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Division

Expert replies
by umaa » Tue Dec 23, 2008 9:50 pm
Don't know how I missed it
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Source: — Problem Solving |

by Stuart@KaplanGMAT » Tue Dec 23, 2008 9:54 pm
Is the integer k divisible by 4?

(1) 8k is divisible by 16

in other words:

8k/16 is an integer

dividing top and bottom by 8:

k/2 is an integer

Well, we know that k must be divisible by 2, but it may or may not be divisible by 4: insufficient.

(2) 9k is divisible by 12

in other words:

9k/12 is an integer

dividing top and bottom by 3:

3k/4 is an integer

Well, there's no way the "4" could come from the "3" part of "3k", so for 3k/4 to be an integer, k/4 must be an integer. Therefore, k is definitely divisible by 4: sufficient.

(2) is sufficient but (1) isn't: choose (B).
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by vishubn » Tue Dec 23, 2008 9:58 pm
stem is K is divisible by 4 sooo k can ne multiple of 2,4

A) 8k/16
in other wrds k/2.....

k is multiple of 2

k=2 T
k=4 F

Insufff

B) 9k/12 -- > 3k/4

that is plugin values

k=1, No
k=2 no
k-3 no
4=4 only
YEs
SO B is suff

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by ronniecoleman » Tue Dec 23, 2008 10:00 pm
is k divisible by 4
8k diviisible by 16
9k is divisible by 12

8 k == 2*2*2* k
16= 2*2*2*2

So k should have atleast one more 2

Not suff


9K = 3*3*k

12 = 4*3

So k has to have atleast one 4

Suff
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by vittalgmat » Wed Dec 24, 2008 3:29 am
ronniecoleman wrote:is k divisible by 4
8k diviisible by 16
9k is divisible by 12

8 k == 2*2*2* k
16= 2*2*2*2

So k should have atleast one more 2

Not suff


9K = 3*3*k

12 = 4*3

So k has to have atleast one 4

Suff
Great job explaining Ronnie.
Thank you for being accomodative.
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