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If m and n are positive integer, and 1800m = n3, what is

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If m and n are positive integer, and 1800m = n3, what is [#permalink] New post 09 Feb 2011, 16:32
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If m and n are positive integer, and 1800m = n^3, what is the least possible value of m?
(A) 2
(B) 3
(C) 15
(D) 30
(E) 45
[Reveal] Spoiler: OA

Last edited by Narenn on 21 Aug 2013, 08:38, edited 1 time in total.
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Re: If m and n are positive integer, and 1800m = n3, what is [#permalink] New post 09 Feb 2011, 16:55
1800*m=n^3
m=n^3/1800
=n^3/(3^2*5^2*2^3)

so n has to be 3*5*2 so that n^3 is divisible by 1800.

So m= (2*3*5)^3/(3^2*5^2*2^3)
m=15
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Re: If m and n are positive integer, and 1800m = n3, what is [#permalink] New post 09 Feb 2011, 16:57
Expert's post
loveparis wrote:
89. If m and n are positive integer, and 1800m = n3, what is the least possible value of m?
(A) 2
(B) 3
(C) 15
(D) 30
(E) 45


1800m=2^3*3^2*5^2*m=n^3 the least value of m for which 2^3*3^2*5^2*m is a perfect cube is 3*5=15 (m must complete the powers of 3 and 5 to cubes): in this case 2^3*3^2*5^2*m=2^3*3^3*5^3=(2*3*5)^3=n^3

Answer: C.
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Re: If m and n are positive integer, and 1800m = n3, what is   [#permalink] 09 Feb 2011, 16:57
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If m and n are positive integer, and 1800m = n3, what is

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