Algebra | have 1/y+1/z=a/x,1/x+1/z=b1/y,1/y+1/x=c1/z .This leads -askIITians

aritra nandy Grade: 10

        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.

7 years ago

Answers : (1)

SAGAR SINGH - IIT DELHI

879 Points

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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Sagar Singh

B.Tech, IIT Delhi