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(25/2)+(25/4)+(25/8)+(25/16) = 12+6+3+1 = 22

A
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Bunuel
If 2^z is a factor of 25!, which of the following could be the value of z?

I. 22
II. 23
III. 24

A. I ​ only
B. II only
C. I​ and II
D. I, II, and III
E. None

25/2 = 12
12/6 = 3
3/2 = 1

12+3+1 = 22

So, Answer will be (A) Only I
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1) To paraphrase the question we need to find the total number of prime factors 2 in 25!
2) We need to count all the multiples of 2, add an additional two for every multiple of 2 that is also a multiples of 4, add an additional two for everything multiple of 2 that is also a multiple of 4 and multiple of 8, and add an additional two for every multiple of 2 that is also a multiple of 4, 6, 8, and 24.
3) 25/2=12; 25/4=6; 25/8=3; 25/24=1; 12+6+3+1=22
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25/2 = 12, 25/4 = 6, 25/8 = 3, 25/16 = 1

12+6+3+1 = 22
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