101! has more factors of 2 than 5, so its rightmost nonzero digit is one of $2,4,6,8$. Notice that if the rightmost nonzero digit of 101 ! is $2 k(1 \leq k \leq 4)$, then 102 ! has rightmost nonzero digit $102(2 k) \equiv 4 k(\bmod 10)$, and 103 ! has rightmost nonzero digit $103(4 k) \equiv 2 k(\bmod 10)$. Hence $n=103$.
The rightmost nonzero digit in the decimal expansion of 101 ! is the same as the rightmost nonzero digit of $n$ !, where $n$ is an integer greater than 101. Find the smallest possible value of $n$.