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I do not agree with the OA..

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I do not agree with the OA.. [#permalink] New post 12 Nov 2006, 12:48
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I do not agree with the OA...
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 [#permalink] New post 12 Nov 2006, 13:32
WEIRED :roll:
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 [#permalink] New post 12 Nov 2006, 18:17
well.... weird question (certainly not from a real GMAT prep), but it has a clear answer (A).

st1 is sufficient. if k is a multiple of 3 then t and 12 has 3 as a common factor:

k/3 is integer n. so m/4+k/3 = m/4+n = (3m+12n)/12. hence t=3m+12n.
t is a multiple of 3 and so is 12.

st2 insuff: for example m=3 k=1 (then t=13 and has only 1 as common factor with 12). m=3 k=3 has more factors. hence insufficient.

note that the questions doesn't state that t and 12 have indeed 12 as common factor. it just looks for common factors of t and 12 that are not 1 or 12. in fact, if 12 is a common factor of any two numbers, so are 2,3,4 and 6.
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 [#permalink] New post 12 Nov 2006, 23:19
Yes Hobbit

You're right...

The fact that T is not necessarily multiple of 12 made this question hard to understand

That's why I did not agree with the OA

This question commes from 800 score

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 [#permalink] New post 12 Nov 2006, 23:52
12 has factors 1,2,3,4,6,12.

(m/4) + (k/3) = (t/12)
3m + 4k = t

St1:
k is a multiple of 3, so 3m + 4k = 3m + 4(3)(n) where n = 1,2,3,4...etc

So 3(3m + 4n) --> which is a multipe of 3. So t and 12 will have at least 3 in common apart from 1 and 12.

St2:
m is a multiple of 3, so 3m + 4k = 3(3)j + 4k = 9j + 4k --> we can't tell since j and k has a number of possibilites.

Ans A
  [#permalink] 12 Nov 2006, 23:52
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I do not agree with the OA..

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