If 33! is divisible by 2^n,where n in N,then | vedclass

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(C) To find the highest power of a prime $p$ dividing $n!$,we use Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor$.Fo

Vedclass · Accepted Answer

$496$

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(C) To find the highest power of a prime $p$ dividing $n!$,we use Legendre's formula: $E_p(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} floor$.For $n=33$ and $p=2$,the exponent of $2$ in $33!$ is:$E_2(33!) = \lfloor \frac{33}{2} floor + \lfloor \frac{33}{4} floor + \lfloor \frac{33}{8} floor + \lfloor \frac{33}{16} floor + \lfloor \frac{33}{32} floor$$E_2(33!) = 16 + 8 + 4 + 2 + 1 = 31$.Thus,$33!$ is divisible by $2^n$ for all $n \in \{1, 2, 3, \ldots, 31\}$.The sum of all possible values of $n$ is the sum of the first $31$ natural numbers:$S = \frac{31(31+1)}{2} = \frac{31 	imes 32}{2} = 31 	imes 16 = 496$.

If $33!$ is divisible by $2^n$,where $n \in N$,then the sum of all possible values of $n$ is:

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