Large Factorials

Firstly, lets find the largest value of ${k}$ such that $\ 10^{k} | 100!$ . Now 10 = 2 $\times$ 5 and 2 appears far more often than 5, so we just need to find the highest power of 5 that divides 100! (I believe this is known as a one-to-one correspondence). Now $Max{k} = \displaystyle\sum{i=1}^{\infty} \lfloor \frac{100}{5^i} \rfloor = 20 + 4 = 24$ $\Rightarrow\ 10^{24} | 100! \rightarrow\therefore\ 100^{12} | 100!$

We know that for any number $N$ which is a multiple of 10, when raised to any number, let's say $n$ , the resulting number ends with zeroes. For $N!$ , when $( 5\leq N\leq\infty )$ , the resulting numbers end with trailing zeroes. So for $100!$ , there are 24 trailing zeroes which has the same for $100^{12}$ .

100 factorial has 24 trailing zeroes it is given that 100^n is factor of 100 factorial so n should be equal to 12

100! will have 24 zeros. So $100^n$ should have 12 zeros. So n is 12