16

There are $100$ integers on the board: the first number is between $0$ and $99$ (inclusive), the second is between $-1$ and $98$, ...., the $100$th is between $-99$ and $0$. Prove that the sum of some of these numbers is equal to $0$.

So, first we choose one number from each interval $[0,99], [-1,98]...,[-99,0]$, (chosen numbers DON'T have to be different, we can choose two equal numbers) let's call that set $A$. We have to prove there is a nonempty subset $B\subset A$ such that the sum of its elements is $0$.

So far, I got this idea. Define $a_0\in[0,99],a_1\in[-1,98],...,a_{99}\in[-99,0]$ to be the written numbers. Then, let $b_i=a_i+i, \forall 0\le i\le99$.

Then, the intervals now become $b_0\in[0,99],b_1\in[0,99],...,b_{99}\in[0,99]$

in other words, all the $b_i$ are in the same interval.

That's what I got so far, I feel like this is useful, but I don't know how to continue. I've been stuck at this for a while...

edit: as the comments suggested, here's the backstory of this problem; my professor (who is by the way an ex IMO contestant) got this problem from the prepperations for the IMO, so the source is unknown. This summer camp, he decided to give us this problem, and no one could solve it, so he told us to move on. To this day, still no one solved it :(.

RobPratt
  • 50,938
Peter
  • 582
  • 2
  • 15
  • 2
    I assume you're talking about integers, not real numbers. Because this is obviously false if the numbers aren't required to be integers. – Robert Shore Aug 15 '24 at 00:02
  • 1
    What is the source of this problem, please? – Gerry Myerson Aug 15 '24 at 00:52
  • 1
    I have an impression that the $100$ integers may have duplicates, though your later paragraph uses the words "set" and "subset". – peterwhy Aug 15 '24 at 02:24
  • It would be better if the question were edited to refer to "multisets" and "submultisets". – Gerry Myerson Aug 15 '24 at 03:22
  • 1
    Here's a hint: is the number $100$ really crucial here ? If not, try to work out much smaller examples to get a feel for what the punchline might be. – François Gatine Aug 15 '24 at 06:33
  • I still cannot solve it... can I please get a full solution? – Peter Aug 15 '24 at 11:27
  • 5
    WHAT IS THE SOURCE OF THIS QUESTION, PLEASE? – Gerry Myerson Aug 15 '24 at 12:49
  • I really don't know... I am on a math summer camp, and our professor did not state the source anywhere on the paper. – Peter Aug 15 '24 at 13:03
  • 1
    Per @GerryMyerson 's line of (repeated) questioning: Is this a problem that your math summer camp professor wants you to do, and/or is it one that can be comfortably outsourced to the internet e.g. MSE? – Benjamin Dickman Aug 15 '24 at 13:22
  • The problem can be comfortably outsourced, our professor (btw he is still in college) didn't come up with it and it is not from a 2024 shortlist of any contests. Also, and I qoute: "the solution is really dumb and you don't have to spend a whole lot of time solving it". All in all, it can be outsourced, it is no problem (I double checked with my professor, he said I can share it) – Peter Aug 15 '24 at 13:32
  • Also, I apologize I didn't answer earlier, I simply forgot... – Peter Aug 15 '24 at 13:33
  • And also, if you're wondering, my professor is an ex IMO contestant, so last year they had a preperation for IMO, and that's where he found this problem. So that's why the source is unknown. – Peter Aug 15 '24 at 13:35
  • All that information should be edited into the body of the question, not buried down here in the comments, please. – Gerry Myerson Aug 15 '24 at 22:42
  • Maybe the solution of this related problem can be adapted to work for this question. – Fabius Wiesner Aug 16 '24 at 20:18
  • @FabiusWiesner: this problem is likely another application of the pigeonhole principle, but I’m skeptical it’s going to be as straightforward as the linked question (unless I’m even rustier than I thought…). – Aphelli Aug 17 '24 at 19:49
  • 1
    @Peter You've reduced this problem to this other one, where my solution is presented in graph theory language is equivalent to abacaba's approach. You define a directed graph and show that there is a loop, which gives $ \sum b_i = \sum i$ and hence $ \sum a_i = 0 $ as desired. – Calvin Lin Sep 05 '24 at 22:37
  • @CalvinLin Yes, you are absolutely right! I never thought it could be solved using graph theory. Thanks for mentioning the post! – Peter Sep 06 '24 at 20:44
  • @CalvinLin Thanks for pointing this out! – abacaba Oct 20 '24 at 19:28

1 Answers1

23

Let $b_0, b_1, \cdots, b_{99}$ be integers, such that $b_i \in [-i, 99-i]$ for every $i$.

Consider the sequence $S_0, S_1, \cdots$ defined recursively by $$S_0 = 0, S_{i + 1} = S_i + b_{S_i}.$$ Since $b_{S_i} \in [-S_i, 99 - S_i]$, we have $S_{i + 1} \in [0, 99]$, so the sequence $(S_i)_{i\in[0, 100]}$ is well-defined and always takes integer values between $0$ and $99$ inclusive. Therefore, by the pigeonhole principle, there exists $i < j$ such that $S_i = S_j$. Take the smallest $j$ such that this happens. Then we have $$S_{j} = S_i + b_{S_i} + b_{S_{i + 1}} + \cdots + b_{S_{j-1}}$$ so $$b_{S_i} + b_{S_{i + 1}} + \cdots + b_{S_{j-1}} = 0$$ and by minimality of $j$, the indices $S_i, \cdots, S_{j - 1}$ are distinct. We have found the desired subset of the integers that sum to $0$.

Some thought process behind the proof.

In the context of math competitions, there are only two common tricks to show statements like "a subset-sum of $S$ equals $0$". (In modern additive combinatorics there are far more tricks.) The first trick is to use the generating function $$\prod_{s \in S} (1 + x^s)$$ which is unlikely to work here. The second way is to arrange the elements of $S$ as a sequence $s_1, \cdots, s_{n}$, and use the pigeonhole principle on the subsequence sums $$S_i = s_1 + \cdots + s_i.$$ For pigeonhole to work, we need to keep the values of $S_i$ in a short interval. This, combined with the strange condition given by the problem, leads naturally to the choice $s_{i + 1} = b_{S_i}$.

As @CalvinLin has pointed out in the comment section, this solution is equivalent to their solution in this post.

abacaba
  • 11,210