If m and n are positive integer, and 1800m = n^3, what is

Question

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

Bunuel · Accepted Answer

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.

geturdream · Answer

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

GMatAspirerCA · Answer

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

stonecold · Answer

Here N must contain at-least one 3 and one 5 for m to be an integer 
hence C

Mathivanan Palraj · Answer

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.

JeffTargetTestPrep · Answer

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

GMATinsight · Answer

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

akadiyan · Answer

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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If m and n are positive integer, and 1800m = n^3, what is

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If m and n are positive integer, and 1800m = n^3, what is

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