Thread: Interesting set/divisibility/counting question

A set S, with only a finite number of integers as members, has the property that the average of any 3 of its members is not an integer. What is the maximum number of elements that S may have?

Hello, Chris11!

There is no formula for this problem (that I know of).
I had to reason it out . . .

A set with only a finite number of integers as members,
has the property that the average of any 3 of its members is not an integer.
What is the maximum number of elements that may have?

There are three types of integers:

. . (1) A multiple of 3: .
. . (2) One more than a multiple of 3: .
. . (3) One less than a multiple of 3: .


We must not have 3 of type (1), or 3 of type (2), or 3 of type (3).
. . Their averages would be integers.

We can have at most two of one type.


We must not have one of each type.
. . Their average would be an integer.

We can have at most two types of numbers.


Hence, we can have (at most) two each of two types of numbers.

Therefore, set can have a maximum of 4 elements.

I believe the answer is 4.

If you look at the integers modulo 3, you have three possibilities: 0, 1, and -1. The set S cannot contain one of the following:

1) 0, 1, and -1
2) At least three repeating elements.

If the order of S is at least five, it's going to have one of the two.

Yep, that's exactly what I did, Black. Soroban, your solution is interesting in that it introduces notation that I didn't think of. Now try it in the general case!

Originally Posted by Chris11

Yep, that's exactly what I did, Black. Soroban, your solution is interesting in that it introduces notation that I didn't think of.
Now try it in the general case!

If the equivalence classes modulo three are note that one of the three must empty. WHY?
None of the three can have three members.

What do you mean by the general case?

Let S be a subset of the integers with the property that the average of any n of its members is not an integer. What is the maximum number of elements that S may have?