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Algebra Difficulty 6.5 National Olympiad Prove it Croatia

Ten chairs are arranged around a round table and marked with numbers 11 to 1010 successively (in such a way that chairs 11 and 1010 are also adjacent), and a knight is sitting in each chair. In the beginning, every knight has an even number of coins. Simultaneously, each knight gives half of his coins to his left neighbour, and the other half to his right neighbour. After that, the knight sitting in chair 11 has 2222 coins, and each succeeding knight has two more coins, up until the knight in chair 1010 that has 4040 coins.
How many coins did the knight that ended up with 3636 coins have in the beginning?
(Hong Kong)

Solution

Let us denote by 2x1,2x2,…,2x102x_1, 2x_2, \dots, 2x_{10} the number of coins that the knights sitting in chairs 1,2,…,101, 2, \dots, 10 had in the beginning, respectively. We have to determine 2x82x_8.

We have a system of equations: x10+x2=22x_{10} + x_2 = 22, x1+x3=24x_1 + x_3 = 24, x2+x4=26x_2 + x_4 = 26, x3+x5=28x_3 + x_5 = 28, …\dots, x8+x10=38x_8 + x_{10} = 38, x9+x1=40x_9 + x_1 = 40.

By combining those equations we get
x10=38−x8,x2=22−x10=22−(38−x8)=x8−16,x4=26−x2=26−(x8−16)=42−x8,x6=30−x4=30−(42−x8)=x8−12,x8=34−x6=34−(x8−12)=46−x8. \begin{aligned} 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{aligned}
From the last equation it follows that 2x8=462x_8 = 46, so the knight that ended up with 3636 coins had 4646 coins in the beginning.

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