# 1966 AHSME Problems/Problem 26

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## Problem

Let $m$ be a positive integer and let the lines $13x+11y=700$ and $y=mx-1$ intersect in a point whose coordinates are integers. Then m can be: $\text{(A) 4 only} \quad \text{(B) 5 only} \quad \text{(C) 6 only} \quad \text{(D) 7 only} \\ \text{(E) one of the integers 4,5,6,7 and one other positive integer}$

## Solution

Substitute the second equation into the first one, we have $13x+11mx-11=700$.

So $(13+11m)x=711$. So $13+11m$ is a factor of $711$. $711=3*3*79$, so the factors of $711$ are: $1, 3, 9, 79, 237, 711$.

Clearly, because $m\ge1$, so $13+11m\ge24$. So we only need to check whether $m$ is an integer when $11m+13=79, 237, 711$.

When $11m+13=79$, $m=6$.

Checking the other two choices, $m$ dosen't yield to be an integer. So $m=6$ is the only option. Select $\boxed{C}$.

~hastapasta

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