GMATBeast 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