Problem 232

Solution:

From $3|x|+5|y|=100$, we have $3|x|=100-5|y|=5(20-|y|)$. This implies that $20-|y|$ must be divisible by $3$ and it must be nonnegative. Thus, $y= \pm 2, \pm 5, \pm 8, \pm 11, \pm 14, \pm 17, \pm 20$. Since each possible value of $y$ has two possible $x$ values except for $y= \pm 20$, then the number of ordered pairs is given by $12 \times 2+2=26$.