Difference between revisions of "2015 AIME II Problems"

Revision as of 19:27, 26 March 2015

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, rulers and compasses are permitted. In particular, graph paper, protractors, 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

In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$ , and the altitude to these bases has length $\log 16$ . The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$ , where $p$ and $q$ are positive integers. Find $p + q$ .

Solution

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 $\frac{ab+1}{a+b} < \frac{3}{2}$ . 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

Call a permutation $a_1, a_2, \ldots, a_n$ of the integers $1, 2, \ldots, n$ quasi-increasing if $a_k \leq a_{k+1} + 2$ for each $1 \leq k \leq n-1$ . For example, 53421 and 14253 are quasi-increasing permutations of the integers $1, 2, 3, 4, 5$ , but 45123 is not. Find the number of quasi-increasing permutations of the integers $1, 2, \ldots, 7$ .

Solution

Problem 11

Problem 12

There are $2^{10} = 1024$ possible 10-letter strings in which each letter is either an A or a B. Find the number of such strings that do not have more than 3 adjacent letters that are identical.

Solution

Problem 13

Problem 14

Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$ .

Solution

Problem 15

2015 AIME II (Problems • Answer Key • Resources)
Preceded by 2014 AIME I, 2014 AIME II	Followed by 2016 AIME
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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

@@ Line 60: / Line 60: @@
 ==Problem 14==
+Let <math>x</math> and <math>y</math> be real numbers satisfying <math>x^4y^5+y^4x^5=810</math> and <math>x^3y^6+y^3x^6=945</math>. Evaluate <math>2x^3+(xy)^3+2y^3</math>.
+[[2015 AIME II Problems/Problem 14 | Solution]]
 ==Problem 15==

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Difference between revisions of "2015 AIME II Problems"

Revision as of 19:27, 26 March 2015

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