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# If the LCM of a amd 12 is 36 what could be the possible values of a?

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Manager
Joined: 16 May 2011
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Kudos [?]: 87 [0], given: 37

LCM and GCD- can som1 suppose any safe and easy method? [#permalink]

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07 Jun 2011, 00:22
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If the LCM of A and 12 is 36, what are the possible values of A?
Veritas Prep GMAT Instructor
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Re: LCM and GCD- can som1 suppose any safe and easy method? [#permalink]

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07 Jun 2011, 10:44
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Expert's post
dimri10 wrote:
If the LCM of A and 12 is 36, what are the possible values of A?

Think of what LCM means before going ahead. If I say LCM of two numbers is $$36 (= 4*9 = 2^2 * 3^2)$$, it means that at least one of them must have a $$2^2$$ and at least one of them must have a $$3^2$$ (It is not possible that both numbers have just 3 because then, the LCM would have just 3, not 9)

If one number is $$12 (= 2^2 * 3)$$, the other number A must have $$3^2$$ since 12 doesn't have it.
So minimum value of A will be 9. A can also have a 2 or a $$2^2$$ so other possible values are $$18 (=9*2)$$ and $$36 (= 9*2^2)$$
Also, A cannot have any other factors since if it did, then the LCM would have to have that factor too.
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Kudos [?]: 3989 [1] , given: 360

Re: LCM and GCD- can som1 suppose any safe and easy method? [#permalink]

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07 Jun 2011, 11:36
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Expert's post
Here is what I do in my head (mental tricks):

LCM (12,36) --> (12/12,36/12) --> (1,3) --> 3*12 or 1*36 --> 36
LCM (12, 30) --> (12/6,30/6) --> (2,5) --> 5*12 or 2*30 --> 60
LCM (10, 28) --> (10/2,28/2) --> (5,14) --> 14*10 or 5*28 --> 140

GCD (12,36) --> (12/12,36/12) --> (1,3) --> 12
GCD (12, 30) --> (12/6,30/6) --> (2,5) --> 6
GCD (10, 28) --> (10/2,28/2) --> (5,14) --> 2

LCM and GCD have a nice property: LCM(x,y)*GCD(x,y) = xy

For example, LCM(12,30)*GCD(12,30) = 60*6 = 12 * 30
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Re: LCM and GCD- can som1 suppose any safe and easy method? [#permalink]

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08 Jun 2011, 02:24
i thank both of you, but i did not get walker's approach.can you please sharpen your explenation?

let's say that there are 3 numbers:
440,120 and 80.how can you use your shortcut
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Re: LCM and GCD- can som1 suppose any safe and easy method? [#permalink]

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08 Jun 2011, 03:03
LCM (80, 120, 440) --> (80/40, 120/40, 440/40) --> (2, 3, 11) --> 3*11*80 --> 2640

GCD (80, 120, 440) --> (80/40, 120/40, 440/40) --> (2, 3, 11) --> 40

40 here is the greatest common divisor.
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If the LCM of a amd 12 is 36 what could be the possible values of a? [#permalink]

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25 Jun 2011, 13:45
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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Re: If the LCM of a amd 12 is 36 what could be the possible values of a? [#permalink]

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25 Jun 2011, 14:40
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.
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Re: LCM and GCD- can som1 suppose any safe and easy method? [#permalink]

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15 Jan 2016, 01:24
VeritasPrepKarishma wrote:
dimri10 wrote:
If the LCM of A and 12 is 36, what are the possible values of A?

Think of what LCM means before going ahead. If I say LCM of two numbers is $$36 (= 4*9 = 2^2 * 3^2)$$, it means that at least one of them must have a $$2^2$$ and at least one of them must have a $$3^2$$ (It is not possible that both numbers have just 3 because then, the LCM would have just 3, not 9)

If one number is $$12 (= 2^2 * 3)$$, the other number A must have $$3^2$$ since 12 doesn't have it.
So minimum value of A will be 9. A can also have a 2 or a $$2^2$$ so other possible values are $$18 (=9*2)$$ and $$36 (= 9*2^2)$$
Also, A cannot have any other factors since if it did, then the LCM would have to have that factor too.

Thanks Karishma. Your solution is very clear and in simple way. I didn't understand the complicated solution prepared by Manhattan. Their prime columns technique is good but solution explanation was confusing.
Re: LCM and GCD- can som1 suppose any safe and easy method?   [#permalink] 15 Jan 2016, 01:24
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