Question

If $(201)!$ is divided by $24^k$ then the largest value of $k$ is

Answer 1

Answer: 65

In $201!$ number of $2^{\prime }s=100+50+25+12+6+3+1$
$=197$
$\therefore$ number of $8^{\prime }s=\left[\frac{197}{3}\right]=65$
number of $3^{\prime }s$ $=67+22+7+2=98$
$\therefore$ number of $24^{\prime }s=65$