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a, b, k, s are different non-zero integers, \(\frac{2}{a}\) +\(\frac{4}{b}\) = \(\frac{k}{s}\) and \(\frac{k}{s}\) is maximally reduced. Does s = ab?

1) a and b are primes

2) a and b have gcd = 1

Sorry don't have the OA.

Hi...

It's all about what values can a and b take....

\(\frac{2}{a}\) +\(\frac{4}{b}\) = \(\frac{k}{s}\)... \(\frac{2b+4a}{ab}=\frac{k}{s}\).. I tells us a and b are prime.. If any of a or b is 2, it will cancel out from NUMERATOR, so k =ab/2... No If they are prime other than 2, k=ab. Yes.. Different answers Insufficient.. SAME will stand for II, where they are coprimes. Ans E
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a, b, s, k are different non-zero integers... [#permalink]

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03 Aug 2017, 11:24

chetan2u wrote:

Limara1 wrote:

a, b, k, s are different non-zero integers, \(\frac{2}{a}\) +\(\frac{4}{b}\) = \(\frac{k}{s}\) and \(\frac{k}{s}\) is maximally reduced. Does s = ab?

1) a and b are primes

2) a and b have gcd = 1

Sorry don't have the OA.

Hi...

It's all about what values can a and b take....

\(\frac{2}{a}\) +\(\frac{4}{b}\) = \(\frac{k}{s}\)... \(\frac{2b+4a}{ab}=\frac{k}{s}\).. I tells us a and b are prime.. If any of a or b is 2, it will cancel out from NUMERATOR, so s =ab/2... No If they are prime other than 2, s=ab. Yes.. Different answers Insufficient.. SAME will stand for II, where they are coprimes. Ans E

You say that if a and b are primes other than 2 then s=ab, but is this always true? Can you prove this (i.e. prove that neither of primes a or b is a factor in b+2a)?

Anyway, for the purposes of this DS question it's enough to show that primes a and b CAN also be such that s=ab (which is pretty obvious, e.g. take a=3, b=5).

given : {2b + 4a} is not divisible by ab or there is no common factor of {2b + 4a} and ab

I have a question here ; Isn't the even value of a or b is out of scope(already discarded) due to the given statement that \(\frac{k}{s}\) is maximally reduced.

In that case we can pick only odd primes from statement 1 which will give us answer yes to the given question
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a, b, s, k are different non-zero integers... [#permalink]

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03 Aug 2017, 17:57

Leo8 wrote:

given : {2b + 4a} is not divisible by ab or there is no common factor of {2b + 4a} and ab

I think you're mistaken here. The prompt says \(\frac{s}{k}\) is maximally reduced, NOT that \(\frac{2}{a}\) + \(\frac{4}{b}\) = \(\frac{2b+4a}{ab}\) is maximally reduced too.

In fact, the whole point of the question is to prove / disprove whether the latter is always true in this situation.