Question

Find the number of zeros at the end of $100!$

Answer 1

In terms of prime factors $100!$ Can be written as $2^a{3}^b{5}^c{7}^d\dots$

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 $E_5(100!)=\left[\frac{100}{5}\right]+\left[\frac{100}{5^2}\right]$

$=20+4=24$

$\therefore 100!=2^{97}\times 3^b\times 5^{24}\times 7^d\times ...$

$=2^{73}\times {(2\times 5)}^{24}\times 7^d\times ...$

$={10}^{24}\times 2^{73}\times 3^b\times 7^d\times ....$

Thus, the number of zeros at the end of $100!$ is $24$.