2017 AIME I Problems/Problem 2

Problem 2

When each of $702$ , $787$ , and $855$ is divided by the positive integer $m$ , the remainder is always the positive integer $r$ . When each of $412$ , $722$ , and $815$ is divided by the positive integer $n$ , the remainder is always the positive integer $s \neq r$ . Find $m+n+r+s$ .

Solution

Let's work on both parts of the problem separately. First, $855 \equiv 787 \equiv 702 \equiv r \pmod{m}.$ We take the difference of $855$ and $787$ , and also of $787$ and $702$ . We find that they are $85$ and $68$ , respectively. Since the greatest common divisor of the two differences is $17$ (and the only one besides one), it's safe to assume that $m = 17$ .

Then, we divide $855$ by $17$ , and it's easy to see that $r = 5$ . Dividing $787$ and $702$ by $17$ also yields remainders of $5$ , which means our work up to here is correct.

Doing the same thing with $815$ , $722$ , and $412$ , the differences between $815$ and $722$ and $412$ are $310$ and $93$ , respectively. Since the only common divisor (besides $1$ , of course) is $31$ , $n = 31$ . Dividing all $3$ numbers by $31$ yields a remainder of $9$ for each, so $s = 9$ . Thus, $m + n + r + s = 17 + 5 + 31 + 9 = \boxed{062}$ .

2017 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2017 AIME I Problems/Problem 2

Problem 2

Solution

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