Different Sums (Posted on 2004-07-22)

	Submitted by Brian Smith
Rating: 3.2500 (4 votes)
Solution:	(Hide)
The simplest method is to simply list out all the possible sums: 1+2+997 to 1+499+500 498 sums with 1 2+3+995 to 2+498+500 496 sums with 2 3+4+993 to 3+498+499 495 sums with 3 4+5+991 to 4+497+499 493 sums with 4 . . . 329+330+341 to 329+335+336 6 sums with 329 330+331+339 to 330+334+336 4 sums with 330 331+332+337, 331+333+336, 331+334+335 3 sums with 331 332+333+335 1 sum with 332 The sum of the subtotals is 1 + 3 + 4 + 6 + ... + 493 + 495 + 496 + 498 = 82834. Oskar gives a formula for any integer N here. Richard goes more in depth with an equivalent formula here.

	Subject	Author	Date
	re: With a little help from awk	Vee-Liem Veefessional	2004-07-30 07:08:55
	re(2): My solution	Richard	2004-07-24 14:54:36
	re: My solution	John	2004-07-24 06:17:39
	My solution	John	2004-07-24 06:15:21
	re(3): Curious second differences	Richard	2004-07-23 21:00:08
	Analytical Solution	Richard	2004-07-22 23:07:39
	re(2): Curious second differences	Oskar	2004-07-22 13:04:49
	blunt force with c - same ans as first	vectorboy	2004-07-22 11:18:51
	re: Curious second differences	Federico Kereki	2004-07-22 11:14:31
	Curious second differences	Federico Kereki	2004-07-22 11:09:54
	Ugly method with excel	Juggler	2004-07-22 11:06:05
	Human Approach	Nick Williams	2004-07-22 11:02:15
	re(3): With a little help from awk	Charlie	2004-07-22 10:46:56
	re(2): With a little help from awk	Charlie	2004-07-22 10:45:04
	re(2): With a little help from awk	Federico Kereki	2004-07-22 10:42:59
	re: With a little help from awk	Sing4TheDay	2004-07-22 10:25:15
	With a little help from awk	Federico Kereki	2004-07-22 10:00:34