Olympiad Maths Prep

Track / Stage 6 / 87 of 400 #1087 of 2000

Problem 1087

National olympiad, first round
Combinatorics Difficulty 6.1 Prove it

Example: 317×17317 \times 17 coins are arranged in a square. Initially, all coins are heads up. Each time, five adjacent coins (either horizontally, vertically, or diagonally) are flipped.
After a finite number of such flips, is it possible for all the coins to be tails up?

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

It is impossible. As follows, each coin is marked, then:
(1) The five coins flipped each time are AA, BB, CC, DD, EE.
\begin{tabular}{cccccccccc}
AA & BB & CC & DD & EE & AA & BB & \cdots & AA & BB \\
DD & EE & AA & BB & CC & DD & EE & \cdots & DD & EE \\
BB & CC & DD & EE & AA & BB & CC & \cdots & BB & CC \\
EE & AA & BB & CC & DD & EE & AA & \cdots & EE & AA \\
CC & DD & EE & AA & BB & CC & DD & \cdots & CC & DD \\
AA & BB & CC & DD & EE & AA & BB & \cdots & AA & BB \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & & \vdots & \vdots \\
AA & BB & CC & DD & EE & AA & BB & \cdots & AA & BB \\
DD & EE & AA & BB & CC & DD & EE & \cdots & DD & EE
\end{tabular}
(2) At the same time, for each mark AA, BB, DD, EE, there are 58 coins, and for CC there are 57 coins.
(3) If all the coins are finally face down, then each coin must have been flipped an odd number of times.

From (2) and (3), it is known that all the coins marked AA have been flipped an even number of times, and all the coins marked CC have been flipped an odd number of times. Therefore, the number of times the coins marked AA are flipped is different from the number of times the coins marked CC are flipped, which contradicts (1). Therefore, it is impossible.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.