The largest value of n, for which 40^n divide | vedclass

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(D) We need to find the largest $n$ such that $40^n$ divides $60!$. $40^n = (2^3 	imes 5)^n = 2^{3n} 	imes 5^n$. Using Legendre's formula, the expon

Vedclass · Accepted Answer

$14$

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(D) We need to find the largest $n$ such that $40^n$ divides $60!$. $40^n = (2^3 	imes 5)^n = 2^{3n} 	imes 5^n$. Using Legendre's formula, the exponent of a prime $p$ in $m!$ is $E_p(m!) = \sum_{k=1}^{\infty} \lfloor \frac{m}{p^k} floor$. For $p=2$: $E_2(60!) = \lfloor \frac{60}{2} floor + \lfloor \frac{60}{4} floor + \lfloor \frac{60}{8} floor + \lfloor \frac{60}{16} floor + \lfloor \frac{60}{32} floor = 30 + 15 + 7 + 3 + 1 = 56$. For $p=5$: $E_5(60!) = \lfloor \frac{60}{5} floor + \lfloor \frac{60}{25} floor = 12 + 2 = 14$. We require $3n \le 56$ and $n \le 14$. From $3n \le 56$, we get $n \le \lfloor \frac{56}{3} floor = 18$. From $n \le 14$, the limiting value is $n = 14$.

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

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