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# LCM

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Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 536
Location: United Kingdom
GMAT 1: 730 Q49 V45
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 74

Kudos [?]: 2987 [0], given: 217

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25 Jun 2011, 12:45
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If the LCM of a & 12 is 36 what could be the possible values of a?

The LCM of 12 and a contains two 2's. Since the LCM contains each prime factor to the power it appears the MOST, we know that a cannot contain more than two 2's.

LCM of 12 and a contain two 3's. But 12 only contains one 3. The 3^2 factor in the LCM must have come from prime factorization of a. Thus we know that a contains exactly two 3's.

Since a must contain exactly two 3's nd can contain no 2's, one 2 or two 2's a could be

3*3=9
3*3*2=18
3*3*2**2=36.

Thus 9,18 and 36 are three values.

I am struggling to understand the concept.
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Best Regards,
E.

MGMAT 1 --> 530
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Director
Joined: 01 Feb 2011
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25 Jun 2011, 13:40
12 = (2^2)*3
a = (2^2 or 2^1 or 2^0)*(3^2)

L.C.M of a & 12 = 36 = (2^2 )*(3^2) = product of Maximum powers of each prime factor

This can be easily understood by comparing 12 with 36 .

12 has one 3 , but L.C.M which is the maximum powers of each prime factor has two 3's

=> a must have two 3's in it = 3^2 ------1

12 has two 2's. L.c.m which is the maximum powers of each prime factor has two 2's as well.

=> a could have zero 2's or one 2 or two 2's------2

clubbing above two statements marked 1 *2 , we have a = (2^2 or 2^1 or 2^0)*(3^2)
=(4 or 2 or 1)*9
=> a could be 36 or 18 or 9.

Hope its clear now.
Re: LCM   [#permalink] 25 Jun 2011, 13:40
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