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?

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- May 3rd 2010, 05:59 PMChris11Interesting 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?

- May 3rd 2010, 09:31 PMSoroban
Hello, Chris11!

There is no formula for this problem (that I know of).

I had to reason it out . . .

Quote:

A set with only a finite number of integers as members,

has the property that the average of any 3 of its members isan integer.*not*

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 musthave 3 of type (1), or 3 of type (2), or 3 of type (3).*not*

. . Their averages would be integers.

We can have*at most two*of one type.

We musthave one of each type.*not*

. . 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.

- May 3rd 2010, 09:34 PMBlack
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. - May 4th 2010, 08:45 AMChris11
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!

- May 4th 2010, 09:01 AMPlato
- May 4th 2010, 10:54 AMChris11
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?