2017 AIME I Problems/Problem 2

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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 $m+n+r+s = 17+31+5+9 = \boxed{62}$

~IYN~

See Also

2017 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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