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# Arithmetic Mean | AIME I, 2015 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean. You may use sequential hints.

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean.

## Arithmetic Mean of Number Theory – AIME 2015

Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

• is 107
• is 431
• is 840
• cannot be determined from the given information

### Key Concepts

Inequalities

Algebra

Number Theory

AIME, 2015, Question 12

Elementary Number Theory by David Burton

## Try with Hints

Each 1000-element subset ${ a_1, a_2,a_3,…,a_{1000}}$ of ${1,2,3,…,2015}$ with $a_1<a_2<a_3<…<a_{1000}$ contributes $a_1$ to sum of least element of each subset and set ${a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$. $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<…<a_{1000}+1$ ($k$ can be anything from $1$ to $a_1$ inclusive

Thus, the number of ways to choose the set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is equal to the sum. But choosing a set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is same as choosing a 1001-element subset from ${1,2,3,…,2016}$!

average =$\frac{2016}{1001}$=$\frac{288}{143}$. Then $p+q=288+143={431}$