All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re: hint: how to prove it My cue - just continue | Comment 6 of 7 |
(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 2016-11-03 22:28:54
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information