South African Mathematics Olympiad

6

We can re-write $\frac{x}{20}+\frac{y}{15}=1$ in the form $3x+4y=60$. Then we can see that $3x$ must be divisible by $4$, so $x$ must be, and trying successive possible values we see that only the following combinations of $x$- and $y$-values are acceptable: $(0;15)$, $(4;12)$, $(8;9)$, $(12;6)$, $(16;3)$, $(20;0)$.