Question

What is the smallest natural number n such that n! is divisible by 990

Answer 1

Factors of `990 = 11**3^2**2**5`
Smallest number `n` such that `n!` is divisible by `11` is `11`.
`11!` is also divisible by `9,2 and 5`.
So, `11!` will have all these factors `11,9,5 and 2`.
So, smallest number that is divisible by `990` is `11!`.