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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
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