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Limara1
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a, b, s, k are different non-zero integers... 02 Aug 2017, 16:28
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

85% (hard)

Question Stats:

48% (02:18) correct 52% (01:53) wrong based on 79 sessions

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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
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E
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a, b, s, k are different non-zero integers... 03 Aug 2017, 09:21
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Limara1
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.
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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... 03 Aug 2017, 11:24
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Limara1
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
Show more

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).
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a, b, s, k are different non-zero integers... 03 Aug 2017, 11:54
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\(\frac{2}{a}\) + \(\frac{4}{b}\) = \(\frac{k}{s}\)

\(\frac{2b + 4a}{ab}\) = \(\frac{k}{s}\)

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... 03 Aug 2017, 17:57
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given : {2b + 4a} is not divisible by ab or there is no common factor of {2b + 4a} and ab

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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.
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a, b, s, k are different non-zero integers... 26 Mar 2025, 21:09
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