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Which of the following could be the sum of the reciprocals

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Which of the following could be the sum of the reciprocals [#permalink] New post 31 Mar 2009, 21:29
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Which of the following could be the sum of the reciprocals of two different prime numbers?

A. 7/13
B. 10/1
C. 11/30
D. 23/50
E. 19/77
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Re: sum of reciprocals. [#permalink] New post 31 Mar 2009, 22:07
10/12 it is
10/12=5/6=(1/2) + (1/3) :)
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Re: sum of reciprocals. [#permalink] New post 02 Jan 2011, 22:23
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...
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Re: sum of reciprocals. [#permalink] New post 03 Jan 2011, 01:41
are there any theories for these kinds of problems? haven't came across any yet, i think hit and trial is the best one!
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Re: sum of reciprocals. [#permalink] New post 03 Jan 2011, 02:53
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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.
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Re: sum of reciprocals. [#permalink] New post 03 Jan 2011, 11:00
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: sum of reciprocals.   [#permalink] 03 Jan 2011, 11:00
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