The $25$th digit of $100!$ I want to find The $25$th digit of $100!$.
My attempt:It is easy to know it has $24$ zeroes.Because:
$\lfloor {\frac{100}{5}} \rfloor+\lfloor {\frac{100}{25}} \rfloor =24$
By getting the fist digits(after deleting all $5$ factors and $24$,$2$ factors)and multiplying them to each other we get the answer $4$ but I want an easier way.

If we exclude the numbers that are divisible by 5, we see a cycle that repeats.
$4! = 24\\
9!/6!\equiv 24 \mod 100$
$100! = \frac {100!}{5^{20} 20!} (5^{20} 20!) = \frac {100!}{5^{20} 20!}\frac {20!}{5^4 4!} (4!)(5^{24})$
$(24^{25})(5^{24}) = (12^{25})(10^{24})(2)$
$12^{25} \equiv 12^5 \mod 100\equiv 32 \mod 100$
The last $2$ non-zero digits of 100! are $64.$
