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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]

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23 Feb 2005, 20:07
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If k, m, and t are positive integers and k/6 + m/4 = t/12 , 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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23 Feb 2005, 20:15
we can simplify the equation to 2K+3M = t

(1) tells us nothing about M. - not sufficient.
(2) tells us nothing about K. - not sufficine.t

Using (1) and (2),
we know t = 2K + 3M
3M is going ot be a multiple of 3 (3 * multiple of 3)
2K is also going ot be a multple of 3 (2* multiple of 3)

So 12 and t do have at least one more common factor. That is 3.

(C)

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23 Feb 2005, 20:20
It does tell m is a positive integer...
ie. m>0

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23 Feb 2005, 20:24
2K+3M = t

(1) 2k is a multiple of 3 (2*multiple of 3) but we know M is greater than 1. And thus 3M is also a multiple of 3. So we can tell that t does have more than one common factor.

Same with (2).

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23 Feb 2005, 20:24
k/6 + m/4 = t/12
simplifying,
2k+3m = t

1) k is a multiple of 3.

2*3*x + 3m = t

3(2x + 3m)= t

3 is a factor of t
=> t and 12 have 3 as a common factor - yes

2) m is a multiple of 3.
2k + 3*3*y = t

can't derive much with this about factors of t.

A) it is

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23 Feb 2005, 20:26
nice method nocilis.. i ought to hammer myself. Only god knoww what i'm thinking of making mistakes twice in a row.

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23 Feb 2005, 20:30
"A" it is.

Although I solved it by substituition , takes lot longer.
_________________

"Forums are meant to benefit all. No one is interested in knowing what your guesses are. Please explain the reasoning behind the answer you chose. This will also help you organize your thoughts quickly during the exam."

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23 Feb 2005, 20:36
ywilfred also has a nice method, you just have to be a bit more careful. (I'm often the same as you. )

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24 Feb 2005, 08:20
"A"

simplify we get 2k+3m = t

state I......k = 3n.......6n+3m = t.....3(2n+m) = t.....ans is yes as t has 3 as a factor....suff

state II.....m = 3n.....2k+9n = t.....for k= 1 and n=1.....t = 11....ans NO.....for k= 1 and n = 2.....t = 20...ans YES...insuff

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24 Feb 2005, 12:12
A.

I) k is a mutilple of 3 and 3m is also a multiple of 3 thus 2k + 3m that is t is also a multiple of 3... suff
II) Doesn't give us anyhting common to derive.. insuff

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26 Feb 2005, 19:42
One more A

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26 Feb 2005, 19:42
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