How many pairs of positive integers $(x,y)$ satisfy $3x+5y=100$?

Rearranged, $3x=100-5y$, so $100-5y$ must be a positive multiple of 3. Testing $y=1$, 2, 3, … the values that work are $y=2$, 5, 8, 11, 14 and 17, giving $x=30$, 25, 20, 15, 10 and 5 — six pairs, after which $y=20$ forces $x=0$. The trap is 7: $x=0$ is not a positive integer, so that last pair must be thrown out.