jmc

There are $\lfloor \frac{46}{3}\rfloor =15$ numbers divisible by $3$, $\lfloor \frac{46}{5}\rfloor =9$ numbers divisible by $5$, so at first we have $15+9=24$ numbers that are divisible by $3$ or $5$, except we counted the multiples of $\text{LCM}(3,5)=15$ twice, once for $3$ and once for $5$. There are $\lfloor \frac{46}{15}\rfloor =3$ numbers divisible by $15$, so there are $24-3=21$ numbers divisible by $3$ or $5$. $\to \boxed{21}$