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Common factor

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Common factor

by helix » Sun Mar 09, 2008 10:13 am
If k, m, and t are positive integers and 6k + 4m = 12t, do 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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Source: — Data Sufficiency |
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by soumitb » Tue Apr 08, 2008 10:06 am
Question stem states 6k + 4m = 12t
=> t = (k/2) + (m/3)

Therefore in order for t and 12 to have a common factor, t should be either even or any multiple of 3.

St 1 - k is a multiple of 3
k = 3,6,9,12....
(k/2) = 3/2, 3, 9/2, 4....

But we don't know m, So insufficient.


St 2 - m is a multiple of 3
m = 3,6,9,12,15....
(m/3) = 1,2,3,4,5... (Integer)

But we don't know k, So insufficient.

St 1 + St 2:
From St 2, we have (m/3) = 1,2,3,4,5... (an integer)
From St 1, we have (k/2) = 3/2, 3, 9/2, 4....
Since t = (k/2) + (m/3) and t is an integer, we can ignore the
fractional values of (k/2)

Therefore, t could be 4, 6, 8, 9, 10...(have common factor with 12)
But t can also be 5, 7, 13 ... (do not have common factor with 12)

Therefore answer is E
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