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Re: The LCM of three numbers is four times their GCF. Which of [#permalink]
12 Mar 2013, 20:29

3

This post received KUDOS

Expert's post

avinashrao9 wrote:

Hi Bunuel,

Is there any general approach to solve this problem?? Finding a set of 3 numbers might be difficult somtimes

Consider a generic approach. LCM - L GCF - G L = 4*G

The numbers must be of the form: Gn, Gm, Gk. Since other than G, L contains only 4, n, m and k must be of one of the three numbers: 1, 2, 4

Mind you, there must be one number which is G because otherwise, the GCF will change. GCF of 2G, 2G, 4G will be 2G, not G. There must be one number which is 4G because otherwise, the LCM will change. LCM of G, 2G, 2G is 2G, not 4G

So the three numbers could be of the form: G, 2G, 4G or G, G, 4G or G, 4G, 4G

If G is even, no number will be odd. Hence I and II needn't be true. III must be true.

Re: The LCM of three numbers is four times their GCF. Which of [#permalink]
27 Mar 2013, 13:09

VeritasPrepKarishma wrote:

avinashrao9 wrote:

Hi Bunuel,

Is there any general approach to solve this problem?? Finding a set of 3 numbers might be difficult somtimes

Consider a generic approach. LCM - L GCF - G L = 4*G

The numbers must be of the form: Gn, Gm, Gk. Since other than G, L contains only 4, n, m and k must be of one of the three numbers: 1, 2, 4

Mind you, there must be one number which is G because otherwise, the GCF will change. GCF of 2G, 2G, 4G will be 2G, not G. There must be one number which is 4G because otherwise, the LCM will change. LCM of G, 2G, 2G is 2G, not 4G

So the three numbers could be of the form: G, 2G, 4G or G, G, 4G or G, 4G, 4G

If G is even, no number will be odd. Hence I and II needn't be true. III must be true.

Answer (B)

Hi karishma, I could not understand anything out of this explanation. Could you please explain it to me in a more detailed manner??

Re: The LCM of three numbers is four times their GCF. Which of [#permalink]
25 Dec 2013, 17:28

Jlgdr,

I guess what Karishma is saying is this :

G*N, G*M, G*K are the 3 numbers whose GCF is G. LCM of these 3 numbers can be written as LCM(G*N,G*M,G*K) = G*LCM(N,M,K) = 4 *G (given). So LCM(N,M,K) = 4. This would mean {N,M,K} can be {1,1,4} or {1,2,4} or {1,4,4} since the LCM of the 3 numbers is 4. So the only possible solution is number B.