Derivation.

We approach the game in reverse and first consider how a situation could look from which Mr. Precise can win in a single move. This is actually only possible if the teacup is exactly opposite the microwave door, i.e., "at the back," which is only possible at all for even $n$. In this case, Mr. Precise enters $n/2$ seconds, and the microwave has no choice but to turn the cup either clockwise or counterclockwise to "front."

We "see" that from all other positions, the microwave always has at least one option to not turn the cup to "front," but we will deal with this in more detail later.

To prevent the microwave from winning, it must avoid placing the cup at the "front" or "back" positions at any time.

We take another step back and consider how Mr. Precise could force the microwave to do exactly that. This is certainly possible when the teacup is exactly between these two positions, i.e., at $n/4$ or $3n/4$, if we label the positions clockwise with the numbers from 1 to $n$, such that $n$ is "front." In this case (which can only occur if $n$ is divisible by 4), Mr. Precise enters $n/4$, which forces the microwave to either place the cup at "front" and immediately lose, or place it at "back" and lose one move later.

We suspect again that for all positions other than these four, the microwave always has the option to not place the cup at "front" or "back," but we defer the proof to later.

We take another step back and see that Mr. Precise can achieve this from the positions $n/8, 3n/8, 5n/8,$ and $7n/8$ (if these are integers, i.e., if $n$ is divisible by 8) by entering $n/8$ each time, and again land on one of the four positions from which we already know the winning strategy.

This can be continued: Let $k$ be the largest natural number such that $2^k$ is a divisor of $n$, i.e., $n$ has the form $n = m \cdot 2^k$ with a suitable odd positive integer $m$. Then Mr. Precise can use the described winning strategy from all positions of the form

$$a \cdot \frac{n}{2^k} = a \cdot m$$

with a positive integer $1 \leq a \leq 2^k$. In the case where $n$ is a power of two (i.e., $m=1$), these are all $n$ positions, so Mr. Precise can always enforce a win for such $n$.

We now suspect that for every $n$ that is not a pure power of two (i.e., $m>1$), the microwave can prevent a win. We already know that it must avoid all positions that can be represented as in (2), i.e., those divisible by $m$. It is also sufficient to show that from every position that is not divisible by $m$, at least one of the two rotation directions leads to a position that is not divisible by $m$. This, in particular, also avoids "front" $= n = 2^k \cdot m$ permanently, preventing Mr. Precise from winning. This claim can be quickly shown with modulo considerations, which we will write down neatly in the following.

Since we do not want to give Mr. Precise and the microwave the entire derivation as game instructions, we will now write down the respective strategies concretely:

## Strategy.

The positions are labeled from 1 to $n$, and the "front" position has the number $n$.

In the case $n=1$, there is only the "front" position, so Mr. Precise always wins.

Let $n$ be a power of two. If the teacup is initially on an odd position, Mr. Precise enters one second, which ensures that the teacup will be on an even position regardless of the rotation direction. (If the teacup is already on an even position, he skips this step.)

If the teacup is now at a position divisible by 2 but not by 4, he enters 2, otherwise he skips this step again.

Thus, he can enforce that after at most $\ell$ moves, the teacup will be at a position divisible by $2^\ell$ (by entering $2^{\ell-1}$ in the $\ell$-th move or doing nothing, depending on whether the current position is already divisible by $2^\ell$ or not).

Now let $n$ not be a power of two, i.e., $n$ has the form $n = m \cdot 2^k$, where $m>1$ is a suitable odd integer. The microwave can now prevent the current position from ever being congruent to 0 modulo $m$. To do this, it initially turns the teacup to position 1 (or any other position not divisible by $m$).

Thereafter, in each move, it has the option to turn the teacup from the current position $a$ to either $a+b$ or $a-b$ modulo $n$ (where $b$ is the number of seconds entered by Mr. Precise). Suppose both were divisible by $m$, then their difference $2b$ would also be divisible by $m$. Since $m$ is odd, $b$ would also be divisible by $m$. But then $(a-b)+b=a$ would also be divisible by $m$, contradicting the fact that the teacup was on a position not divisible by $m$ at the beginning of the move. Therefore, the microwave has at least one option in each move that is not divisible by $m$ (and thus in particular not the number $n$ itself).

(Birgit Vera Schmidt)