In an election there are 1080 votes, 5 electoral candidates but only 3 available seats. Calculate minimum number of votes needed to guarantee a seat.

Question

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!

Answer 1

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 $${1080\over4}+1=271$$

Answer 2

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.