Question

Which among the following options are correct for $N$= $50!$

• A. Exponent of $2$ in $N$ is $47$
• B. Exponent of $3$ in $N$ is $22$
• C. Exponent of $24$ in $N$ is $15$
• D. Number of zeroes in the end of $N$will be $12$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Exponent of $2$ in $N$ is $47$
B Exponent of $3$ in $N$ is $22$
C Exponent of $24$ in $N$ is $15$
D Number of zeroes in the end of $N$will be $12$

Exponent of $2$ in $50!$ is $=\left[\frac{50}{2}\right]+\left[\frac{50}{2^2}\right]+......+\left[\frac{50}{2^5}\right]$
$=25+12+6+3+1=47$

Exponent of $3$ in $50!$ is $=\left[\frac{50}{3}\right]+\left[\frac{50}{3^2}\right]+\left[\frac{50}{3^3}\right]$
$=16+5+1=22$

Exponent of $2$ in $50!$ is $47$
$\therefore$ Exponent of $2^3$ in $50!$ is $15$
So, exponent of $24=2^3\times 3$ in $50!$ is minimum of exponent of $\{2^3,3\}$ $=15$

Exponent of $5$ in $50!$ is $=\left[\frac{50}{5}\right]+\left[\frac{50}{5^2}\right]$
$=10+2=12$

Now, number of zeroes in the end of $N$ is equal to the exponent of $10$ in $N$

$\therefore$ Exponent of $10=2\times 5$ in $50!$ is minimum of exponent of $\{2,5\}$ $=12$