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Updated on: 26 May 2013, 09:16
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.
Last edited by Bunuel on 26 May 2013, 09:16, edited 1 time in total.
Edited the question.
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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!
(1) m is the product of the first positive 40 numbers .
(2) n > 8*m/40!
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.
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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mvictor
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19 Nov 2015, 20:49
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oh man..completely forgot about trailing zeroes..
did the long way to see how many numbers of factors have 5 and 2...
did the long way to see how many numbers of factors have 5 and 2...
