Since $2021$ is an odd number and $0$ is an even number, the answer is no.

Since $2021,$ and $0,$ have different parity, the answer is no.

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.

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)

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

2021 is odd and 0 is even so that would mean $\boxed{\text{no}}.$