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I've received this grade school math problem but unfortunately algebra cannot be used. I am somewhat stumped and not convinced with my own answers. Here goes:

At a grade school election, there were 1080 voters. 5 students are running to be officers however there are only 3 seats available. What is the minimum number of votes needed to guarantee a seat?

Assume that everyone votes and that each person can only vote once. I also think the voting system might be culturally different, so the way it works here is that the 3 people with the highest votes are automatically officers. Basically the 1080 votes are spread over the 5 candidates, only the highest 3 are guaranteed positions.

I would appreciate both algebraic and non-algebraic answers.

Thank you very much!

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  • $\begingroup$ What is your own answer? $\endgroup$ – saulspatz Mar 9 at 15:28
  • $\begingroup$ What voting system is applied? $\endgroup$ – celtschk Mar 9 at 15:31
  • $\begingroup$ @celtschk plurality I suppose. It's an elementary math problem $\endgroup$ – user652360 Mar 9 at 15:33
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This is a trick question, if I understand it correctly. You must get at least $1$ vote. Two other candidates each get no votes, and the remaining two candidates split the other $1079$ votes between them.

If you get a quarter of the votes, it's not possible for three people to get more votes than you, because that would add up to more than four quarters, but since $1080$ is divisible by $4$ the selection could end in a four-way tie. The answe therefore is $${1080\over4}+1=271$$

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  • $\begingroup$ Hello! Thank you for this answer! I've recently updated the question with the assumption that everyone has to vote and can only vote once, it also looks like 2 people getting zero votes is not a possibility. There is an answer to the problem but it is unfortunately not yet available to me. $\endgroup$ – user652360 Mar 9 at 15:41
  • $\begingroup$ @user652360 Are you assuming everyone votes for himself? That isn't stated in the problem. If you want to assume that, then the minimum is $2$. $\endgroup$ – saulspatz Mar 9 at 15:43
  • $\begingroup$ I think the question is asking 'How many votes will guarantee a seat' $\endgroup$ – Daniel Mathias Mar 9 at 15:44
  • $\begingroup$ @DanielMathias That makes a lot more sense than my reading. $\endgroup$ – saulspatz Mar 9 at 15:45
  • $\begingroup$ Assuming each candidate is also a voter and will vote for self, $270$ votes will secure a seat. The remaining $810$ votes being distributed among four candidates with each getting at least one vote means at least two candidates will have fewer than $270$ votes. $\endgroup$ – Daniel Mathias Mar 9 at 15:55
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Assuming draws are allowed (in which case the higher index is then prefered), we have that $$ \left\{ \matrix{ 0 \le x_{\,1} \le x_{\,2} \le x_{\,3} \le x_{\,4} \le x_5 \hfill \cr x_{\,1} + x_{\,2} + x_{\,3} + x_{\,4} + x_5 = n \hfill \cr} \right. $$

Clearly the minimum for $x_3$ will occur when $x_1=x_2=0$, in which case $1$ vote (or also none, according to the preference rule) for $x_3$ is sufficient.

The challenge comes when $x_2$ get "many" votes.
The maximum he can reach without succeeding is when $$ \left\{ \matrix{ 0 = x_{\,1} \hfill \cr x_{\,2} \le x_{\,3} \le x_{\,4} \le x_5 \hfill \cr} \right.\quad \Rightarrow \,\quad \left\lfloor {n/4} \right\rfloor = x_{\,2} \le x_{\,3} $$

So the answer is that $x_3$ shall get minimum $\left\lfloor {n/4} \right\rfloor$ votes to be sure of being elected.

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