Make it equal (Posted on 2010-05-26)

	Submitted by Ady TZIDON
Rating: 3.0000 (2 votes)
Solution:	(Hide)
Let us show that once a solution exist for (1,2,3,... n), creating a 3E-Partition for (1,2,3,...n+2, n+3) is trivial . My proof: Let the 3E-Partition for sum(n) be S1, S2, S3. We have to add 3 additional numbers n+1 , n+2 ,n+3 i.e. to increase each subset by n+2. We will replace number 1 (present in one of the subsets, say S2) by n+3; the number n+2 will go to another subset (e.g. S1) without replacement. Two unassigned numbers 1 and n+1 go to the last subset (S3 in our example). Since existence of a solution for n=5 and 4=6 was shown de facto, the process of full induction is concluded. Ans:5,6.8.9.11,12.....(3n-1),(3n),....

	Subject	Author	Date
	re: Full proof.	Ady TZIDON	2010-05-27 13:51:04
	Full proof.	Jer	2010-05-27 11:32:44
	re: No problems	Ady TZIDON	2010-05-27 06:19:31
	No problems	Steve Herman	2010-05-27 01:32:23
	re: half of it SOME HINTS(a partial spoiler)	Ady TZIDON	2010-05-26 19:29:38
	Doesn't work for n = 10	Steve Herman	2010-05-26 18:20:31
	half of it	Daniel	2010-05-26 16:52:29