Combinatorics Challenge: Finding Equal Age Sums with 10 People in a Room

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Discussion Overview

The discussion revolves around a combinatorial problem involving 10 people in a room, each with an age between 1 and 60. The challenge is to prove that it is always possible to find two distinct groups of people whose ages sum to the same total. The scope includes theoretical exploration and mathematical reasoning related to combinatorics.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant expresses confusion about the problem's definition of "group" and suggests that a group consists of at least one person.
  • The same participant notes the potential for (2^10)-1 sums, ranging from 10 to 600, and speculates about the distribution of these sums.
  • Another participant mentions an existing thread on the same question, implying that the problem has been discussed previously.
  • A different participant references the pigeonhole principle, stating that with 1023 possible groups and only 591 possible sums, there must be at least two groups with the same sum. They also suggest that if groups share members, those can be removed without affecting the sum.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the problem's definitions or the implications of the pigeonhole principle. There are varying levels of understanding and interpretation of the problem, indicating that multiple views remain.

Contextual Notes

The discussion lacks clarity on the definition of "group" and the specific conditions under which the sums are calculated. There are also unresolved assumptions regarding the distribution of ages and the implications of the pigeonhole principle.

Sick0Fant
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I rarely care enough about one problem to ask for help, but there are a million problems that are similar to this one and I don't really understand any of them.

The problem I'm looking at reads:

In a room there are 10 people, none of whom are older than 60 (ages are considered as whole numbers only) but each of whom is at least 1 year old. Prove that one can always find two groups of people (with no common person) the sum of whose ages is the same.

The author failed to define "group" (which makes me even more confused), but I would assume that a group consists of at least one person.

I observe that there is a possibility of (2^10)-1 sums, each of which add up to between 10 and 600. I also believe that there are fewer ways to add up to 10 or 600 and that there are a lot of ways to add to 300.

If anyone can help me out in any way at all, I'd be appreciative.
 
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hahha there's already a thread on that exact same question. Sorry i Don't have the link...use the search option for combinatorics.
 
Must be in my class!
 
hmm, interesting question :smile:

https://www.physicsforums.com/showthread.php?t=130586

Basically, we have 1023 groups (or possible sums) and only 591 possible values for the groups (sums), so by the pigeonhole principle there are two groups with the same value. If the groups have a common person(s) just remove that person(s) from the group (the sum of the ages will still be the same).
 
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