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 exponent of $12$ in $100!$ is -

• A. 32
• B. 48
• C. 97
• D. none of these

Answer 1

Answer: (B)

As $12=2^2.3$, here we have to calculate exponent of $2$ and exponent of $3$ in $100!$ exponent of $2$
$=\left[\frac{100}{2}\right]+\left[\frac{50}{2}\right]+\left[\frac{25}{2}\right]+\left[\frac{12}{2}\right]+\left[\frac{6}{2}\right]+\left[\frac{3}{2}\right]=97$
exponent of $3=\left[\frac{100}{3}\right]+\left[\frac{33}{3}\right]+\left[\frac{11}{3}\right]+\left[\frac{3}{3}\right]=48$
Now, $12=223$
we require two $2^{\prime }s$ & one $3$
$\therefore$ exponent of $3$ will give us the exponent of $12$ in $100!$ i.e. $48$