## December 2009 riddle

This is a question I first heard from Ori Pomerantz.

#### Part 1

Consider all naturals from 2 and up, and divide them into two sets, A and B. Prove that in at least one of these sets there are three distinct numbers, a, b and c, such that ab=c.

#### Part 2

Show how to construct such a division into two infinite sets A and B so that only one of them will have such an a, b, c triplet.

Solve both questions to appear on the solvers' list.

### List of solvers:

James Ge (1 December 23:59)
Itsik Horovitz (2 December 10:23)
Gaoyuan Chen (2 December 20:42)
Li Wei (3 December 05:19)
Oded Margalit (3 December 23:09)
Omer Angel (4 December 05:57)
Shmuel Menachem Spiegel (4 December 15:15)
Øyvind Grotmol (6 December 06:50)
Ori Pomerantz (6 December 15:24)
Anurag Anshu (6 December 20:08)
Miao Hua (8 December 17:58)
Pei Wu (9 December 13:46)
Dan Dima (10 December 22:49)
Bojan Bašić (12 December 04:17)
Ante Turudić (14 December 23:36)
Harsha HS (16 December 21:38)
Ante Kovačić (17 December 12:10)
Victor Chang (17 December 13:48)
Hongcheng Zhu (26 December 18:51)

Elegant and original solutions can be submitted to the puzzlemaster at riddlesbrand.scso.com. Names of solvers will be posted on this page. Notify if you don't want your name to be mentioned.

The solution will be published at the end of the month.

Enjoy!