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If k, m, and t are positive integers and k/6 + m/4 = t/12 , [#permalink]
 07 Oct 2008, 23:07
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If k, m, and t are positive integers and k/6 + m/4 = t/12 , does t and 12 have a common factor greater than 1 ? (1) k is a multiple of 3. (2) m is a multiple of 3.
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Re: Highest Common Factor [#permalink]
07 Oct 2008, 23:13
leonidas wrote: If k, m, and t are positive integers and k/6 + m/4 = t/12 , does t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3. k/6 + m/4 = t/12 => (2k + 3m)/12 = t/12 (1) Sufficient. If k=3*n, the equation becomes (6n + 3m)/12 => 3 is a common factor of t and 12 (2) Insufficient. If m=3*n, the equation becomes (2k + 9m)/12 => no common factor visible Hence (A)
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Re: Highest Common Factor [#permalink]
07 Oct 2008, 23:23
incognito1 wrote: leonidas wrote: If k, m, and t are positive integers and k/6 + m/4 = t/12 , does t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3. k/6 + m/4 = t/12 => (2k + 3m)/12 = t/12 (1) Sufficient. If k=3*n, the equation becomes (6n + 3m)/12 => 3 is a common factor of t and 12 (2) Insufficient. If m=3*n, the equation becomes (2k + 9m)/12 => no common factor visible Hence (A) If there is no common factor visible, shouldn't (B) also be sufficient? as this is a YES or NO type question.
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Re: Highest Common Factor [#permalink]
07 Oct 2008, 23:29
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leonidas wrote: incognito1 wrote: leonidas wrote: If k, m, and t are positive integers and k/6 + m/4 = t/12 , does t and 12 have a common factor greater than 1 ?
(1) k is a multiple of 3.
(2) m is a multiple of 3. k/6 + m/4 = t/12 => (2k + 3m)/12 = t/12 (1) Sufficient. If k=3*n, the equation becomes (6n + 3m)/12 => 3 is a common factor of t and 12 (2) Insufficient. If m=3*n, the equation becomes (2k + 9m)/12 => no common factor visible Hence (A) If there is no common factor visible, shouldn't (B) also be sufficient? as this is a YES or NO type question. (B) does not guarantee of a common factor (A, on the other hand, does). It's possible (in B) for t to have something in common with 12 even when no common factor is obvious. For instance, if k=3, m=4, then common factor = 2 However, k=5, m=3 does not give any common factor
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Re: Highest Common Factor [#permalink]
07 Oct 2008, 23:36
Got it.... I didn't try the other condition....  Thank you.
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Re: Highest Common Factor [#permalink]
17 Dec 2009, 02:19
This is a good question. Especially when it comes to attacking the second statement
2K + 3M / 12 = T/12
Statement 1-----------SUFFICIENT
K=3N
Try N =1, (M CAN TAKE ANY VALUES AND STILL BE A MUTILPLE OF 3 AS IT HAS 3 AS ITS COEFFECIENT)
6 + 3M /12 =T/12
3(2+M)/12 =T/12
Therefore 3 is a factor of T and 12 other than 1
Hence A
Statement 2------------INSUFFICIENT
M=3N
Try N=1
2K + 9/ 12 = T/12
No common factor other than 1
But if we put K =6, AND N=1
We will get
12 + 9 / 12 =T/12
=3 (4+3)/12 = T/12
Common factor 3
Two different values. Hence Insufficient
Hence A
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Re: Highest Common Factor
[#permalink]
17 Dec 2009, 02:19
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