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.
(In reply to
hint: how to prove it My cue  just continue by Ady TZIDON)
To comply with the condition that ab/c, ac/b and bc/a all have a remainder of 1 and that they be positive, a, b,and c must all be different and greater than 1.
From Ady's cue:
1/c + 1/b + 1/a  1/abc = k
As k is a positive integer and a, b, c different positive integers greater than 1, the sum of the first three terms above can be at most less than 2 and k must equal 1. The greatest of a, b, c, must be less than 6 or the sum of the first three terms before being reduced by the fourth term won't exceed 1, hence a, b, c, must be three of the numbers {2, 3, 4, 5}
(2, 3, 5) is the only solution.
Thank you, Ady!

Posted by ken
on 20161103 22:28:54 