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# If n is a positive integer and 10^n is a factor of m, what

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If n is a positive integer and 10^n is a factor of m, what [#permalink]

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Updated on: 26 May 2013, 09:16
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95% (hard)

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48% (01:36) correct 52% (01:27) wrong based on 337 sessions

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If n is a positive integer and 10^n is a factor of m, what is the value of n?

(1) m is the product of the first positive 40 numbers .
(2) n > 8*m/40!

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Originally posted by atalpanditgmat on 26 May 2013, 08:56.
Last edited by Bunuel on 26 May 2013, 09:16, edited 1 time in total.
Edited the question.
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Re: If n is a positive integer and 10n is a factor of m, what is [#permalink]

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26 May 2013, 09:18
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If n is a positive integer and 10^n is a factor of m, what is the value of n?

Given that $$m=10^n*k$$ for some positive integer k.

(1) m is the product of the first positive 40 numbers --> $$m=40!$$ --> $$40!=10^n*k$$. The number of trailing zeros of 40! is 40/5+40/25=8+1=9 (check here: everything-about-factorials-on-the-gmat-85592.html). So, 40! ends with 9 zeros, which means that n could be from 1 to 9, inclusive. Not sufficient.

(2) n > 8*m/40!. Clearly not sufficient.

(1)+(2) From (1) $$m=40!$$, thus from (2) $$n>\frac{8*40!}{40!}=8$$. Since from (1) we also have that $$1\leq{n}\leq{9}$$, then n=9. Sufficient.

Similar question to practice: if-d-is-a-positive-integer-and-f-is-the-product-of-the-first-126692.html

Hope it helps.
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Re: If n is a positive integer and 10^n is a factor of m, what [#permalink]

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19 Nov 2015, 20:49
oh man..completely forgot about trailing zeroes..
did the long way to see how many numbers of factors have 5 and 2...
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Re: If n is a positive integer and 10^n is a factor of m, what [#permalink]

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05 Apr 2018, 09:00
Bunuel wrote:
If n is a positive integer and 10^n is a factor of m, what is the value of n?

Given that $$m=10^n*k$$ for some positive integer k.

(1) m is the product of the first positive 40 numbers --> $$m=40!$$ --> $$40!=10^n*k$$. The number of trailing zeros of 40! is 40/5+40/25=8+1=9 (check here: http://gmatclub.com/forum/everything-ab ... 85592.html). So, 40! ends with 9 zeros, which means that n could be from 1 to 9, inclusive. Not sufficient.

(2) n > 8*m/40!. Clearly not sufficient.

(1)+(2) From (1) $$m=40!$$, thus from (2) $$n>\frac{8*40!}{40!}=8$$. Since from (1) we also have that $$1\leq{n}\leq{9}$$, then n=9. Sufficient.

Similar question to practice: http://gmatclub.com/forum/if-d-is-a-pos ... 26692.html

Hope it helps.

Awesome explanation Bunuel, DS is always painful for me.
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Re: If n is a positive integer and 10^n is a factor of m, what   [#permalink] 05 Apr 2018, 09:00
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