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A is the product of the first 100 multiples of 8.

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A is the product of the first 100 multiples of 8.  [#permalink]

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New post 29 Sep 2018, 19:18
3
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

58% (01:48) correct 43% (02:23) wrong based on 41 sessions

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A is the product of the first 100 multiples of 8. How many zeros would be there at the end of A?

a) 10
b) 11
c) 12
d) 24
e) 25
Spoiler: OA

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Re: A is the product of the first 100 multiples of 8.  [#permalink]

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New post 29 Sep 2018, 19:43
Afc0892 wrote:
A is the product of the first 100 multiples of 8. How many zeros would be there at the end of A?

a) 10
b) 11
c) 12
d) 24
e) 25


So \(8*(8*2)*(8*3)*(8*4).....(8*99)(8*100)=8^{100}*(1*2*3..99*100)=8^{100}*100!\)

So zeroes would depend on numbers of 5s in 100! = \(\frac{100}{5}\frac{100}{5^2}=20+4=24\)

So D
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Re: A is the product of the first 100 multiples of 8.  [#permalink]

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New post 29 Sep 2018, 20:48
this math problem is a common pattern in gmat.

Here, there is a minor error of the question. That is, 0 can be the multiple of 8. I believe that there should be a modification on this question.

The answer is simple itself. A = 100! * 8^100. We look for the number of factors 5 => D is the answer
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Re: A is the product of the first 100 multiples of 8.   [#permalink] 29 Sep 2018, 20:48
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