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.

$7x+11y=100$
$7x=100-11y$
$x=\frac{100-11y}7=14-2y+\frac{2+3y}7$
$a=\frac{2+3y}7$
$7a=2+3y$
$3y=-2+7a$
$y=\frac{-2+7a}3=-1+2a+\frac{1+a}3$
$b=\frac{1+a}3$
$3b=1+a$
$a=3b-1$
$y=\frac{-2+7(3b-1)}3=\frac{-9+21b}3=-3+7b$
$x=\frac{100-11(-3+7b)}7=\frac{133-77b}7=19-11b$
$\begin{matrix}
x\gt 0&\to&19-11b\gt 0&\to&11b\lt 19&\to&b\lt\frac{19}{11}&\to&b\le 1&\\
&&&&&&&&&b=1\\
y\gt 0&\to&-3+7b\gt 0&\to&7b\gt 3&\to&b\gt\frac37&\to&b\ge 1&\\
\end{matrix}$
$1$ is the only integral value of $b$ for which both $x$ and $y$ yield positive values, as required by the problem. So, $x=8$ and $y=4$.
