First round – City competition

1.1. Let us denote by $2x_1,2x_2,\dots ,2x_{10}$ the number of coins that the knights sitting in chairs 1, 2, ..., 10 had in the beginning, respectively. We have to determine $2x_8$. We have a system of equations: $x_{10}+x_2=22,x_1+x_3=24,x_2+x_4=26,x_3+x_5=28,\dots ,x_8+x_{10}=38,x_9+x_1=40$. By combining those equations we get $$\begin{array}{rl}{x}_{10}& =38-x_8,\\ x_2& =22-x_{10}=22-(38-x_8)=x_8-16,\\ x_4& =26-x_2=26-(x_8-16)=42-x_8,\\ x_6& =30-x_4=30-(42-x_8)=x_8-12,\\ x_8& =34-x_6=34-(x_8-12)=46-x_8.\end{array}$$ From the last equation it follows that $2x_8=46$, so the knight that ended up with 36 coins had 46 coins in the beginning.