On a table are 10 coins, all with "heads" facing up. A move consists of flipping exactly 4 coins.

a) What is the minimum number of moves required for all to be "tails" facing up?

b) If there were 11 coins, would it be possible to have all of them with tails facing up?
Solution

a) With 2 moves it is not possible, as we can only change $2 \cdot 4=8$ coins and we need to change 10. We will now show that it is possible with 3 moves, analyzing the sequence in the figure, where $k$ represents heads and $C$, tails.

![](https://cdn.mathpix.com/cropped/2024_05_01_3fc4f3a6ef7302a792fcg-08.jpg?height=613&width=1317&top_left_y=1465&top_left_x=272)

b) We always flip 4 coins (an even number) with each move. Initially, the number of heads (10) and tails (0) is even. Each time we flip 4 coins, the number is either even in both heads and tails or odd in both, since the sum of the quantities flipped is even (4). Thus, after each move, the number of heads and tails has the same parity (both even or both odd). Therefore, we conclude that it is not possible to have all 11 coins with "tails" facing up, as the number of tails (11) would be odd and the number of heads (0) would be even.