2015 AIME II Problems

2015 AIME II (Answer Key) Printable version \| AoPS Contest Collections • PDF
Instructions 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. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
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Problem 1

Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$ .

Solution

Problem 2

In a new school, $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

Problem 3

Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$ . Find $m$ .

Solution

Problem 4

Problem 5

Two unit squares are selected at random without replacement from an $n \times n$ grid of unit squares. Find the least positive integer $n$ such that the probability that the two selected unit squares are horizontally or vertically adjacent is less than $\frac{1}{2015}$ .

Solution

Problem 6

Problem 7

Problem 8

Let $a$ and $b$ be positive integers satisfying $\frac{ab+1}{a+b} \lt \frac{3}{2}$ (Error compiling LaTeX. Unknown error_msg). The maximum possible value of $\frac{a^3b^3+1}{a^3+b^3}$ is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Solution

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

2015 AIME II (Problems • Answer Key • Resources)
Preceded by 2014 AIME I, 2014 AIME II	Followed by 2016 AIME
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All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2015 AIME II Problems

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15