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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.
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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
26 Feb 2005, 19:42
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