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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
1) a and b are primes
2) a and b have gcd = 1
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.
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
03 Aug 2017, 11:24
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chetan2u
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.
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).
