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A peculiar triplet (Posted on 2016-11-01) Difficulty: 3 of 5
This triplet of positive integers has this peculiarity:
A product of any its two numbers divided by the 3rd number
has 1 as a remainder.

Find it.
Show that no other exist.

See The Solution Submitted by Ady TZIDON    
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Hints/Tips hint: how to prove it My cue - just continue | Comment 5 of 7 |

2,3,5  is the ONLY solution

To prove it :

ab+ac+bc-1 obviously is divisible both by a, b, c and therefore is divisible  by abc   i.e. ab+ac+bc = -1+kabc, k being an integer

divide by abc:    .....   .......

analyze the possible values of k


Edited on November 3, 2016, 6:15 am
  Posted by Ady TZIDON on 2016-11-02 14:55:51

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