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If k, m, and t are positive integers and k/6 + m/4 = t/12 ,

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If k, m, and t are positive integers and k/6 + m/4 = t/12 , [#permalink] New post 07 Oct 2008, 22:07
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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] New post 07 Oct 2008, 22:13
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] New post 07 Oct 2008, 22:23
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] New post 07 Oct 2008, 22:29
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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] New post 07 Oct 2008, 22:36
Got it.... I didn't try the other condition.... :(

Thank you.
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Re: Highest Common Factor [#permalink] New post 17 Dec 2009, 01:19
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
Re: Highest Common Factor   [#permalink] 17 Dec 2009, 01:19
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If k, m, and t are positive integers and k/6 + m/4 = t/12 ,

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