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Difference between revisions of "2013 AIME II Problems"
 
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==Problem 1==
 
==Problem 1==
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Suppose that the measurement of time during the day converted to the metric system so that each day has 10 metic hours, and each metric hour has 100 metric minutes. Digital clocks would then be produced that would read 9:99 just before midnight, 0:00 at midnight, 1:25 at the former 3:00 AM, and 7:50p at the former 6:00. After the conversion, a person who wanted to wake up at the equivalent to the former 6:36AM would set his new digital alarm clock for A:BC, where A, B, and C are digits. Find 100A +10B + C.
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Suppose that the measurement of time during the day is converted to the metric system so that each day has <math>10</math> metric hours, and each metric hour has <math>100</math> metric minutes. Digital clocks would then be produced that would read <math>\text{9:99}</math> just before midnight, <math>\text{0:00}</math> at midnight, <math>\text{1:25}</math> at the former <math>\text{3:00}</math> AM, and <math>\text{7:50}</math> at the former <math>\text{6:00}</math> PM. After the conversion, a person who wanted to wake up at the equivalent of the former <math>\text{6:36}</math> AM would set his new digital alarm clock for <math>\text{A:BC}</math>, where <math>\text{A}</math>, <math>\text{B}</math>, and <math>\text{C}</math> are digits. Find <math>100\text{A}+10\text{B}+\text{C}</math>.
  
 
[[2013 AIME II Problems/Problem 1|Solution]]
 
[[2013 AIME II Problems/Problem 1|Solution]]
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<cmath>\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.</cmath>
 
<cmath>\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.</cmath>
 
Find the sum of all possible values of <math>a+b</math>.
 
Find the sum of all possible values of <math>a+b</math>.
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[[2013 AIME II Problems/Problem 2|Solution]]
  
 
==Problem 3==
 
==Problem 3==
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[[2013 AIME II Problems/Problem 4|Solution]]
 
[[2013 AIME II Problems/Problem 4|Solution]]
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==Problem 5==
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In equilateral <math>\triangle ABC</math> let points <math>D</math> and <math>E</math> trisect <math>\overline{BC}</math>. Then <math>\sin(\angle DAE)</math> can be expressed in the form <math>\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is an integer that is not divisible by the square of any prime. Find <math>a+b+c</math>.
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[[2013 AIME II Problems/Problem 5|Solution]]
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==Problem 6==
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Find the least positive integer <math>N</math> such that the set of <math>1000</math> consecutive integers beginning with <math>1000\cdot N</math> contains no square of an integer.
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[[2013 AIME II Problems/Problem 6|Solution]]
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==Problem 7==
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A group of clerks is assigned the task of sorting <math>1775</math> files. Each clerk sorts at a constant rate of <math>30</math> files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in <math>3</math> hours and <math>10</math> minutes. Find the number of files sorted during the first one and a half hours of sorting.
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[[2013 AIME II Problems/Problem 7|Solution]]
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==Problem 8==
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A hexagon that is inscribed in a circle has side lengths <math>22</math>, <math>22</math>, <math>20</math>, <math>22</math>, <math>22</math>, and <math>20</math> in that order. The radius of the circle can be written as <math>p+\sqrt{q}</math>, where <math>p</math> and <math>q</math> are positive integers. Find <math>p+q</math>.
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[[2013 AIME II Problems/Problem 8|Solution]]
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==Problem 9==
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A <math>7\times 1</math> board is completely covered by <math>m\times 1</math> tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let <math>N</math> be the number of tilings of the <math>7\times 1</math> board in which all three colors are used at least once. For example, a <math>1\times 1</math> red tile followed by a <math>2\times 1</math> green tile, a <math>1\times 1</math> green tile, a <math>2\times 1</math> blue tile, and a <math>1\times 1</math> green tile is a valid tiling. Note that if the <math>2\times 1</math> blue tile is replaced by two <math>1\times 1</math> blue tiles, this results in a different tiling. Find the remainder when <math>N</math> is divided by <math>1000</math>.
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[[2013 AIME II Problems/Problem 9|Solution]]
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==Problem 10==
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Given a circle of radius <math>\sqrt{13}</math>, let <math>A</math> be a point at a distance <math>4 + \sqrt{13}</math> from the center <math>O</math> of the circle. Let <math>B</math> be the point on the circle nearest to point <math>A</math>. A line passing through the point <math>A</math> intersects the circle at points <math>K</math> and <math>L</math>. The maximum possible area for <math>\triangle BKL</math> can be written in the form <math>\frac{a - b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers, <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>.
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[[2013 AIME II Problems/Problem 10|Solution]]
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==Problem 11==
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Let <math>A = \{1, 2, 3, 4, 5, 6, 7\}</math>, and let <math>N</math> be the number of functions <math>f</math> from set <math>A</math> to set <math>A</math> such that <math>f(f(x))</math> is a constant function. Find the remainder when <math>N</math> is divided by <math>1000</math>.
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[[2013 AIME II Problems/Problem 11|Solution]]
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==Problem 12==
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Let <math>S</math> be the set of all polynomials of the form <math>z^3 + az^2 + bz + c</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are integers. Find the number of polynomials in <math>S</math> such that each of its roots <math>z</math> satisfies either <math>|z| = 20</math> or <math>|z| = 13</math>.
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[[2013 AIME II Problems/Problem 12|Solution]]
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==Problem 13==
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In <math>\triangle ABC</math>, <math>AC = BC</math>, and point <math>D</math> is on <math>\overline{BC}</math> so that <math>CD = 3\cdot BD</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that <math>CE = \sqrt{7}</math> and <math>BE = 3</math>, the area of <math>\triangle ABC</math> can be expressed in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
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[[2013 AIME II Problems/Problem 13|Solution]]
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==Problem 14==
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For positive integers <math>n</math> and <math>k</math>, let <math>f(n, k)</math> be the remainder when <math>n</math> is divided by <math>k</math>, and for <math>n > 1</math> let <math>F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)</math>. Find the remainder when <math>\sum\limits_{n=20}^{100} F(n)</math> is divided by <math>1000</math>.
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[[2013 AIME II Problems/Problem 14|Solution]]
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==Problem 15==
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Let <math>A,B,C</math> be angles of an acute triangle with
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<cmath> \begin{align*}
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\cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\
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\cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9}
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\end{align*} </cmath>
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There are positive integers <math>p</math>, <math>q</math>, <math>r</math>, and <math>s</math> for which <cmath> \cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s}, </cmath> where <math>p+q</math> and <math>s</math> are relatively prime and <math>r</math> is not divisible by the square of any prime.  Find <math>p+q+r+s</math>.
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[[2013 AIME II Problems/Problem 15|Solution]]
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{{AIME box|year=2013|n=II|before=[[2013 AIME I Problems]]|after=[[2014 AIME I Problems]]}}
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{{MAA Notice}}

Latest revision as of 11:17, 13 March 2020

2013 AIME II (Answer Key)
Printable version | AoPS Contest Collections • PDF

Instructions

  1. This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
  2. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Problem 1

Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$, where $\text{A}$, $\text{B}$, and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$.

Solution

Problem 2

Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$.

Solution

Problem 3

A large candle is $119$ centimeters tall. It is designed to burn down more quickly when it is first lit and more slowly as it approaches its bottom. Specifically, the candle takes $10$ seconds to burn down the first centimeter from the top, $20$ seconds to burn down the second centimeter, and $10k$ seconds to burn down the $k$-th centimeter. Suppose it takes $T$ seconds for the candle to burn down completely. Then $\tfrac{T}{2}$ seconds after it is lit, the candle's height in centimeters will be $h$. Find $10h$.

Solution

Problem 4

In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$. Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant. Let $P=(x,y)$ be the center of $\triangle ABC$. Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$, where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime. Find $p+q+r$.

Solution

Problem 5

In equilateral $\triangle ABC$ let points $D$ and $E$ trisect $\overline{BC}$. Then $\sin(\angle DAE)$ can be expressed in the form $\frac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is an integer that is not divisible by the square of any prime. Find $a+b+c$.

Solution

Problem 6

Find the least positive integer $N$ such that the set of $1000$ consecutive integers beginning with $1000\cdot N$ contains no square of an integer.

Solution

Problem 7

A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting.

Solution

Problem 8

A hexagon that is inscribed in a circle has side lengths $22$, $22$, $20$, $22$, $22$, and $20$ in that order. The radius of the circle can be written as $p+\sqrt{q}$, where $p$ and $q$ are positive integers. Find $p+q$.

Solution

Problem 9

A $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find the remainder when $N$ is divided by $1000$.

Solution

Problem 10

Given a circle of radius $\sqrt{13}$, let $A$ be a point at a distance $4 + \sqrt{13}$ from the center $O$ of the circle. Let $B$ be the point on the circle nearest to point $A$. A line passing through the point $A$ intersects the circle at points $K$ and $L$. The maximum possible area for $\triangle BKL$ can be written in the form $\frac{a - b\sqrt{c}}{d}$, where $a$, $b$, $c$, and $d$ are positive integers, $a$ and $d$ are relatively prime, and $c$ is not divisible by the square of any prime. Find $a+b+c+d$.

Solution

Problem 11

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$, and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$.

Solution

Problem 12

Let $S$ be the set of all polynomials of the form $z^3 + az^2 + bz + c$, where $a$, $b$, and $c$ are integers. Find the number of polynomials in $S$ such that each of its roots $z$ satisfies either $|z| = 20$ or $|z| = 13$.

Solution

Problem 13

In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.

Solution

Problem 14

For positive integers $n$ and $k$, let $f(n, k)$ be the remainder when $n$ is divided by $k$, and for $n > 1$ let $F(n) = \max_{\substack{1\le k\le \frac{n}{2}}} f(n, k)$. Find the remainder when $\sum\limits_{n=20}^{100} F(n)$ is divided by $1000$.

Solution

Problem 15

Let $A,B,C$ be angles of an acute triangle with \begin{align*} \cos^2 A + \cos^2 B + 2 \sin A \sin B \cos C &= \frac{15}{8} \text{ and} \\ \cos^2 B + \cos^2 C + 2 \sin B \sin C \cos A &= \frac{14}{9} \end{align*} There are positive integers $p$, $q$, $r$, and $s$ for which \[\cos^2 C + \cos^2 A + 2 \sin C \sin A \cos B = \frac{p-q\sqrt{r}}{s},\] where $p+q$ and $s$ are relatively prime and $r$ is not divisible by the square of any prime. Find $p+q+r+s$.

Solution

2013 AIME II (Problems • Answer Key • Resources)
Preceded by
2013 AIME I Problems 		Followed by
2014 AIME I Problems
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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