Question

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

• A. 36
• B. 18
• C. 24
• D. None of these

Answer 1

Answer: (C)

In terms of prime factors $100!$ can be written as $2^a\cdot 3^b\cdot 5^c\cdot 7^d\cdots$
Now, $E_2(100!)$
$=\left[\frac{100}{2}\right]+\left[\frac{100}{2^2}\right]+\left[\frac{100}{2^3}\right]+\left[\frac{100}{2^4}\right]+\left[\frac{100}{2^5}\right]+\left[\frac{100}{2^6}\right]$
$=50+25+12+6+3+1=97$
and, $E_5(100!)=\left[\frac{100}{5}\right]+\left[\frac{100}{5^2}\right]$
$=20+4=24$
$100!=2^{97}\cdot 3^b\cdot 5^{24}\cdot 7^d\dots$
$=2^{73}\cdot 3^b\cdot (2\times 5{)}^{24}\cdot 7^d\dots$
$=2^{73}\cdot 3^b\cdot (10{)}^{24}\cdot 7^d\dots$
Hence, number of zeros at the end of $100!$ is $24.$