A number theory problem by Prithwish Roy

Though there are many formal methods yet, I think this can be one of the solutions.

We notice that $100!$ has 24 trailing zeroes and hence the last non zero digit is the answer.

We notice that in $100!$ , the last non zero term of $10!$ is repeated again and again in the last non zero place.
I mean to say that the last non zero place in $11 \times 12 \times 13\times ......\times 20$ is same as that of the last non zero place in $1 \times 2 \times 3\times ......\times 10$ i.e. $10!$

Hence this will be repeated 10 times from 1 - 100 and hence the last digit will be the Last non zero digit in $(10!)^{10}$

Thus, $10! = 3628800$ i.e the last non zero digit is 8
Hence last digit of $8^{10} = 2 ^{30} = 2^{2} = 4$

The answer is $\boxed{4}$

Friends if you don't like this you can find a more generalised and better method at (http://www.campusgate.co.in/2013/10/finding-right-most-non-zero-digit-of.html)

Number of Digit at end of 100! will be 24 by using [100/5] +[100/25] .. now it will be canceled by denominator Now it is only depend on first non zero digit of 100! . 100! =10^20 . 20! . 2 ^20 now first non zero depend only of 20! .2^20 by observation last digit will be 4.