Write 100 as the sum of two positive integers 
Write $100$ as the sum of two positive integers, one of them being a multiple of $7$, while the other is a multiple of $11$.

Since $100$ is not a big number, I followed the straightforward reasoning of writing all multiples up to $100$ of either $11$ or $7$, and then finding the complement that is also a multiple of the other. So then 
$100 = 44 + 56 = 4 \times 11 + 8 \times 7$. 
But is it the smart way of doing it? Is it the way I was supposed to solve it? I'm thinking here about a situation with a really large number that turns my plug-in method sort of unwise.

$7a+11b=100$ so modding both sides by $11$ you get $11 \mid (7a-1)$. Let $11a'=7a-1$. Then the problem is transformed to finding $a',b$ such that $a'+b=9$ and $7 \mid (11a'+1)$, which (in my opinion) is somewhat easier to test for.
