1. From 37 coins arranged in a row, 9 are "tails" up, and 28 are "heads" up. In one step, any 20 coins are flipped. Is it possible after several steps for all coins to be "tails"? "heads"? In how few steps is this possible?
Solution. Let the first step be to flip $x$ "tails" to "heads". Then $20-x$ "Heads" will be flipped to "tails". After this step, we have $9-x+(20-x)=29-2x$ "tails" and $x+(28-(20-x))=8+2x$ "heads". After the first step, we will have an odd number of "tails" and an even number of "heads". Clearly, this rule will remain in every subsequent step. Therefore, we will always have an odd number of "tails", meaning we will never get all coins to be "heads". Since the number of heads remains even after the first step, we can choose $x$ such that $2x+8=20$, i.e., $x=6$. After the first step, we will have 17 "tails" and 20 "heads". In the second step, we flip the 20 "heads" to "tails", and all coins will be flipped to "tails" up. Therefore, all coins will be flipped to "tails" up in at least 2 steps.