Let the number of numbers divisible by 2 be denoted by $n(2);$ Numbers divisible by $3$ by $n(3);$ and the number of numbers divisible by $5$ by $n(5).$

Numbers divisible by 4 will be fully divisible by 2, so no need to consider those numbers separately.

$n(2) = 51$ $\left(\lfloor \frac{100}{2} \rfloor +1\right)$ [300 also included]