The number of positive integers which divide (25)! are -

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Question

The number of positive integers which divide $(2^{5})!$ are -

Options:

A .

2^{13} {.3}^{3} {.5}^{2}

B .

2^{8} {.3}^{2} {.5}^{2}

C .

2^{11} {.3}^{2} .5

D .

2^{8} {.3}^{3} {.5}^{3}

Answer: Option A
:
A

$32! = 2^{31} {.3}^{14} . 5^{7} {.7}^{4} {.11}^{2} {.13}^{2} .17 .19 .23 .29 .31$ (on prime factorization)

So it has $(31 + 1) (14 + 1) (7 + 1) (4 + 1) (2 + 1) (2 + 1) (1 + 1) (1 + 1) (1 + 1) (1 + 1) (1 + 1) = 2^{13} . 3^{3} {.5}^{2}$ factors.

Hence, choice (a) is the right answer.

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