It is currently 21 Jan 2018, 16:41

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43348

Kudos [?]: 139709 [0], given: 12794

If y is an integer, is y^3 divisible by 12? [#permalink]

### Show Tags

10 Dec 2017, 00:20
Expert's post
2
This post was
BOOKMARKED
00:00

Difficulty:

35% (medium)

Question Stats:

71% (00:33) correct 29% (00:41) wrong based on 38 sessions

### HideShow timer Statistics

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

(1) y is divisible by 4.
(2) y is divisible by 6.
[Reveal] Spoiler: OA

_________________

Kudos [?]: 139709 [0], given: 12794

PS Forum Moderator
Joined: 25 Feb 2013
Posts: 826

Kudos [?]: 384 [1], given: 42

Location: India
GPA: 3.82
If y is an integer, is y^3 divisible by 12? [#permalink]

### Show Tags

10 Dec 2017, 02:16
1
KUDOS
Bunuel wrote:
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

Kudos [?]: 384 [1], given: 42

Intern
Joined: 04 Aug 2017
Posts: 10

Kudos [?]: 1 [0], given: 802

If y is an integer, is y^3 divisible by 12? [#permalink]

### Show Tags

24 Dec 2017, 04:36
niks18 wrote:
Bunuel wrote:
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!

Kudos [?]: 1 [0], given: 802

PS Forum Moderator
Joined: 25 Feb 2013
Posts: 826

Kudos [?]: 384 [0], given: 42

Location: India
GPA: 3.82
Re: If y is an integer, is y^3 divisible by 12? [#permalink]

### Show Tags

24 Dec 2017, 04:44
imranmah wrote:
niks18 wrote:
Bunuel wrote:
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!

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.

Kudos [?]: 384 [0], given: 42

Re: If y is an integer, is y^3 divisible by 12?   [#permalink] 24 Dec 2017, 04:44
Display posts from previous: Sort by