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 tackle the first part of the problem first. We can safely assume: $702 = xm + r$ $787 = ym + r$ $855 = zm + r$ Now, if we subtract two values: $787-702 = 85 = 17\cdot5$ which also equals $(ym+r)-(xm+r) = m\cdot(y-x)$ Similarly, $855-787 = 68 = 17\cdot4; (zm+r)-(ym+r) = m\cdot(z-y)$ Since $17$ is the only common factor, we can assume that $m=17$ , and through simple division, that $r=5$ .

Using the same method on the second half: $412 = an + s$ $722 = bn + s$ $815 = cn + s$ Then. $722-412 = 310 = 31\cdot10; (bn+s)-(an+s) = n\cdot(b-a)$ $815-722 = 93 = 31\cdot3; (cn+s)-(bn+s) = n\cdot(c-b)$ The common factor is $31$ , so $n=31$ and through division, $s=9$ .

The answer is $m+n+r+s = 17+31+5+9 = \boxed{062}$

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2017 AIME I Problems/Problem 2

Problem 2

Solution