Assume $x,y\in\mathbb{Z}^+$ and $25x+9y=2014$. What is the minimum value of $x+y$ ?
To solve this problem I noticed that we are looking for large values of $x$ since it has coefficient $25$ so I started from $x=80,79,78,\ldots$ After trying these values I found out $x=73$ works and $y$ should be $21$. Hence $\min(x+y)=94$.
But is it possible to solve this problem without try and error? From $25x+9y=2014$ we can write $25\vert2014-9y\quad$ or $\quad9\vert 2014-25x$. But I don't know how to continue.
The system
$$25x \equiv 2014 \mod9 \equiv 6 \mod 9 $$
$$9y \equiv 2014 \mod25 \equiv 14 \mod 25$$
has a unique solution modulo $225$ as $(25,9)=1$. Now observe that
$$25(4) \equiv 1 \mod 9$$
$$9(14) \equiv 1 \mod 25$$
Thus
$$25(6) \equiv 6 \mod 9$$
$$9(21) \equiv 14 \mod 25$$
when we multiply both sides of each equation by $2014$.
Now a solution to the above system is $(x,y) = (73,21)$ thus all solutions are of the form $(73+9t, 21 -25t)$ where $t \in \mathbb{Z}$ if $(x,y) \in (\mathbb{Z}^{+})^2$ then $-8\leq t \leq 0$. The sum $x+y$ is $94-16t$ which is minimized at $t =0$. Thus the minimum is $94$.