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If y is an integer, is y^3 divisible by 12? 10 Dec 2017, 01:20
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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25% (medium)

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78% (01:04) correct 22% (01:02) wrong based on 63 sessions

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If y is an integer, is y^3 divisible by 12?

(1) y is divisible by 4.
(2) y is divisible by 6.
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Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
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If y is an integer, is y^3 divisible by 12? 10 Dec 2017, 03:16
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Bunuel
If y is an integer, is y^3 divisible by 12?

(1) y is divisible by 4.
(2) y is divisible by 6.
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For \(y^3\) to be divisible by \(12\), \(y\) must contain at least one \(2\) & one \(3\).

Statement 1: \(y =4p =>y^3=64p^3\). \(p\) can be a multiple of \(3\) or not. Hence Insufficient

Statement 2: \(y=6k =>y^3=(2*3)^3k^3\). Clearly divisible by \(12\). Sufficient

Option B
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If y is an integer, is y^3 divisible by 12? 24 Dec 2017, 05:36
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niks18
Bunuel
If y is an integer, is y^3 divisible by 12?

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

For \(y^3\) to be divisible by \(12\), \(y\) must contain at least one \(2\) & one \(3\).

Statement 1: \(y =4p =>y^3=64p^3\). \(p\) can be a multiple of \(3\) or not. Hence Insufficient

Statement 2: \(y=6k =>y^3=(2*3)^3k^3\). Clearly divisible by \(12\). Sufficient

Option B
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Would you be able to explain why y has to have least one \(2\) & one \(3\) and not two \(2s\) & one \(3\) Thank you!
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If y is an integer, is y^3 divisible by 12? 24 Dec 2017, 05:44
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imranmah
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Bunuel
If y is an integer, is y^3 divisible by 12?

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

For \(y^3\) to be divisible by \(12\), \(y\) must contain at least one \(2\) & one \(3\).

Statement 1: \(y =4p =>y^3=64p^3\). \(p\) can be a multiple of \(3\) or not. Hence Insufficient

Statement 2: \(y=6k =>y^3=(2*3)^3k^3\). Clearly divisible by \(12\). Sufficient

Option B

Would you be able to explain why y has to have least one \(2\) & one \(3\) and not two \(2s\) & one \(3\) Thank you!
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Hi imranmah

y can have more than one 2 and 3 but it has to have AT LEAST one 2 & 3.

so let's assume \(y=2*3=6\) then \(y^3=6*6*6\). therefore \(\frac{y^3}{12}=\frac{6*6*6}{12}\), divisible by 12

and as per your example if \(y=2*2*3\), then \(\frac{y^3}{12}=\frac{2*2*2*2*2*2*3*3*3}{12}\), divisible by 12

but it cannot be that y has only one 2 and no 3 or vice-versa.
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If y is an integer, is y^3 divisible by 12? 24 Jun 2023, 07:28
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