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Integer n

by vinayreguri » Sun Jun 12, 2011 7:35 am
if an integer is divisible by both 9 & 12, then it must be divisible by all of the following except
A. 36
B. 27
C. 18
D. 12
E. 9

[spoiler]OA : B[/spoiler]

My answer is B since 27 is not a factor of the number which can be divisible by 9 & 12.
All others are multiple of 9 & 12 except 27.

Any other different approach here.
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by manpsingh87 » Sun Jun 12, 2011 7:39 am
vinayreguri wrote:if an integer is divisible by both 9 & 12, then it must be divisible by all of the following except
A. 36
B. 27
C. 18
D. 12
E. 9

[spoiler]OA : B[/spoiler]

My answer is B since 27 is not a factor of the number which can be divisible by 9 & 12.
All others are multiple of 9 & 12 except 27.

Any other different approach here.
LCM of 9 and 12 is 36, so number will be of the type 36k...!!! hence answer should be B
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by galaxian » Sun Jun 12, 2011 11:53 am
Also,looking at the options, only B doesnt have any factor of 2, it is 3*3*3 while the no if divisible by 12 & 9 should have at least 3*2*2 or 3*3.
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by breakkgmat » Sun Jun 12, 2011 12:06 pm
If we divide the no in Prime Factorization we get, 9=3*3 & 12=3*2*2..
Only 27 & 9 do not have any 2's in the prime factorization...But 9 is the number itself..
So,ans indeed B..
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by Roy@MasterGmat » Mon Jun 13, 2011 4:18 am
Only 27 & 9 do not have any 2's in the prime factorization...But 9 is the number itself..
So,ans indeed B..
Right answer, wrong reason ;)

The correct answer doesn't need to include a '2', it simply must not include any additional factors on top of those present in a number divisible by 12 and 9.

As long as our integer is divisible by 9 and 12, then it must contain at least two '3's and at least two '2's. Therefore, it is also divisible by any number consisting of no more than those two '3's and two '2's.

The right approach here would be factoring the numbers in the answer choices, and choosing the first one found to contain any building blocks beyond 2*2*3*3. In this case, that means answer B, since 27=3*3*3.
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