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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.
```
6 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
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```
6 years ago
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