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If the least common multiple of a positive integer m and n [#permalink]
 13 Jun 2008, 00:09
If the least common multiple of a positive integer m and n is 120 and m:n is 3:4 what is the greatest common factor of m and n 3 5 6 10 12 Please provide detailed explanations on how to solve this many thanks
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Re: If the least common multiple of a positive integer [#permalink]
13 Jun 2008, 00:26
vdhawan1 wrote: If the least common multiple of a positive integer m and n is 120 and m:n is 3:4 what is the greatest common factor of m and n
3
5
6
10
12
Please provide detailed explanations on how to solve this
many thanks D for me! x is the least common factor of m and n m*n=3x*4x=120*x, so x=10, because x can not be 0!
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Re: If the least common multiple of a positive integer [#permalink]
13 Jun 2008, 04:56
sondenso wrote: D for me!
x is the least common factor of m and n
m*n=3x*4x I don't get this (where does it come from ?) sondenso wrote: 3x*4x=120*x This is just false sondenso wrote: =120*x, so x=10, because x can not be 0! I don't get this either. Can you explain ?
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Re: If the least common multiple of a positive integer [#permalink]
13 Jun 2008, 05:08
I m not sure that i understand the solution correctly
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Re: If the least common multiple of a positive integer [#permalink]
13 Jun 2008, 05:49
And I'm sure I don't 
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Re: If the least common multiple of a positive integer [#permalink]
13 Jun 2008, 06:37
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Dfist way: LCM=120=3*2^3*5 1. LCM contains prime number 5, so m or n or both also contain 5. 2. if m:n=3:4 then only both m and m contain 5. Therefore, GCD is 5 or 10. 3. LCM contains 2^3, but in ratio m:n we have only 4=2^2. So, both m and m contain 2. GCD=10. second way: m*n=LCM*GCD - it is a formula. m*n=3x*4x=120*GCD ---> GCD=x^2/10 ---> only 10 works.
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Re: If the least common multiple of a positive integer [#permalink]
13 Jun 2008, 06:45
walker wrote: D
fist way:
LCM=120=3*2^3*5
1. LCM contains prime number 5, so m or n or both also contain 5.
2. if m:n=3:4 then only both m and m contain 5. Therefore, GCD is 5 or 10.
3. LCM contains 2^3, but in ratio m:n we have only 4=2^2. So, both m and m contain 2. GCD=10. I like this. Thanks ! walker wrote: second way:
m*n=LCM*GCD - it is a formula.
m*n=3x*4x=120*GCD ---> GCD=x^2/10 ---> only 10 works. Thanks for the refresh on the formula, I did not remember. But why is m*n = 3x*4x ?
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Re: If the least common multiple of a positive integer [#permalink]
13 Jun 2008, 06:48
Oski wrote: But why is m*n = 3x*4x ? m:n=3:4 --> m=3x, n=4x where x is an integer (m/n=3x/4x=3/4)
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Re: If the least common multiple of a positive integer [#permalink]
13 Jun 2008, 06:53
walker wrote: Oski wrote: But why is m*n = 3x*4x ? m:n=3:4 --> m=3x, n=4x where x is an integer (m/n=3x/4x=3/4) Yes, sure, but why is this x necessarily the GCD ? Edit : Okay, I got it. This is because there is no common divisors in 3 and 4... (I guess this should be part of the demonstration ^^)
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Re: If the least common multiple of a positive integer [#permalink]
13 Jun 2008, 19:16
Oski wrote: sondenso wrote: D for me!
x is the least common factor of m and n
m*n=3x*4x I don't get this (where does it come from ?) sondenso wrote: 3x*4x=120*x This is just false sondenso wrote: =120*x, so x=10, because x can not be 0! I don't get this either. Can you explain ? Many thanks Walker, You have thorough explaination!
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Re: If the least common multiple of a positive integer
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13 Jun 2008, 19:16
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