meters per second and the remaining $300$ meters at an average speed of $5$ meters per second. How much time did Sofia take running the $5$ laps?
$\qquad\textbf{(A)}\ \text{5 minutes a... | If Sofia ran the first $100$ meters of each lap at $4$ meters per second and the remaining $300$ meters of each lap at $5$ meters per second, then she took $\frac{100}{4}+\frac{300}{5}=25+60=85$ seconds for each lap. Because she ran $5$ laps, she took a total of $5 \cdot 85=425$ seconds, or $7$ minutes and $5$ seconds.... | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $N=123456789101112\dots4344$ be the $79$-digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$ | We only need to find the remainders of N when divided by 5 and 9 to determine the answer.
By inspection, $N \equiv 4 \text{ (mod 5)}$.
The remainder when $N$ is divided by $9$ is $1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4$, but since $10 \equiv 1 \text{ (mod 9)}$, we can also write this as $1+2+3 +\cdots +10+11+1... | 9 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$. What is the value of $\frac{x+3y}{3x-y}$?
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$ | Rearranging, we find $3x+y=-2x+6y$, or $5x=5y\implies x=y$.
Substituting, we can convert the second equation into $\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}$.
More step-by-step explanation:
$\frac{3x+y}{x-3y}=-2$
$3x+y=-2\left(x-3y\right)$
$3x+y=-2x+6y$
$5x=5y$
$x=y$
$\frac{x+3y}{3x-y}=\frac{1+3\left(1\r... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the largest number of solid $2\text{-in} \times 2\text{-in} \times 1\text{-in}$ blocks that can fit in a $3\text{-in} \times 2\text{-in}\times3\text{-in}$ box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | We find that the volume of the larger block is $18$, and the volume of the smaller block is $4$. Dividing the two, we see that only a maximum of four $2$ by $2$ by $1$ blocks can fit inside the $3$ by $3$ by $2$ block. Drawing it out, we see that such a configuration is indeed possible. Therefore, the answer is $\boxed... | 4 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations?
\begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*}
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this:
[asy] draw((... | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For how many (not necessarily positive) integer values of $n$ is the value of $4000\cdot \left(\tfrac{2}{5}\right)^n$ an integer?
$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$ | Note that \[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.\]
Since this expression is an integer, we need:
$5+n\geq0,$ from which $n\geq-5.$
$3-n\geq0,$ from which $n\leq3.$
Taking the intersection gives $-5\leq n\leq3.$ So, there are $3-(-5)+1=\bo... | 9 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Joe has a collection of $23$ coins, consisting of $5$-cent coins, $10$-cent coins, and $25$-cent coins. He has $3$ more $10$-cent coins than $5$-cent coins, and the total value of his collection is $320$ cents. How many more $25$-cent coins does Joe have than $5$-cent coins?
$\textbf{(A) } 0 \qquad \textbf{(... | Let $x$ be the number of $5$-cent coins that Joe has. Therefore, he must have $(x+3) \ 10$-cent coins and $(23-(x+3)-x) \ 25$-cent coins. Since the total value of his collection is $320$ cents, we can write
\begin{align*} 5x + 10(x+3) + 25(23-(x+3)-x) &= 320 \\ 5x + 10x + 30 + 500 - 50x &= 320 \\ 35x &= 210 \\ x &= 6. ... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In the rectangular parallelepiped shown, $AB$ = $3$, $BC$ = $1$, and $CG$ = $2$. Point $M$ is the midpoint of $\overline{FG}$. What is the volume of the rectangular pyramid with base $BCHE$ and apex $M$?
$\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3... | Consider the cross-sectional plane and label its area $b$. Note that the volume of the triangular prism that encloses the pyramid is $\frac{bh}{2}=3$, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is $\frac{bh}{3}$, so the answer is $\boxed{(E) ... | 2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\]
What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | Note that, because of Fermat's Little Theorem, for any integer $a$ relatively prime to 6, we have $a^2 \equiv 1 \pmod{6} \implies a^3 \equiv a \pmod{6}$. This is also trivially true if $a$ is not relatively prime to 6 because $a \equiv 0 \pmod{6}$. Therefore,
\[a_1+a_2+\cdots+a_{2018} \equiv a_1^3+a_2^3+\cdots+a_{2018}... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In rectangle $PQRS$, $PQ=8$ and $QR=6$. Points $A$ and $B$ lie on $\overline{PQ}$, points $C$ and $D$ lie on $\overline{QR}$, points $E$ and $F$ lie on $\overline{RS}$, and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon ... | Let $AP=BQ=x$. Then $AB=8-2x$.
Now notice that since $CD=8-2x$ we have $QC=DR=x-1$.
Thus by the Pythagorean Theorem we have $x^2+(x-1)^2=(8-2x)^2$ which becomes $2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}$.
Our answer is $8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}$. (Mudkipswims42) | 7 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age i... | Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that \[\frac{n+c}{n+1}=1+\frac{c-1}{n+1}\] is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. The prime ... | 11 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the value of \[2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?\]
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$ | $2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9$
$= 1+1 = \boxed{2}$ which corresponds to $\boxed{\text{C}}$. | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the hundreds digit of $(20!-15!)?$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }4\qquad\textbf{(E) }5$ | Because we know that $5^3$ is a factor of $15!$ and $20!$, the last three digits of both numbers is a 0, this means that the difference of the hundreds digits is also $\boxed{\textbf{(A) }0}$. | 0 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Ana and Bonita were born on the same date in different years, $n$ years apart. Last year Ana was $5$ times as old as Bonita. This year Ana's age is the square of Bonita's age. What is $n?$
$\textbf{(A) } 3 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 9 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 15$ | Solution 1
Let $A$ be the age of Ana and $B$ be the age of Bonita. Then,
\[A-1 = 5(B-1)\]
and
\[A = B^2.\]
Substituting the second equation into the first gives us
\[B^2-1 = 5(B-1).\]
By using difference of squares and dividing, $B=4.$ Moreover, $A=B^2=16.$
The answer is $16-4 = 12 \implies \boxed{\textbf{(D)}}$
Solut... | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?
a square
a rectangle that is not a square
a rhombus that is not a square
a parallelogram that is not a rectangle or a rhombus
an isosceles tr... | This question is simply asking how many of the listed quadrilaterals are cyclic (since the point equidistant from all four vertices would be the center of the circumscribed circle). A square, a rectangle, and an isosceles trapezoid (that isn't a parallelogram) are all cyclic, and the other two are not. Thus, the answer... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
What is the sum of all real numbers $x$ for which the median of the numbers $4,6,8,17,$ and $x$ is equal to the mean of those five numbers?
$\textbf{(A) } -5 \qquad\textbf{(B) } 0 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } \frac{15}{4} \qquad\textbf{(E) } \frac{35}{4}$ | The mean is $\frac{4+6+8+17+x}{5}=\frac{35+x}{5}$.
There are three possibilities for the median: it is either $6$, $8$, or $x$.
Let's start with $6$.
$\frac{35+x}{5}=6$ has solution $x=-5$, and the sequence is $-5, 4, 6, 8, 17$, which does have median $6$, so this is a valid solution.
Now let the median be $8$.
$\frac{... | -5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$, where $T$, $M$, and $H$ denote digits that are not given. What is $T+M+H$?
$\textbf{(A) }3 \qquad\textbf{(B) }8 \qquad\textbf{(C) }12 \qquad\textbf{(D) }14 \qquad\textbf{(E) } 17$ | We can figure out $H = 0$ by noticing that $19!$ will end with $3$ zeroes, as there are three factors of $5$ in its prime factorization, so there would be 3 powers of 10 meaning it will end in 3 zeros. Next, we use the fact that $19!$ is a multiple of both $11$ and $9$. Their divisibility rules (see Solution 2) tell us... | 12 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$. What is the sum of the digits of $n$?
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | Solution 1
\[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]
Solving by the quadratic formula, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{3... | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$-axis, and reflection across the ... | [asy] size(10cm); Label f; f.p=fontsize(6); xaxis(-6,6,Ticks(f, 2.0)); yaxis(-6,6,Ticks(f, 2.0)); filldraw(origin--(4,0)--(0,3)--cycle, gray, black+linewidth(1)); [/asy]
First, any combination of motions we can make must reflect $T$ an even number of times. This is because every time we reflect $T$, it changes orie... | 12 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Assuming $a\neq3$, $b\neq4$, and $c\neq5$, what is the value in simplest form of the following expression?
\[\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}\]
$\textbf{(A) } {-}1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \frac{abc}{60} \qquad \textbf{(D) } \frac{1}{abc} - \frac{1}{60} \qquad \textbf{(E) }... | If $x\neq y,$ then $\frac{x-y}{y-x}=-1.$ We use this fact to simplify the original expression:
\[\frac{\color{red}\overset{-1}{\cancel{a-3}}}{\color{blue}\underset{1}{\cancel{5-c}}} \cdot \frac{\color{green}\overset{-1}{\cancel{b-4}}}{\color{red}\underset{1}{\cancel{3-a}}} \cdot \frac{\color{blue}\overset{-1}{\cancel{c... | -1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the value of
\[1-(-2)-3-(-4)-5-(-6)?\]
$\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21$ | We know that when we subtract negative numbers, $a-(-b)=a+b$.
The equation becomes $1+2-3+4-5+6 = \boxed{\textbf{(D)}\ 5}$.
~quacker88 | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Steve wrote the digits $1$, $2$, $3$, $4$, and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$rd, $6$th, $9$th, $\ldots$ digits from the left), then erased every fourth digit from the resultin... | Note that cycles exist initially and after each round of erasing.
Let the parentheses denote cycles. It follows that:
Initially, the list has cycles of length $5:$ \[(12345)=12345123451234512345\cdots.\]
To find one cycle after the first round of erasing, we need one cycle of length $\operatorname{lcm}(3,5)=15$ befor... | 11 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$. What is the digit $A$?
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\te... | $158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}$
We're looking for the amount of ways we can get $10$ cards from a deck of $52$, which is represented by $\binom{52}{10}$.
$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cd... | 2 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\](Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$.)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | We can first consider the equation without a floor function:
\[\dfrac{n+1000}{70} = \sqrt{n}\]
Multiplying both sides by 70 and then squaring:
\[n^2 + 2000n + 1000000 = 4900n\]
Moving all terms to the left:
\[n^2 - 2900n + 1000000 = 0\]
Now we can determine the factors:
\[(n-400)(n-2500) = 0\]
This means that for $n = ... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$, where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 11$ | Since the three angles of a triangle add up to $180^{\circ}$ and one of the angles is $90^{\circ}$ because it's a right triangle, $a^{\circ} + b^{\circ} = 90^{\circ}$.
The greatest prime number less than $90$ is $89$. If $a=89^{\circ}$, then $b=90^{\circ}-89^{\circ}=1^{\circ}$, which is not prime.
The next greatest pri... | 7 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many positive even multiples of $3$ less than $2020$ are perfect squares?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$ | Any even multiple of $3$ is a multiple of $6$, so we need to find multiples of $6$ that are perfect squares and less than $2020$. Any solution that we want will be in the form $(6n)^2$, where $n$ is a positive integer. The smallest possible value is at $n=1$, and the largest is at $n=7$ (where the expression equals $17... | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Points $P$ and $Q$ lie in a plane with $PQ=8$. How many locations for point $R$ in this plane are there such that the triangle with vertices $P$, $Q$, and $R$ is a right triangle with area $12$ square units?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$ | Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$
We construct a circle with diameter $\overline{PQ}.$ All such locations for $R$ are shown below:
We apply casework to the right angle of $\triangle PQR:$
If $\angle P=90^\circ,$ then $R\in\{... | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
What is the value of \[(2^2-2)-(3^2-3)+(4^2-4)\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 12$ | $(4-2)-(9-3)+(16-4)=2-6+12=8.$ This corresponds to $\boxed{\textbf{(D) } 8}.$
-happykeeper | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For which of the following integers $b$ is the base-$b$ number $2021_b - 221_b$ not divisible by $3$?
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~4\qquad\textbf{(C)} ~6\qquad\textbf{(D)} ~7\qquad\textbf{(E)} ~8$ | We have
\begin{align*} 2021_b - 221_b &= (2021_b - 21_b) - (221_b - 21_b) \\ &= 2000_b - 200_b \\ &= 2b^3 - 2b^2 \\ &= 2b^2(b-1), \end{align*}
which is divisible by $3$ unless $b\equiv2\pmod{3}.$ The only choice congruent to $2$ modulo $3$ is $\boxed{\textbf{(E)} ~8}.$
~MRENTHUSIASM | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the volume of tetrahedron $ABCD$ with edge lengths $AB = 2$, $AC = 3$, $AD = 4$, $BC = \sqrt{13}$, $BD = 2\sqrt{5}$, and $CD = 5$ ?
$\textbf{(A)} ~3 \qquad\textbf{(B)} ~2\sqrt{3} \qquad\textbf{(C)} ~4\qquad\textbf{(D)} ~3\sqrt{3}\qquad\textbf{(E)} ~6$ | Drawing the tetrahedron out and testing side lengths, we realize that the $\triangle ACD, \triangle ABC,$ and $\triangle ABD$ are right triangles by the Converse of the Pythagorean Theorem. It is now easy to calculate the volume of the tetrahedron using the formula for the volume of a pyramid. If we take $\triangle ADC... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
What is the least possible value of $(xy-1)^2+(x+y)^2$ for real numbers $x$ and $y$?
$\textbf{(A)} ~0\qquad\textbf{(B)} ~\frac{1}{4}\qquad\textbf{(C)} ~\frac{1}{2} \qquad\textbf{(D)} ~1 \qquad\textbf{(E)} ~2$ | Expanding, we get that the expression is $x^2+2xy+y^2+x^2y^2-2xy+1$ or $x^2+y^2+x^2y^2+1$. By the Trivial Inequality (all squares are nonnegative) the minimum value for this is $\boxed{\textbf{(D)} ~1}$, which can be achieved at $x=y=0$.
~aop2014 | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263$, and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$
$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\tex... | We can start by setting up an equation to convert $\underline{32d}$ base $n$ to base 10. To convert this to base 10, it would be $3{n}^2+2n+d.$ Because it is equal to 263, we can set this equation to 263. Finally, subtract $d$ from both sides to get $3{n}^2+2n = 263-d$.
We can also set up equations to convert $\underli... | 11 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$. What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | Since two parallel chords have the same length ($38$), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is
\[2d + d = 3d\]
and the distance between each... | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The real number $x$ satisfies the equation $x+\frac{1}{x} = \sqrt{5}$. What is the value of $x^{11}-7x^{7}+x^3?$
$\textbf{(A)} ~-1 \qquad\textbf{(B)} ~0 \qquad\textbf{(C)} ~1 \qquad\textbf{(D)} ~2 \qquad\textbf{(E)} ~\sqrt{5}$ | We square $x+\frac{1}{x}=\sqrt5$ to get $x^2+2+\frac{1}{x^2}=5$. We subtract 2 on both sides for $x^2+\frac{1}{x^2}=3$ and square again, and see that $x^4+2+\frac{1}{x^4}=9$ so $x^4+\frac{1}{x^4}=7$. We can factor out $x^7$ from our original expression of $x^{11}-7x^7+x^3$ to get that it is equal to $x^7(x^4-7+\frac{1}... | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Call a positive integer an uphill integer if every digit is strictly greater than the previous digit. For example, $1357, 89,$ and $5$ are all uphill integers, but $32, 1240,$ and $466$ are not. How many uphill integers are divisible by $15$?
$\textbf{(A)} ~4 \qquad\textbf{(B)} ~5 \qquad\textbf{(C)} ~6 \qquad\textbf{(D... | The divisibility rule of $15$ is that the number must be congruent to $0$ mod $3$ and congruent to $0$ mod $5$. Being divisible by $5$ means that it must end with a $5$ or a $0$. We can rule out the case when the number ends with a $0$ immediately because the only integer that is uphill and ends with a $0$ is $0$ which... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
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