Problem 227

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: $(0; 15)$, $(4; 12)$, $(8; 9)$, $(12; 6)$, $(16; 3)$, $(20; 0)$.