Question

Let $p$ be a prime number & $n$ be a positive integer, then exponent of prime $p$ in $n!$ is denoted by $E_p(n!)$ & is given by $E_p(n!)=\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\left[\frac{n}{p^3}\right]+.....+\left[\frac{n}{p^x}\right]$ where $x$ is the largest positive integer such that $p^x\le n<p^{x+1}$ and $[\cdot ]$ denotes the greatest integer

Again every natural number $N$ can be expressed as the product of its prime factors given by $N=P_1^{k_2}{P}_2^{k_2}....P_r^{k_r}$ where $P_1,P_2,P_3,.......P_r$ are prime numbers & $k_1$ are whole numbers.

The greatest integer $n$ for which $77!$ is divisible by $3^n$ is

• A. $27$
• B. $32$
• C. $35$
• D. $40$

Answer 1

The correct option is C $35$

$3^3\le 77\le 3^4$

Hence, $\left[\frac{77}{3}\right]+\left[\frac{77}{3^2}\right]+\left[\frac{77}{3^3}\right]$

$=25+8+2$

$=33+2$

$=35$

Therefore, the exponent of $3$ is $3^{35}$ in $77!$

Hence, $n=35$