A number theory problem by Jose Sacramento

At first, one might attempt to think about possible values of n that makes 7n end in units digit six. But, it is impossible here. The second strategy is to think, how do we get numbers (disregarding m and n) that must add together to produce six? This leads us to think about multiples of 7 ending in one and multiples of 5 ending in 5. So, after trial and error, 5x(5) + 7x(3)= 46. m is 5 and n is 3. 3x5=15.

This is a rather long solution but for those who are new to number theory the beauty of this branch is to incorporate logic. Number theory requires logical deduction to get our answers. This is why I wrote it so long. Cheers.

Using extended euclidean algorithm we find that general integer solutions for $m$ and $n$ are $m = 138 + 7k$ and $n = -92 - 5k$ where $k$ is an integer.

Now, since both $m$ and $n$ are positive, we must have $92 + 5k < 0 < 138 + 7k \implies -20 < k < -18$ . (Remember that $k$ is an integer.)

Hence, $k = -19 \implies m = 5, n = 3 \implies m \times n = \boxed{15}$