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Factors and power problem..........Expert needed

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Source: — Data Sufficiency |
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by GMATGuruNY » Fri Jan 20, 2017 6:03 am
Mo2men wrote:If y and z are factors of x^2, is x divisible by y?

(1) y is a prime number.
(2) z = y^2
Question stem, rephrased:
Does x/y = integer?

Statements combined:
Let y=2 and z=2²=4.

Since x² must be divisible by y=2 and z=4, the smallest possible option for x² is 4.
If x²=4, then x=2, with the result that x/y = 2/2 = 1 = integer.
The next greatest option for x² is 8.
If x²=8, then x=√8, with the result that x/y = √8/2 = (2√2)/2 = √2 = noninteger.

Since the answer is YES in the first case but NO in the second case, the two statements combined are INSUFFICIENT.

The correct answer is E.
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by Jay@ManhattanReview » Fri Jan 20, 2017 11:10 pm
Mo2men wrote:If y and z are factors of x^2, is x divisible by y?

(1) y is a prime number.
(2) z = y^2
The most important aspect of this question is that you need not assume that x is an integer. While x^2 is an integer, x may/may not be. Keeping this in mind, you should choose the test values of x^2: one renders x an integer and the other renders x a non-integer.

Let us take each statement one by one.

S1: y is a prime number.

Case 1: Say x^3 = 9 => x = 3, y=3 (prime), and z = 9. x is divisible by y. Answer is YES.
Case 2: Say x^3 = 27 => x = 3√3, y=3 (prime), and z = 9. x is NOT divisible by y. Answer is NO.

Not sufficient.

S2: z = y^2

Each of the above two cases fit here too. z (9) = y^2 = 3^2. Not sufficient.

Even after combining both the statements, we cannot uniquely answer the question.

Answer: E

Hope this helps!

-Jay
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