What maximum number of elements can be selected from the set $\{1, 2, 3, \dots, 100\}$ so that no sum of any three selected numbers is equal to a selected number? Proposed by A. Golovanov
Problem
Source: Tuymaada 1999, Q3
Tags: combinatorics unsolved, combinatorics
jmerry
01.08.2012 02:44
Two. Because if there are three elements, their sum is bigger than any of the three - and the selected set has a maximum. I think there's a mistake, and this is not the intended question.
HansV
18.09.2016 12:26
33, 34, 35, etc. evidently works, but I can't proof that it's not more.
Lepsvera
18.09.2016 12:40
jmerry wrote: Two. Because if there are three elements, their sum is bigger than any of the three - and the selected set has a maximum. I think there's a mistake, and this is not the intended question. I think there is like this, we colour the numbers, and sum of any 3 coloured numbers must not be other coloured number. and we want to know maximum number of coloured numbers
ThE-dArK-lOrD
18.09.2016 14:16
See here for generalized.