# 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, $m + n + r + s = 17 + 5 + 31 + 9 = \boxed{062}$.

## Solution 2

We know that $702 = am + r, 787 = bm + r,$ and $855 = cm+r$ where $a-c$ are integers.

Subtracting the first two, the first and third, and the last two, we get $85 = (b-a)m, 153=(c-a)m,$ and $68=(c-b)m.$

We know that $b-a, c-a$ and $c-b$ must be integers, so all the numbers are divisible by $m.$

Factorizing the numbers, we get $85 = 5 \cdot 17, 153 = 3^2 \cdot 17,$ and $68 = 2^2 \cdot 17.$ We see that all these have a factor of 17, so $m=17.$

Finding the remainder when $702$ is divided by $17,$ we get $n=5.$

Doing the same thing for the next three numbers, we get $17 + 5 + 31 + 9 = \boxed{062}$

## Solution 3 (Sol. 1 but possibly more clear)

As in Solution 1, we are given $855\equiv787\equiv702\equiv r\pmod{m}$ and $815\equiv722\equiv412\equiv s\pmod{n}.$ Tackling the first equation, we can simply look at $855\equiv787\equiv702\pmod{m}$. We subtract $702$ from each component of the congruency to get $153\equiv85\equiv0\pmod{m}$. Thus, we know that $153$ and $85$ must both be divisible by $m$. The only possible $m$, in this case, become $17$ and $1$; obviously, $m\neq1$, so we know $m=17$. We go back to the original equation, plug in $m$, and we find that $r=5$.

Similarly, we can subtract out the smallest value in the second congruency, $412$. We end up with $403\equiv310\equiv0\pmod{n}$. Again, we find that $n=31$ or $1$, so $n=31$. We also find that $s=9$.

Thus, our answer is $\boxed{062}$.

~Technodoggo

## Video Solution

https://youtu.be/BiiKzctXDJg ~Shreyas S