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

Home > Just Math
Gonna party like it's 1999 (Posted on 2004-09-12) Difficulty: 1 of 5
Find a solution to:
x1^4 + x2^4 + x3^4 + ... + xn^4 = 1999

where each xy is a distinct integer.

(Or prove that it is impossible).

See The Solution Submitted by SilverKnight    
Rating: 3.2500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Logically Forced Solution | Comment 7 of 8 |

Since any X integer raised to the 4th power yields a positive number, X<=6 because X> 6 yields values higher than 1999.

So, -6<= X <=6, or we can use the X =1, to X=6 values twice and still have a distintive integer.

A quick check of these, and one can see that 2004 has distintive X solution, but not 1999.


  Posted by Michael on 2004-11-26 21:01:03
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