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

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atalpanditgmat
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If n is a positive integer and 10^n is a factor of m, what [#permalink] New post   Updated on: 26 May 2013, 09:16
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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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C

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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Bunuel
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Re: If n is a positive integer and 10n is a factor of m, what is [#permalink] New post   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.

Answer: C.

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] New post   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] New post   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: https://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.

Answer: C.

Similar question to practice: https://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] New post   06 Sep 2020, 09:42
What exactly is the second statement telling us? I get that combining is the key, but I dont understand the standalone value of statement 2
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Re: If n is a positive integer and 10^n is a factor of m, what [#permalink] New post   03 Jan 2021, 10:04
testtakerstrategy wrote:
What exactly is the second statement telling us? I get that combining is the key, but I dont understand the standalone value of statement 2


We can only say that it is a "B Trap", honeytrap, geddit? ;)
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Re: If n is a positive integer and 10^n is a factor of m, what [#permalink] New post   16 Jan 2022, 08:25
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Re: If n is a positive integer and 10^n is a factor of m, what [#permalink]
16 Jan 2022, 08:25
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