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have 1/y+1/z=a/x,1/x+1/z=b1/y,1/y+1/x=c1/z .This leads to (a+1)/x=(b+1)/y=(c+1)/z=k, and then to 1/(a+1)+1/(b+1)+1/(c+1)=1.Since x,y,z need to be distinct distinct, so are a,b,c , hence the only solution is 2,3,6. the question was: Find three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer. Plse make me understand the solution fast.need it immediately.

have 1/y+1/z=a/x,1/x+1/z=b1/y,1/y+1/x=c1/z .This leads to (a+1)/x=(b+1)/y=(c+1)/z=k, and then to 1/(a+1)+1/(b+1)+1/(c+1)=1.Since x,y,z need to be distinct
distinct, so are a,b,c , hence the only solution is 2,3,6.


the question was:
Find three distinct positive integers with the least possible sum such that the sum
of the reciprocals of any two integers among them is an integral multiple of the
reciprocal of the third integer.

Plse make me understand the solution fast.need it immediately.

Grade:10

1 Answers

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

The solutionis quite clear..

1/y+1/z=a/x,1/x+1/z=b1/y,1/y+1/x=c1/z .This leads to (a+1)/x=(b+1)/y=(c+1)/z=k, and then to 1/(a+1)+1/(b+1)+1/(c+1)=1.Since x,y,z need to be distinct distinct, so are a,b,c , hence the only solution is 2,3,6

 

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

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