Here there are many possibilities. 

- $A_{2}=\left\lfloor\frac{100}{2}\right\rfloor=50$ numbers less than 100 divisible by 2.

- $A_{3}=\left\lfloor\frac{100}{3}\right\rfloor=33$ numbers less than 100 divisible by 3

- $A_{5}=\left\lfloor\frac{100}{5}\right\rfloor=20$ numbers less than 100 divisible by 5

- $A_{2,3}=\left\lfloor\frac{100}{6}\right\rfloor=16$ numbers less than 100 divisible by 2 and by 3

- $A_{2,5}=\left\lfloor\frac{100}{10}\right\rfloor=10$ numbers less than 100 divisible by 2 and by 5.

- $A_{3,5}=\left\lfloor\frac{100}{15}\right\rfloor=6$ numbers less than 100 divisible by 3 and by 5

- $A_{2,3,5}=\left\lfloor\frac{100}{30}\right\rfloor=3$ numbers less than 100 divisible by 2, 3 and 5.

A first attempt would be to take $A_{2}+A_{3}+A_{5}-A_{2,3}-A_{2,5}-A_{3,5}$. But then we would realize that the numbers divisible by 2, 3, and 5 are not counted. The correct formula is then $A_{2}+A_{3}+A_{5}-A_{2,3}-A_{2,5}-A_{3,5}+A_{2,3,5}=50+33+20-16-10-6+3=74$.

## “Product”
How many numbers less than 100 are divisible by 2, 3, or 5?