The largest value of n, for which 40^n divides 60!, is

Question

The largest value of $n$, for which $40^n$ divides $60!$, is

Accepted Answer

To find largest value of $n$ for which 60 ! is divisible by ${40}^{n}$

This means we need to find exponent of 40 in 60 !

$40={2}^{3}	imes 5$

Key Concept: formula for exponent of prime $p$ in $k!$ is given as $[\frac{k}{p}]+[\frac{k}{{p}^{2}}]+[\frac{k}{{p}^{3}}]+\ldots$ where [] is greatest integer function.

so, exponent of 2 in 60 !

$\begin{gathered} [\frac{60}{2}]+[\frac{60}{{2}^{2}}]+[\frac{60}{{2}^{3}}]+[\frac{60}{{2}^{4}}]+[\frac{60}{{2}^{5}}]+[\frac{60}{{2}^{6}}]+\ldots \ 30+15+7+3+1=56- \end{gathered}$

exponent of 5 in 60 !

$\begin{gathered} [\frac{60}{5}]+[\frac{60}{{5}^{2}}]+[\frac{60}{{5}^{3}}]+\ldots \ 12+2+0\cdots =14-(2) \end{gathered}$

Combining (1) and (2)

60 ! is divisible by ${2}^{56}	imes {5}^{14}	imes N$, where $N$ is natural number.

60 ! is divisible by $4	imes {({2}^{3})}^{18}	imes {5}^{14}	imes N$

60 ! is divisible by $4	imes {({2}^{3})}^{4}	imes {({2}^{3})}^{14}	imes {5}^{14}	imes N$

$60!$ is divisible by ${2}^{14}	imes {40}^{14}	imes N$

So largest value of $n$ is 14

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