Answer to Question #147052 in Discrete Mathematics for Bella Adam

Question

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

Answer 1

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