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If k, m, and t are positive integers and 6k + 4m = 12t, do t [#permalink]

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13 Oct 2008, 08:34

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E

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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.

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.

I am confused between C and E but will go with E

Statement 1: only K is defined , there is no information of M. Not suff. Statement 2: only m is defined, there is no information of K. not suff. now taking both together for some multiples of 3, 12 and T have common factors but for some they don't have a common factor. for instance 6*6+4*6= 12*5 . here 12 and 5 do not have any multiples common other than 1

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.

2(3k+2M)=12t

3k+2M=6t <-- stem

1) K=3N

9N+2M=6t, if N=1 then 9+2M=6t if M and t are positive integers then, 2M=6t-9 or 2M=3(t-3)

then t-3 has to be even, so suppose t=5, then M=3, if t=5 and 12 have no factor greater than 1..

if 2M=3(t-3), suppose 2M=18, the t-3=6 t=9, then 12 and t have factors, 1 and 3 in common..insuff

2)I just picked example above of M=multiple of 3.. 6K+4*3N=12t

3(2K+4N)=12t -> 2k+4N=4t -> 2(K+2n)=4t ; k+2n=2t ; then we know K has to be even,

suppose K=2, n=2, then 2+4=2t or 6=2t t=3 then 12 and 3 have factors 1 and 3 in common

suppose K=4, N=2 4+4=2t, 8, t=4 then 12 and t have 1 and 2 and 4 in common..

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.

t=k/2 + m/3

We know that for t to be an integer, k must be an even integer and m must be a multiple of 3

(1) If k is a multiple of 3, so is k/2. However m/3 could be any positive integer. NOT SUFF (2) No new imformation. NOT SUFF (T) If k=6 and m=3 YES If k=6 and m=6 NO

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.

given

2(3k+2m) =12t ie 3k+2m = 6t

from one

3k = 9f , f can be even or odd

2m is even and so is 6t , as such f is even , rewriting 3k = 9f = 18j

18j+2m = 6t , 9j+m = 3t thus surely m is a multiple of 3

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.

1) Since LHS is a multiple of 12 we can write the LHS as

we have to express it as a multiple of 12

So it takes the form 36a + 12b......

Which takes the form 12(3a +1)

Now, 3a + 1 can never be a multiple of 12...........There fore they are prime to each other.

Taking 2) 6m + 12b =12t

this takes the form 12a +12b = 12t

which is 12(a+b) = 12 t

Now here (a+b) can take any value. It can be prime to 12 as well as have common factor with 12.

I think the question in this post is typed incorrectly, and should have been the same question in the post I linked. That would explain why A) is the correct answer.