## Discrete Mathematics with Applications 4th Edition

We want the smallest integers $m$ and $w$ such that $\frac{2}{3}m$ and $\frac{3}{5}w$ are also integers, $m\geq100$, and $\frac{2}{3}m=\frac{3}{5}w$. For these conditions to hold, $m$ must be divisible by $3$ and $w$ must be divisible by $5$, so we can let $m=3x$ and $w=5y$ for some integers $x$ and $y$. By substitution, we get $\frac{2}{3}(3x)=2x=3y=\frac{3}{5}(5y)$. This implies that $2x$ is divisible by $3$, so $x$ is divisible by $3$. We therefore want the smallest integer less than or equal to $34$ (i.e., the smallest integer $x$ such that $3x\geq100$) that is also divisible by $3$. That number is $36$, so it must be that $m=3(36)=108$ and $w=5y=5(\frac{2}{3}x)=(5)(\frac{2}{3})(36)=120$.