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. 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! 
 A: 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 $${1080\over4}+1=271$$ 
A: 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.
