The Universe of Discourse

Fri, 15 Dec 2017

This math.se question asks how to show that, among any 11 integers, one can find a subset of exactly six that add up to a multiple of 6. Let's call this “Ebrahimi’s theorem”.

This was the last thing I read before I put away my phone and closed my eyes for the night, and it was a race to see if I would find an answer before I fell asleep. Sleep won the race this time. But the answer is not too hard.

1. First, observe that among any five numbers there are three that sum to a multiple of 3: Consider the remainders of the five numbers upon division by 3. There are three possible remainders. If all three remainders are represented, then the remainders are !!\{0,1,2\}!! and the sum of their representatives is a multiple of 3. Otherwise there is some remainder with three representatives, and the sum of these three is a multiple of 3.

2. Now take the 11 given numbers. Find a group of three whose sum is a multiple of 3 and set them aside. From the remaining 8 numbers, do this a second time. From the remaining 5 numbers, do it a third time.

3. We now have three groups of three numbers that each sum to a multiple of 3. Two of these sums must have the same parity. The six numbers in those two groups have an even sum that is a multiple of 3, and we win.

Here is a randomly-generated example:

$$3\quad 17\quad 35\quad 42\quad 44\quad 58\quad 60\quad 69\quad 92\quad 97\quad 97$$

Looking at the first 5 numbers !!3\ 17\ 35\ 42\ 44!! we see that on division by 3 these have remainders !!0\ 2\ 2\ 0\ 2!!. The remainder !!2!! is there three times, so we choose those three numbers !!\langle17\ 35\ 44\rangle!!, whose sum is a multiple of 3, and set them aside.

Now we take the leftover !!3!! and !!42!! and supplement them with three more unused numbers !!58\ 60\ 69!!. The remainders are !!0\ 0\ 1\ 0\ 0!! so we take !!\langle3\ 42\ 60\rangle!! and set them aside as a second group.

Then we take the five remaining unused numbers !!58\ 69\ 92\ 97\ 97!!. The remainders are !!1\ 0\ 2\ 1\ 1!!. The first three !!\langle 58\ 69\ 92\rangle!!have all different remainders, so let's use those as our third group.

The three groups are now !! \langle17\ 35\ 44\rangle, \langle3\ 42\ 60\rangle, \langle58\ 69\ 92\rangle!!. The first one has an even sum and the second has an odd sum. The third group has an odd sum, which matches the second group, so we choose the second and third groups, and that is our answer:

$$3\qquad 42\qquad 60\qquad 58 \qquad 69 \qquad 92$$

The sum of these is !!324 = 6\cdot 54!!.

This proves that 11 input numbers are sufficient to produce one output set of 6 whose sum is a multiple of 6. Let's write !!E(n, k)!! to mean that !!n!! inputs are enough to produce !!k!! outputs. That is, !!E(n, k)!! means “any set of !!n!! numbers contains !!k!! distinct 6-element subsets whose sum is a multiple of 6.” Ebrahimi’s theorem, which we have just proved, states that !!E(11, 1)!! is true, and obviously it also proves !!E(n, 1)!! for all larger !!n!!.

I would like to consider the following questions:

1. Does this proof suffice to show that !!E(10, 1)!! is false?
2. Does this proof suffice to show that !!E(11, 2)!! is false?

I am specifically not asking whether !!E(10, 1)!! or !!E(11, 2)!! are actually false. There are easy counterexamples that can be found without reference to the proof above. What I want to know is if the proof, as given, contains nontrivial information about these questions.

The reason I think this is interesting is that I think, upon more careful examination, that I will find that the proof above does prove at least one of these, perhaps with a very small bit of additional reasoning. But there are many similar proofs that do not work this way. Here is a famous example. Let !!W(n, k)!! be shorthand for the following claim:

Let the integers from 1 to !!n!! be partitioned into two sets. Then one of the two sets contains !!k!! distinct subsets of three elements of the form !!\{a, a+d, a+2d\}!! for integers !!a, d!!.

Then:

Van der Waerden's theorem: !!W(325, 1)!! is true.

!!W()!!, like !!E()!!, is monotonic: van der Waerden's theorem trivially implies !!W(n, 1)!! for all !!n!! larger than 325. Does it also imply that !!W(n, 1)!! is false for smaller !!n!!? No, not at all; this is actually untrue. Does it also imply that !!W(325, k)!! is false for !!k>1!!? No, this is false also.

Van der Waerden's theorem takes 325 inputs (the integers) and among them finds one output (the desired set of three). But this is extravagantly wasteful. A better argument shows that only 9 inputs were required for the same output, and once we know this it is trivial that 325 inputs will always produce at least 36 outputs, and probably a great many more.

Proofs of theorems in Ramsey theory are noted for being extravagant in exactly this way. But the proof of Ebrahimi's theorem is different. It is not only frugal, it is optimally so. It uses no more inputs than are absolutely necessary.

What is different about these cases? What is the source the frugality of the proof of Ebrahimi’s theorem? Is there a way that we can see from examination of the proof that it will be optimally frugal?

Ebrahimi’s theorem shows !!E(11, 1)!!. Suppose instead we want to show !!E(n, 2)!! for some !!n!!. From Ebrahimi’s theorem itself we immediately get !!E(22, 2)!! and indeed !!E(17, 2)!!. Is this the best we can do? (That is, is !!E(16, 2)!! false?) I bet it isn't. If it isn't, what went wrong? Or rather, what went right in the !!k=1!! case that stopped working when !!k>1!!?

I don't know.