How to find the rightmost 25 digits in $100!$? The question is 

Find the rightmost 25 digits in decimal expansion of $100!=1\times 2\times \dotsb \times100$

By counting the number of fives in the prime factorisation of $100!$, I know there are $\lfloor {100 \over 5} \rfloor + \lfloor {100 \over {5^2}} \rfloor =24$ trailing zeros.
But I am really stuck with finding the remaining 25th digit, could someone help me please?

Just figured out this myself.
It is easy to determine that there are 24 trailing zeros, so the only thing left is to determine $\frac{100!}{10^{24}}\pmod {10}$
Writing the fraction as $$\frac{(1 \cdot 3 \cdot 7 \cdot 9 \ \dotsc \cdot 99)(5 \cdot 10 \cdot 15 \cdot \dotsc \cdot 100)(2 \cdot 4 \cdot 6 \cdot 8 \cdot \dotsc \cdot 98) }{10^{24}} \equiv \frac{(1 \cdot 3 \cdot 7 \cdot 9)^{10} \cdot 5^{20} \cdot(1 \cdot 2 \cdot 3 \cdot \dotsc \cdot 20) \cdot 2^{40} \cdot (1 \cdot 2 \cdot 3 \cdot 4)^{10} }{5^{24} \cdot 2^{24}}\equiv \frac{(9)^{10} \cdot(1 \cdot 2 \cdot 3 \cdot \dotsc \cdot 20) \cdot 6 \cdot (4)^{10} }{5^{4}} \equiv 1 \cdot (1 \cdot 2 \cdot 3 \cdot 4)^4 \cdot 6 \cdot 6 \equiv 4$$
So the 25th digit is 4.
