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# factors of 12 are 3,2,2 If i understand correctly, if 3 , 2

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factors of 12 are 3,2,2 If i understand correctly, if 3 , 2 [#permalink]  20 Jun 2011, 06:55
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factors of 12 are 3,2,2

If i understand correctly, if 3 , 2 are factors of 12. certain multiples of 3,2 will also be factors of 12 because you are going up or down a factor tree.

so 2k + 3m = t

even without reading the statements, doesn't the equation above say that t and 12 have 3,2 as common factors simply because those numbers are being multiplied to make t?
[Reveal] Spoiler: OA

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Re: divisibility & primes [#permalink]  20 Jun 2011, 07:21
nvm....figured it out....ill post for those wondering....

if it was like 2k=t or 3m=t, then yes t and 12 would have factors 2,3 accordingly.

but since 2k and 3m are being added, you cannot simply say they 2,3 themselves can be the factors but the factors will be generated by the SUM of the multiples of 2 and 3 here.
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Re: divisibility & primes [#permalink]  07 Aug 2011, 18:30
Are you sure the answer is A? if so could you please post the explanation that was provided along with the question.
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Re: divisibility & primes [#permalink]  07 Aug 2011, 19:48

By simplifying the equation we get 2k + 3m = t
Does this mean that 2 and/or 3 are factors of t? Not necessarily!
Consider k=1 and m=1 => t=5. Does t have 2 and 3 as factors? No!
Alternately, consider k=3 and m=2 => t=12. In this case, t does have both k and m as factors.

Point to note:
If a positive integer is the sum of the multiples of other positive integers, it need not be a multiple of either of the integers!

Carrying on with this question,

Using statement 1: If k is a multiple of 3, then the equation can be written as
2k + 3m = t
=> 2*3n + 3m = t (where n is a positive integer)
=> 3 (2n +m) = t
=> 3 is a factor of t
=> t and 12 have a common factor greater than 1 (i.e. 3)
SUFFICIENT.

Consider statement 2: If m is a multiple of 3, we can write the equation as
2k + 3m = t
=> 2k + 3*3n = t (where n is a positive integer)
=> 2k + 9n = t
If we take n=1 and k=3, we get t=15, which has 3 as a common factor greater than 1 with 12
If we take k=1 and n=1, we get t=11, which has no common factor greater than 1 with 12
Therefore statement 2 alone is insufficient.

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Re: divisibility & primes   [#permalink] 07 Aug 2011, 19:48
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