Counting with at least one

Sets.
- $|A_2|=\lfloor100/2\rfloor=50$
- $|A_3|=\lfloor100/3\rfloor=33$
- $|A_5|=\lfloor100/5\rfloor=20$
- $|A_2\cap A_3|=|A_6|=16$
- $|A_2\cap A_5|=|A_{10}|=10$
- $|A_3\cap A_5|=|A_{15}|=6$
- $|A_2\cap A_3\cap A_5|=|A_{30}|=3$

Union. $50+33+20-16-10-6+3=74$.

Conclusion. $74$ numbers.