Data Sufficiency

NandishSS wrote:Is the positive integer y a multiple of 12?

(1) y^3 is a multiple of 48
(2) y^2 is a multiple of 30

OA:A

Hi NandishSS,

Statement 1: y^3 is a multiple of 48

Let's do the prime factorization of 48.

48 = 2*2*2*2*3

Since y is a positive integer, y^3 must have each of the prime factors in a pair of three. We see that there are four 2's and only one 3, thus, it must need at least two more 2's and two more 3's.

Thus, the minimum value of y^3 = (2^3)*(2^3)*(3^3)

=> Minimum value of y = 2*2*3 = 12

=> y a multiple of 12. Sufficient.

Statement 2: y^2 is a multiple of 30

Let's do the prime factorization of 30.

30 = 2*3*5

As discussed in Statement 1, since y is a positive integer, y^2 must have each of the prime factors in a pair of two. We see that there are only one 2, one 3 and one 5, thus, it must need at least one more 2, one more 3 and one more 5.

=> Thus, the minimum value of y^2 = (2^2)*(3^2)*(5^2)

=> Minimum value of y = 2*3*5 = 30

Since for y to be a multiple of 12 = 2^2*3, there must be two 2's, it is not necessary that y is a multiple of 12. Insufficient.

The correct answer: A

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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NandishSS wrote:Is the positive integer y a multiple of 12?

(1) y^3 is a multiple of 48
(2) y^2 is a multiple of 30

We need to determine whether y/12 is an integer.

Statement One Alone:

y^3 is a multiple of 48.

Since y^3/48 = integer, we can say that the product of 48 and some integer n is equal to a perfect cube. In other words, 48n = y^3.

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 48 into primes to help determine what extra prime factors we need to make 48n a perfect cube.

48 = 16 x 3 = 2^4 x 3^1

In order to make 48n a perfect cube, we need two more 2s and two more 3s. Thus, the smallest perfect cube that is a multiple of 48 is 2^6 x 3^3.

To determine the least possible value of y, we can take the cube root of 2^6 x 3^3 and we have:

2^2 x 3 = 12

Thus, the minimum value of y is 12, so y/12 is an integer. Statement one alone is sufficient to answer the question.

Statement Two Alone:

y^2 is a multiple of 30.

Since y^2/30 = integer, we can say that the product of 30 and some integer m is equal to a perfect square. In other words, 30m = y^2.

We must remember that all perfect squares break down to unique prime factors, each of which has an exponent that is a multiple of 2. So let's break down 30 into primes to help determine what extra prime factors we need to make 30m a perfect square.

30 = 5 x 3 x 2

In order to make 30m a perfect square, we need one more 2, one more 3, and one more 5. Thus, the smallest perfect square that is a multiple of 30 is 2^2 x 3^2 x 5^2.

To determine the least possible value of y, we can take the square root of 2^2 x 3^2 x 5^2 and we have:

2 x 3 x 5 = 30

Thus, the minimum value of y is 30; however if y is 30, then y/12 is not an integer; and if y is 60, then y/12 is an integer. Statement two alone is not sufficient to answer the question.

Answer: A