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

Home > Just Math
Simultaneous Real Settlement (Posted on 2015-07-20) Difficulty: 3 of 5
Determine all quadruplets (P,Q,R,S) of real numbers that satisfy this system of equations:

(Q+R+S)2016 = 3P, and:

(P+R+S)2016 = 3Q, and:

(P+Q+S)2016 = 3R, and:

(P+Q+R)2016 = 3S

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
solution Comment 2 of 2 |
Look at the first equation.  LHS is never negative, so neither is RHS.  Similarly, Q, R, S are >=0 too.

Say you can order the unknowns P<Q<R<S.  Then substituting P for Q in first equation would make LHS<3P.  But the modified LHS=3Q by the second equation giving 3Q<3P, an impossibility.  By extension the same is true for the other unknowns.  Therefore they are all equal and the only solutions are the ones Jer noted:  P=0 and P=1/3.

  Posted by xdog on 2015-07-21 11:27:32
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 (6)
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