# Difference between revisions of "2017 AIME I Problems/Problem 2"

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The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{062}</math> | The answer is <math>m+n+r+s = 17+31+5+9 = \boxed{062}</math> | ||

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## Revision as of 15:53, 8 March 2017

## Problem 2

When each of , , and is divided by the positive integer , the remainder is always the positive integer . When each of , , and is divided by the positive integer , the remainder is always the positive integer . Find .

## Solution

Let's tackle the first part of the problem first. We can safely assume: Now, if we subtract two values: which also equals Similarly, Since is the only common factor, we can assume that , and through simple division, that .

Using the same method on the second half: Then. The common factor is , so and through division, .

The answer is

~IYN~