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APMO 2014 Problem 4:

Prove: A 9 element subset of ${1,2,...,99}$ must have two distinct subsets with the same sum.

I am having a lot of trouble with this problem. The official solution: https://cms.math.ca/Competitions/APMO/sol/apmo2014-sol.pdf, as usual, does not provide motivation, so I can't learn anything from it.

Here are some easy observations:

There are $511$ nonempty subsets of the 9 element set. The problem is solved by pigeonhole if we can show that such a set must generate less than $511$ distinct subset sums.

The weakest bound on the number of distinct subset sums is:

$(99+...+91)-(9+..+1)=810$

If you find a solution, please include the motivation.

RobPratt
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math_lover
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  • There are way more $9$-element subsets than that. $\binom{99}{9}$ is large. There are $512$ subsets of any $9$-element set. – Thomas Andrews Jan 09 '16 at 00:03
  • also, the subset sum can be anywhere from zero to the maximum. – Thomas Andrews Jan 09 '16 at 00:07
  • What exactly do you mean by include the "motivation"? Are you asking for an application to a different area of mathematics for a competition problem? – theREALyumdub Jan 09 '16 at 00:10
  • ^^^ my apologies. I wrote down what I was thinking incorrectly. Fixed now. @theREALyumdub: most official solutions to Olympiad problems are written backwards, so that it appears like it came out of nowhere. By motivation I mean the thought process, not a completed solution which doesn't help me know how to approach similar problems – math_lover Jan 09 '16 at 00:41
  • Are the numbers drawn from $1, 2, \ldots, 99$, or from $1, 2, \ldots, 100$? I'm guessing $99$, but please confirm. – Brian Tung Jan 09 '16 at 00:48
  • It's 99. Was I unclear? – math_lover Jan 09 '16 at 02:28
  • The title says 100... – kccu Jan 09 '16 at 02:34
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    Oh dear. I should not have written this from my phone... I apologize for the mistakes... – math_lover Jan 09 '16 at 02:40

1 Answers1

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NOTE: As mentioned in the comments, the problem I solve below may not be the one given in the exam.

This approach differs from that in the solution manual.

Let $a_i$ be the ordered elements of any 9-element subset.

If $a_1+a_2+\ldots+a_7 \leq 99$, you have at least 128 (= $2^7$) sums less than 99, so there must be a repeat by the piegonhole principle.

We now consider cases:

  • If $a_1+\ldots+a_7 \geq 100$, but $a_1+\ldots+a_6 \leq 99$, then $a_7 \geq 18$

$a_1+\ldots+a_7 \geq 100$ and $a_8+a_9 \geq 19+20$, so total $\geq 139$

$a_1+\ldots+a_6 \leq 99$ and $a_7+\ldots+a_9 \leq 97+98+99$, so total $\leq 393$

255 possible sums, but 512 possible combinations, so pigeonhole.

  • If $a_1+\ldots+a_6 \geq 100$, but $a_1+\ldots+a_5 \leq 99$, then $a_6 \geq 20$

$a_1+\ldots+a_6 \geq 100$ and $a_7+\ldots+a_9 \geq 21+22+23$, so total $\geq 166$

$a_1+\ldots+a_5 \leq 99$ and $a_6+\ldots+a_9 \leq 96+97+98+99$, so total $\leq 489$

324 possible sums, 512 possible combinations, pigeonhole.

  • If $a_1+\ldots+a_5 \geq 100$, but $a_1+..+a_4 \leq 99$, then $a_5 \geq 22$

$a_1+\ldots+a_5 \geq 100$ and $a_6+\ldots+a_9 \geq 23+24+25+26$, so total $\geq 198$

$a_1+\ldots+a_4 \leq 99$ and $a_5+\ldots+a_9 \leq 95+96+97+98+99$, so total $\leq 584$

387 possible sums, 512 possible combinations, pigeonhole.

  • If $a_1+\ldots+a_4 \geq 100$ but $a_1+..+a_3 \leq 99$, then $a_4 \geq 27$

$a_1+\ldots+a_4 \geq 100$ and $a_5+\ldots+a_9 \geq 28+29+30+31$, so total $\geq 218$

$a_1+\ldots+a_3 \leq 99$ and $a_4+\ldots+a_9 \leq 94+95+96+97+98+99$, so total $\leq 678$

461 possible sums, 512 possible combinations, pigeonhole.

  • If $a_1+\ldots+a_3 \geq 100$ but $a_1+a_2 \leq 99$, then $a_3 \geq 35$

$a_1+\ldots+a_3 \geq 100$ and $a_4+\ldots+a_9 \geq 36+37+38+39+40$, so total $\geq 290$

$a_1+a_2 \leq 99$ and $a_3+\ldots+a_9 \leq 93+94+95+96+97+98+99$, so total $\leq 771$

482 possible sums, 512 possible combinations, pigeonhole.

  • If $a_1+a_2 \geq 100$, $a_2 \geq 51$

$a_1+a_2 \geq 100$ and $a_3+\ldots+a_9 \geq 52+53+54+55+56+57+58$, so total $\geq 485$

$a_1+\ldots+a_9 \leq 91+92+93+94+95+96+97+98+99$, so total $\leq 855$

371 possible sums, 512 possible combinations, pigeonhole.

  • I don't think this method of proof is correct. For example, the fact that $\sum_1^9a_i$ lies between $218$ and $678$ does not mean there are $461$ possible sums since the sums can actually range from $a_1$ to $678$. Is there something I'm misunderstanding in your method? –  Dec 09 '19 at 12:54
  • @S.Dolan The sets have exactly 9 elements. Does that help? If not, let me know, I'm not perfect and can make mistakes. –  Dec 09 '19 at 17:10
  • Thanks for getting back to me. However, the elements being added are the elements of subsets of a set with 9 elements. (The OP seems to get this wrong when writing his "weakest" bound.) –  Dec 09 '19 at 17:19
  • @S.Dolan OK, I was relying on the OP's accuracy, but you're right. I think i proved something completely different from what was asked. I'll add a note and maybe see if I can convert my answer to a correct one later. –  Dec 09 '19 at 17:20
  • Yes, you might like to have a go at the question I have posed at https://math.stackexchange.com/questions/3468730/prove-that-an-s-element-subset-of-1-2-n-must-have-two-distinct-subsets-w/3469110#3469110. I should be interested in more input into how to tackle this type of problem. –  Dec 09 '19 at 17:25