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# If n is the smallest positive integer such that n^3/3920 is

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GMAT Instructor
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If n is the smallest positive integer such that n^3/3920 is [#permalink]

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07 Jul 2006, 08:01
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Question Stats:

100% (04:41) correct 0% (00:00) wrong based on 4 sessions

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If n is the smallest positive integer such that n^3/3920 is also an integer, what is the sum of the digits of n?

(A) 5 (B) 7 (C) 9 (D) 11 (E) 13
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07 Jul 2006, 08:08
is it A?
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Re: PS: Smallest Postive Integer [#permalink]

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07 Jul 2006, 15:22
kevincan wrote:
If n is the smallest positive integer such that n^3/3920 is also an integer, what is the sum of the digits of n?

(A) 5
(B) 7
(C) 9
(D) 11
(E) 13

k = (n x n x n) / (2x2x2x2x5x7x7)
k = (n x n x n) / (4^2 x 5 x 7^2)

n should be = (4 x 5 x 7)
so total = 16

am i missing or the question is incorrect???????

laxi, can you elabroate, pls..
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Re: PS: Smallest Postive Integer [#permalink]

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07 Jul 2006, 15:31
MA wrote:
kevincan wrote:
If n is the smallest positive integer such that n^3/3920 is also an integer, what is the sum of the digits of n?

(A) 5
(B) 7
(C) 9
(D) 11
(E) 13

k = (n x n x n) / (2x2x2x2x5x7x7)
k = (n x n x n) / (4^2 x 5 x 7^2)

n should be = (4 x 5 x 7)
so total = 16

n = 140
total = 1+4 =5
Need to calculate the sum of digits of n.
GMAT Instructor
Joined: 04 Jul 2006
Posts: 1263
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07 Jul 2006, 15:32
Indeed n=4*7*5=140

So sum of digits of n is 1+4+0=5

Great work, though. Just read the question more carefully
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Re: PS: Smallest Postive Integer [#permalink]

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07 Jul 2006, 19:09
sgrover wrote:
MA wrote:
kevincan wrote:
If n is the smallest positive integer such that n^3/3920 is also an integer, what is the sum of the digits of n?

(A) 5
(B) 7
(C) 9
(D) 11
(E) 13

k = (n x n x n) / (2x2x2x2x5x7x7)
k = (n x n x n) / (4^2 x 5 x 7^2)

n should be = (4 x 5 x 7)
so total = 16

n = 140
total = 1+4 =5
Need to calculate the sum of digits of n.

loosing focus.
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Joined: 20 Nov 2005
Posts: 2898
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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07 Jul 2006, 22:34
5 it is.

3920 = 5*4*4*4*7*7

So smallest value for n^3 = 5*5*5*4*4*4*7*7*7
So n = 5*4*7 = 140
Sum of digits = 5
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

07 Jul 2006, 22:34
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