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  4. The number of positive integers n such that 2n divides n ! is

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Q. The number of positive integers $ n $ such that $ 2^n $ divides $ n ! $ is

AMUAMU 2011Binomial Theorem

Solution:

Let $E = \frac{2^n}{n!}$
When $ n = 1$
$E = \frac{2}{1 !} = 2$
When $ n = 2$
$ E = \frac{2^2}{2 !} = \frac{4}{2} = 2$
When $ n = 3$
$E = \frac{2^3}{3 !} = \frac{4}{3}$
Hence, two values of $n$ exist.