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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Jeffery Miller

Head of GMAT Instruction

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