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Geometric Integers II (Posted on 2011-08-02) Difficulty: 3 of 5
Consider three positive integers x, y and z in geometric sequence with x < y < z < 2011 and, x+y=z+1 and determine all possible triplets (x,y,z) that satisfy the given conditions.

No Solution Yet Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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re: Fibonacci Comment 3 of 3 |
(In reply to Fibonacci by Math Man)

The puzzle asks for strictly increasing, so (1,1,1) is excluded.
  Posted by Charlie on 2011-08-04 11:46:35

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