How many numbers are divisible by 2, 5, 9, and 13 between 100 and 100,000? How to implement this question in discrete math ?

Since the numbers 2, 5, 9, and 13 are pairwise relatively prime, their the least common multiple is equal to $2\cdot 5\cdot 9\cdot 13=1170,$ and therefore, the number $n\in\mathbb N$ is divisible by 2, 5, 9, and 13 if and only if $n$ is divisible by $1,170.$

The floor function $\lfloor x \rfloor$ is defined to be the greatest integer less than or equal to the real number $x$.

In discrete math is well-known the fact that the number of numbers that do not exceed $n\in \mathbb N$ and are divisible by $d\in\mathbb N$ is equal to $\lfloor \frac{n}{d} \rfloor$. Since each number $n\le1000$ is not divisible by 1170, between 1000 and 100,000 there are $\lfloor \frac{100,000}{1,170} \rfloor=85$ numbers which are divisible by 2, 5, 9, and 13.

Answer: 85