Set J is comprised of all positive integers x such that x^3 is a multi

Question

Set J is comprised of all positive integers x such that x^3 is a multiple of both 72 and 216. Which of the following integers are factors of every member of set J?

I. 6 
II. 9 
III. 12

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

LeoN88 · Answer

x^3 = k1(2^3 * 3^2) = k2( 2^3 * 3^3)

so, x will have at least one 2 and one 3.

IanStewart · Answer

If x = 6, then x^3 = 6^3 = 216 is divisible by 216 (and automatically by 72 too, because 72 is a factor of 216). So the number 6 is in set J, and clearly then we have numbers in set J which are not divisible by 9 or by 12. So the only possible answer is I only.

CrackverbalGMAT · Answer

Since x is a positive integer,  x^3  has to be a perfect cube. 216 is a multiple of 72. So, if  x^3  is a multiple of both 72 and 216, it invariably means it’s a multiple of 216.

In general, we can say,
  x^3  = 216k, where k is a positive integer.

When we prime factorise 216, we can write it as, 216 =  2^3  *  3^3 . Therefore,

x^3  =  2^3  *  3^3  * k.

Clearly, 6 has to be a factor of any number  x^3 , regardless of the value of k. So, 6 also has to be the factor of every member of set J.

9 and 12 may be factors of some elements of the set, which depends on the value of k.

For example, if k =  2^3  *  3^3 , then  x^3  =  2^6 *  3^6 , in which case, x =  2^2  *  3^2 . For this value of x, 9 and 12 are factors of x.

But, if k = 1,  x^3  = 216, then x = 6. For this value of x, 9 and 12 are not factors of x.

So, the correct answer option is A.

Hope this helps!

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Set J is comprised of all positive integers x such that x^3 is a multi

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Set J is comprised of all positive integers x such that x^3 is a multi

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