<s>[INST] 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?
 [/INST] 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.
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