Trushin 5.
100 thimbles are arranged in a circle. Under one of them, a coin is hidden. In one move, it is allowed to flip over four thimbles and check if the coin is under one of them. After this, they are returned to their original position, and the coin is moved to one of the adjacent thimbles. What is the minimum number of moves required to definitely find the coin?
After each of our moves and the movement of the coin, we will turn all the thimbles (along with the coin) one position clockwise. Then we can consider that after each move, the coin either stays in place or moves two positions clockwise.

Let's number all the thimbles counterclockwise in order from 0 to 99. It is clear that the parity of the number of the thimble under which the coin lies does not change. We will call a thimble empty if there cannot be a coin under it at the moment (initially there are no empty thimbles). Obviously, after each of our moves, the number of empty thimbles increases by no more than 4. Note that if after our move there are empty thimbles with numbers of a certain parity that are neither 0 nor 50, then there will be an empty thimble with the same parity.

There are only two situations in which the number of empty thimbles does not increase: the first is when 50 thimbles with numbers of one parity are empty, and 50 of the other parity are all non-empty (we can assume that this situation has only occurred once!), and the second is the final situation, when all thimbles are empty. After each of the other moves and the movement of the coin, the number of empty thimbles increases, but by no more than 3. Evaluation. Suppose no more than 32 moves have been made, but the coin has not yet been found. Then by this time, there are no more than $30 \cdot 3 + 2 \cdot 4 = 98$ empty thimbles, which means there are still at least two thimbles under which the coin can be. Therefore, the search is not over. Thus, at least 33 moves are required.

Algorithm. We will lift the thimbles numbered $0,2,4,6$. If the coin is not under them, then after the movement, it cannot end up under thimbles $0,2,4$, as all their neighbors were empty. (We will always assume that the coin was not found by the next move - otherwise, everything is already done.)

Suppose that after some move (and the subsequent movement of the coin) thimbles with numbers $0,2,4, \ldots, 2s$ became empty. Then the next move we will lift thimbles with numbers $2s+2, 2s+4, 2s+6, 2s+8$; after the movement, the coin cannot end up under thimbles with numbers $0,2,4, \ldots, 2s+6$. Thus, the number of empty thimbles has increased by 3.

So, after 16 moves, $16 \cdot 3 = 48$ even thimbles will be empty. On the 17th move, we will check the remaining two even thimbles: 98 and 100, as well as two odd ones: with numbers 1 and 3. After this move (before the movement of the coin), all even thimbles will be empty and will remain empty until the end. We then proceed according to the same scheme. After the 17th move and the movement of the coin, one odd thimble remains empty; therefore, after another 15 moves, there will be $1 + 3 \cdot 15 = 46$ empty odd thimbles (and all even ones). Then, with the last, 33rd move, we can check the remaining four thimbles and guarantee finding the coin (if, of course, we haven't found it before).

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