Question

The number of zeros at the end of $100!$ is ?

• A. $20$
• B. $22$
• C. $24$
• D. $26$

Answer 1

The correct option is C

$24$

Simplify the given factorial

Given, $100!$

To get a zero at the end a number must be multiplied with $10$

Therefore we need the number of times product of $2\times 5$ occurs to find the number of zeroes.

Calculate the powers of $2$ in $100!$

The power of $2$ is sum of $\left[\frac{100}{2}\right]=50$, $\left[\frac{50}{2}\right]=25$, $\left[\frac{25}{2}\right]=12$, $\left[\frac{12}{2}\right]=6$, $\left[\frac{6}{2}\right]=3$, $\left[\frac{3}{2}\right]=1$, $\left[\frac{1}{2}\right]=0$,

where $\left[\right]$is the Greatest integer function.

The power of $2=50+25+12+6+3+1=97$

Similarly, the power of $5$ is sum of $\left[\frac{100}{5}\right]=20,\left[\frac{20}{5}\right]=4,\left[\frac{4}{5}\right]=0$

The power of $5=20+4=24$

The limiting value to decide the number of zeroes is power of $5$ as it lessens than the power of $2$.

Therefore, the number of $10$ formed $=24$

there are $24$ zeroes at the end of $100!$, so,

Hence option $\left(C\right)$ is the correct .