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The number of positive integers $n$ such that $2^{n}$ di vides n! is

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Correct option is D

The exponent of 2 in n! is given byE=n2+n22+n23+…where [x] denotes greatest integer ≤x, As [x]≤x ∀xn2m=0 after finite number of terms. Thus we getE

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The number of positive integers n such that 2n di vides n! is

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The number of positive integers $n$ such that $2^{n}$ di vides n! is