Question

The number of zeros in $120!$ is

Answer 1

$E_2(120!)=\left[\frac{120}{2}\right]+\left[\frac{120}{4}\right]+\left[\frac{120}{8}\right]+\left[\frac{120}{16}\right]+\left[\frac{120}{32}\right]+\left[\frac{120}{64}\right]+\left[\frac{120}{128}\right]+\cdots$
$=60+30+15+7+3+1+0+\cdots$
$=116$ and
$E_5(120!)=\left[\frac{120}{5}\right]+\left[\frac{120}{25}\right]+\left[\frac{120}{125}\right]$
$=24+4+0=28$
$\therefore$ Number of zeros in $120!=min\{116,28\}=28$