Question

Find the number of zeroes at the end of $100!$?

Answer 1

In terms of prime factors $100!$ can be written as $2^a{.3}^b{.5}^c{.7}^d....$
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 $E_5(100!)=\left[\frac{100}{5}\right]+\left[\frac{100}{5^2}\right]$
$=20+4=24$
$\therefore 100!=2^{97}{.3}^b{.5}^{24}{.7}^d...$
$=2^{73}{.3}^b.(2\times 5{)}^{24}{.7}^d...$
$=2^{73}{.3}^b.(10{)}^{24}{.7}^d...$
Hence number of zeroes at the end of $100!$ is $24$.
Or
Exponent of $10$ in $100!$ $=$ min$(97,24)$ $=$ $24$