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

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loveparis
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If m and n are positive integer, and 1800m = n3, what is [#permalink] New post  Updated on: 21 Aug 2013, 09:38
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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
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Originally posted by loveparis on 09 Feb 2011, 17:32.
Last edited by Narenn on 21 Aug 2013, 09:38, 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, 17:57
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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] New post  09 Feb 2011, 17:55
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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  18 Jun 2014, 21:04
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Hi
This is one of the questions in which plugin the answers will make it much easier.

2.3.15.30.45

a) start with 2
1800 *2 = 3600 not a cube
b) 3
5400 not a cube
c) 1800 * 15 = 27000 = 30 to the power of 3

d) and e) are higher, so no need to go ahead further.
ans : C
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Re: If m and n are positive integer, and 1800m = n3, what is [#permalink] New post  16 Mar 2016, 01:39
Here N must contain at-least one 3 and one 5 for m to be an integer
hence C
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Re: If m and n are positive integer, and 1800m = n3, what is [#permalink] New post  16 Mar 2016, 02:04
loveparis wrote:
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

1800*m is a cube number'
The first cube number is 27000 when m = 15
Tip: Prime factor 1800:2^3*3^2*5^2; we require one more 3 and one more 5.
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Re: If m and n are positive integer, and 1800m = n3, what is [#permalink] New post  02 May 2017, 16:44
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loveparis wrote:
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


Since 1800m = n^3, we can say that the product of 1800 and some integer m is equal to a perfect cube.
We must remember that all perfect cubes break down to unique prime factors, each of which has an exponent that is a multiple of 3. So, let’s break down 1800 into primes to help determine what extra prime factors we need to make 1800m a perfect cube.

1800 = 18 x 100 = 2 x 9 x 10 x 10 = 2 x 3 x 3 x 2 x 5 x 2 x 5 = 2^3 x 3^2 x 5^2

In order to make 1800n a perfect cube, we need one more 3 and one more 5. Thus, the least value of m is 3 x 5 = 15.

Answer: C
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Re: If m and n are positive integer, and 1800m = n3, what is [#permalink] New post  24 Jan 2020, 22:47
Expert Reply
loveparis wrote:
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



\(1800m=2^3*3^2*5^2*m=n^3\)

\(n^3\) represents a perfect cube and for a number to be a perfect cube the exponents (powers) of all distinct prime factors of number must be multiples of 3

therefore, we need minimum one 3 and one 5 to make their powers multiples of 3

the least value of \(m\) for which \(2^3*3^2*5^2*m\) is a perfect cube is 3*5=15

for m = 15, \(2^3*3^2*5^2*m=2^3*3^3*5^3=(2*3*5)^3=n^3\)

Answer: Option C
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Re: If m and n are positive integer, and 1800m = n3, what is [#permalink] New post  14 Nov 2020, 18:11
If m and n are positive integer, and 1800m=n^3 what is the least possible value of m?

Prime Factorization of 1800 = \(2^3\)*\(3^2\)*\(5^2\)

\(2^3\)*\(3^2\)*\(5^2\) * M = \(N^3\)

To have M equal to cube power of N, we need to have 3 and 5 in cube power, so that we have \((2*3*5)^3\)

\(3^1\)*\(5^1\) = M

M = 15

Ans: C
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