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Re: sum of reciprocals. [#permalink]
03 Jan 2011, 01:53

7

This post received KUDOS

Expert's post

jullysabat wrote:

nitya34 wrote:

10/12 it is 10/12=5/6=(1/2) + (1/3)

Was it a guess ??? or how do we approach to this kind of problem...

prab wrote:

are there any theories for these kinds of problems? haven't came across any yet, i think hit and trial is the best one!

Which of the following could be the sum of the reciprocals of two different prime numbers? A. 7/13 B. 10/12 C. 11/30 D. 23/50 E. 19/77

Let \(x\) and \(y\) be two different primes, then their sum will be \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\). So, the denominator of the reduced fraction must be the product of two different primes: only B, \(\frac{10}{12}=\frac{5}{6}=\frac{5}{2*3}\), and E, \(\frac{19}{77}=\frac{19}{7*11}\), have such denominators and nominator must be the sum of the primes in denominator, thus only B is left: \(\frac{10}{12}=\frac{5}{6}=\frac{2+3}{2*3}\).

Re: sum of reciprocals. [#permalink]
03 Jan 2011, 10:00

Bunuel wrote:

jullysabat wrote:

nitya34 wrote:

10/12 it is 10/12=5/6=(1/2) + (1/3)

Was it a guess ??? or how do we approach to this kind of problem...

prab wrote:

are there any theories for these kinds of problems? haven't came across any yet, i think hit and trial is the best one!

Which of the following could be the sum of the reciprocals of two different prime numbers? A. 7/13 B. 10/12 C. 11/30 D. 23/50 E. 19/77

Let \(x\) and \(y\) be two different primes, then their sum will be \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\). So, the denominator of the reduced fraction must be the product of two different primes: only B, \(\frac{10}{12}=\frac{5}{6}=\frac{5}{2*3}\), and E, \(\frac{19}{77}=\frac{19}{7*11}\), have such denominators and nominator must be the sum of the primes in denominator, thus only B is left: \(\frac{10}{12}=\frac{5}{6}=\frac{2+3}{2*3}\).

Answer: B.

Hope it's clear.

Hey Bunnel,

Thank u so much.. for this explanation... very very simple and robust way of explaning things...

Re: Which of the following could be the sum of the reciprocals [#permalink]
09 Jul 2014, 08:19

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