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

Home > Just Math
A Multi-solution Puzzle (Posted on 2006-08-03) Difficulty: 3 of 5
Consider the equation:

(3P–2Q)² = 24PQ/(3P+2Q-1)

where P and Q are positive integers. It can be verified that (P=5, Q=5) and (P=26, Q=33) are two solutions.

(A) Can you give at least three other solutions to the above equation?

(B) Determine whether or not the equation admits of an infinite number of solutions.

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some answers w/o derivations | Comment 1 of 3
First of all, there is the trivial solution P=Q=0. This is part of the family of solutions P= 2j+24j^2, Q=36j^2-3j, j any integer. This family also includes P=26, Q=33.

There is also another family of solutions that includes P=Q=5, namely P=5-22k+24k^2, Q=5-27k+36k^2, k any integer.

The key to solving was .... (later, maybe).  Let's see if some other  solvers can get off their duffs.

  Posted by Richard on 2006-08-03 16:41:51
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information