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?
 A: Like To find the right most non zero digit, 
For $\prod_{r=1}^{10}(10a+r)\equiv\prod_{r=1}^{10}r\pmod5$
Now if we exclude the multiples of $5,$ which have been consumed in forming $24$ trailing zeros
$1\cdot2\cdot3\cdot4\cdot6\cdot7\cdot8\cdot9\cdot2\equiv2\pmod5$
$\prod_{a=0}^9\left(\prod_{r=1}^{10}(10a+r)\right)\equiv\left(\prod_{r=1}^{10}r\right)^{10}\pmod5\equiv2^{10}\equiv2^2\equiv4$
Now $2^{24}$ have been consumed in forming $24$ trailing zeros
As $2^4\equiv1\pmod5,2^{24}\equiv1^6\equiv1$
So, the last non-zero digit in $100!$ will be $4\cdot1^{-1}\equiv4\pmod5\  \ \  \  (1)$
and it is already $0\pmod2\  \ \  \  (2)$
Using Chinese Remainder Theorem on $(1),(2)$ 
or by observation as $x\equiv4\pmod5\implies x\equiv4,4+5\pmod{}10,$
the last non-zero digit in $100!$ will be $4\pmod{10}$
A: 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.
