jmc

First, we find $150=2\cdot 3\cdot 5^2.$ The prime factorization of a divisor of 150 must take the form $2^m{3}^n{5}^p$ for nonnegative integers $m\le 1$, $n\le 1$, and $p\le 2$. The requirement that the divisor is not divisible by 5 means that we must have $p=0$. Therefore, there are $2$ possibilities for $m$ (namely, 0 or 1) and the same for $n$ for a total of $2\cdot 2=\boxed{4}$ such factors.