Summing Squares Sequentially! (Posted on 2004-10-01)

	Submitted by Old Original Oskar!
Rating: 2.0000 (2 votes)
Solution:	(Hide)
We can express the sum as (n-4999)²+(n-4998)²+...+(n+4999)² and we wish to show that it cannot be B to the P for integer B, P (P>1). The sum is 9999n²+2(1²+2²+...+4999²)= 9999n²+2x4999x5000x9999/6 = 3333(3n²+4999x5000) If this is B to the P, then 3 must divide B, and thus 3 to the P divides B to the P. However, 3333(3n²+4999x5000) can only be divided by 3, so P=1.

	Subject	Author	Date
	Puzzle Solution	K Sengupta	2007-05-23 05:26:49
	Correct Solution	np_rt	2004-10-03 10:49:21
	re(4): Solution (need more work)	np_rt	2004-10-02 10:38:36
	re(2): Solution	Ken Haley	2004-10-02 00:00:37
	re: Solution	Ken Haley	2004-10-01 23:57:21
	re(3): Solution (slight correction)	Federico Kereki	2004-10-01 16:11:28
	re(2): Solution (slight correction)	np_rt	2004-10-01 15:35:04
	re: Solution	Federico Kereki	2004-10-01 15:24:10
	Solution	np_rt	2004-10-01 14:15:05
	I am really smart	RandyOrton	2004-10-01 13:56:53