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 $\displaystyle S$ 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 $\displaystyle S$ may have?
There are three types of integers:
. . (1) A multiple of 3: .$\displaystyle 3a$
. . (2) One more than a multiple of 3: .$\displaystyle 3b+1$
. . (3) One less than a multiple of 3: .$\displaystyle 3c-1$
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 $\displaystyle S$ 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.