Question

Let $p$ be a prime number and $n$ be a positive integer, then exponent of $p$ is $n!$ is denoted by $E_p(n!)$ and is given by
$E_p(n!)=\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\left[\frac{n}{p^3}\right]+\dots .+\left[\frac{n}{p^k}\right]$
where $p^k<n<p^{k+1}$
and $[x]$ denotes the integral part of $x$.
If we isolate the power of each prime contained in any number $N$, then $N$ can be written as
where ${\alpha }_i$ are whole numbers.
The number of zeros at the end of $108!$ is -

• A. 10
• B. 13
• C. 25
• D. 26

Answer 1

Answer: (C)

Product of 5's & 2's constitute 0's at the end of a number
$\Rightarrow$ No. of 0 's in $108!$ = exponent of 5 in $108!$
(Note that exponent of 2 will be more than exponent of 5 in $108!)$
$\Rightarrow \left[\frac{108}{5}\right]+\left[\frac{21}{5}\right]=21+4=25$