As usual, let $n$ ! denote the product of the integers from 1 to $n$ inclusive. Determine the largest integer $m$ such that $m$ ! divides $100 !+99 !+98 !$.
The answer is $m=98$. Set

$$
N=98 !+99 !+100 !=98 !(1+99+99 \cdot 100)
$$

Hence $N$ is divisible by 98 !. But

$$
\frac{N}{98 !}=1+99 \cdot 101
$$

is not divisible by 99 . Hence $N$ is not divisible by 99 !.