Let S consist of 16 elements of the set (1,2,3, ... , 106) so
that no two elements of S differ by 6; 9; 12; 15; 18; or 21.
Prove that at least two of
the elements of S must differ by 3.
Divide the 16 elements into three groups with similar remainders mod 3. At least one such group must have at least 6 members (If all groups had at most 5 then there could only be 15 total which is too few.)
These 6 elements must all have differences that are multiples of three. Suppose none differ by exactly 3. Then the smallest difference for each pair is 24. In particular, if we sort the 6 elements numerically, the difference between each successive pair would have to be >= 24. If the smallest number is N, then the 6th number would have to be at least N + 24*5 = N + 120. Even if N =1, this last number would be too big to fit in the set [1..106]. Accordingly, the assumption that no pair difference is 3 must be false, and there must be at least one pair that differs by exactly 3.

Posted by Paul
on 20151027 11:19:17 