Question

If 100! is divided by $(72{)}^k$, where $k\in N),$ then the maximum value of $k$ is

Answer 1

Exponent of $2$ in $100!$
$=\left[\frac{100}{2}\right]+\left[\frac{100}{2^2}\right]+\left[\frac{100}{2^3}\right]+.....+\left[\frac{100}{2^6}\right]=50+25+12+6+3+1=97$
So, the exponent of $2^3$ in $100!$ is $32$.

Exponent of $3$ in $100!$
$=\left[\frac{100}{3}\right]+\left[\frac{100}{3^2}\right]+\left[\frac{100}{3^3}\right]+\left[\frac{100}{3^4}\right]=33+11+3+1=48$
So, the exponent of $3^2$ in $100!$ is $24$.

$\Rightarrow$ Exponent of $(72=2^3\times 3^2)$ in $100!$ is min $\{32,24\}=24$
$\Rightarrow$ Exponent of $72$ in $100!$ is $=24$
$\Rightarrow$ maximum value of $k$ is $24$.