If y is an integer, is y^3 divisible by 9?

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If y is an integer, is y^3 divisible by 9?

by VJesus12 » Thu Nov 23, 2017 7:00 am
If y is an integer, is y^3 divisible by 9?

(1) y is divisible by 4.
(2) y is divisible by 6.

The OA is B .

Experts, how can I show that statement (1) is not sufficient?

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by Jay@ManhattanReview » Thu Nov 23, 2017 9:01 pm
VJesus12 wrote:If y is an integer, is y^3 divisible by 9?

(1) y is divisible by 4.
(2) y is divisible by 6.

The OA is B .

Experts, how can I show that statement (1) is not sufficient?
(1) y is divisible by 4.

Say y = 4k; where k is an integer

Thus, y^3 = 64k^3

If k^3 is divisible by 9, the answer is Yes, else No. Insufficient.

(2) y is divisible by 6.

Say y = 6k; where k is an integer

Thus, y^3 = 6^3k^3 = (2*3)^3k^3 = 2^3*3^3k^3

Since 3^3 is divisible by 9, the answer is Yes, else No. Sufficient.

The correct answer: B

Hope this helps!

-Jay
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