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  • The number of positive integers such that <img alt="\mathrm{2^{n}}" src="https://entrancecorner.oncod

The number of positive integers \mathrm{n} such that \mathrm{2^{n}} divides \mathrm{n!} is

Option: 1

exactly 1


Option: 2

exactly 2


Option: 3

infinite


Option: 4

 none of these


Answers (1)

best_answer

The exponent of 2 in $n$ !  is given by

\mathrm{E=\left[\frac{n}{2}\right]+\left[\frac{n}{2^{2}}\right]+\left[\frac{n}{2^{3}}\right]+\ldots}

where \mathrm{[x]} denotes greatest integer \mathrm{\leq x, As\, [x] \leq x \, \forall \, x, \left[\frac{n}{2^{m}}\right]=0} after finite number of terms. Thus we get
\mathrm{E<\frac{n}{2}+\frac{n}{2^{2}}+\frac{n}{2^{3}}+\ldots=\frac{n / 2}{1-1 / 2}=n}
Thus, there is no positive integer for which \mathrm{2^{n}} divides \mathrm{n!}

Posted by

Pankaj

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