If a set N
_{9} = {1, 2, 3, 4, 5, 6, 7, 8, 9} of 9 numbers is split into two subsets, then at least one of them contains three terms in arithmetic progression.
The statement is not true for a set N
_{8} of 8 integers.
Seems obvious?
Prove it.
The
N_{9} partition must have a subset with at least 5 numbers. If I didn't make any mistakes, there are four possibilities where this subset has no APs. In each case the other subset has an AP in its 4 numbers instead.
12489 / 3567
12679 / 3458
12689 / 3457
13489 / 2567
The N_{8} partition is easy. Here are the ways I found
1256 / 3478
1368 / 2457
1458 / 2367
Edited on March 25, 2017, 9:30 am

Posted by Jer
on 20170324 12:39:01 