Trushin B.
There are 100 thimbles arranged in a circle. Under one of them, a coin is hidden. In one move, you are allowed to flip over four thimbles and check if the coin is under one of them. After this, the thimbles 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 rotate 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 the coin cannot be 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 the number of empty thimbles with a certain parity is neither 0 nor 50, then there will be an empty thimble with the same parity, one position away from which (counterclockwise) there is a non-empty thimble, and therefore, after moving the coin, this thimble will also become non-empty. Clearly, during the search, only two situations can arise in which the number of empty thimbles does not increase: the first - 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 occurs only once!), and the second - 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.

Estimate. Suppose no more than 32 moves have been made, but the coin has not yet been found. Then at this point, the number of empty thimbles is no more than $30 \cdot 3 + 2 \cdot 4 = 98$, 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. (From now on, we will always assume that the coin was not found with the current move - otherwise, everything is already done.)

Suppose after some move (and the subsequent movement of the coin) thimbles numbered $0, 2, 4, \ldots, 2s$ became empty. Then the next move we will lift thimbles numbered $2s+2, 2s+4, 2s+6, 2s+8$; after the movement, the coin cannot end up under thimbles numbered $0, 2, 4, \ldots, 2s+6$. 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: numbered 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 will 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, $1 + 3 \cdot 15 = 46$ odd thimbles (and all even ones) will be empty. Then, with the last, 33rd move, we can check the remaining four thimbles and guarantee finding the coin (if, of course, we did not find it earlier).