There is a regular 2021-gon. We put a coin with heads up on every vertex of it. Every time, you can choose one vertex, and flip the coin on the vertices adjacent to it. Can you make all the coin tails up?
Problem
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Tags: combinatorics
30.12.2021 14:02
Since 2021 is an odd number and 0 is an even number, the answer is no.
12.09.2023 13:50
Since 2021, and 0, have different parity, the answer is no.
12.09.2023 14:00
mathmax12 wrote: Since 2021, and 0, have different parity, the answer is no. What's a parity? The only parity I know is special cases in the 4x4 or 6x6/ even numbered cubes that can't be solved with regular algorithms.
12.09.2023 14:24
when you make a move, there are only three possible outcomes: 1) you flip two coins with tails up, which subtracts 2 from total count of coins facing tails up(-2) ; 2)you flip one coin with heads up and one with tails up, this doesn't change total count of coins which are facing up tails(+0); 3) you flip two coins with heads up, which adds two to the total count of the coins with tails facing up(+2). starting from zero coins which are with tails facing up, we can either add two, subtract two or add zero, and we can't get 2021. (sorry for bad English)
21.09.2023 14:22
YIYI-JP wrote: mathmax12 wrote: Since 2021, and 0, have different parity, the answer is no. What's a parity? The only parity I know is special cases in the 4x4 or 6x6/ even numbered cubes that can't be solved with regular algorithms. Even and Odd
28.09.2023 14:15
2021 is odd and 0 is even so that would mean no.