Evie and Odette are playing a game. Three pebbles are placed on the number line; one at -2020 , one at 2020, and one at $n$, where $n$ is an integer between -2020 and 2020. They take it in turns moving either the leftmost or the rightmost pebble to an integer between the other two pebbles. The game ends when the pebbles occupy three consecutive integers.
Odette wins if their sum is odd; Evie wins if their sum is even. For how many values of $n$ can Evie guarantee victory if:
(a) Odette goes first;
(b) Evie goes first?
Solution
First note that the game must end after a finite number of moves, because the difference of the positions of the outer pebbles is a strictly decreasing sequence of positive integers until it is impossible to make a move (and so one player will win).

Claim 1: If Odette goes first she can force a win for any $n$.
Claim 2: If Evie goes first she can force a win if and only if $n$ is odd.
These claims show the answer to the questions are 0 and 2020 respectively.
Proof of claim 1
Odette can play in such a way that after every turn Evie takes, the two outer pebbles are even.
If the configuration before Os turn is even, even, even $\mathrm{O}$ can move to another such configuration, or, if that is impossible, to a winning configuration for her.
If the configuration is even, odd, even and O has not already won, then she can move an outer pebble to an even number directly adjacent to the odd pebble. This will force $E$ to move the odd pebble back between the two even ones.
This strategy ensures that the only way either player can end the game is by moving to consecutive numbers where the middle one is odd: a win for $\mathrm{O}$.

Proof of claim 2
If the initial configuration is even, even, even before Es turn, then the argument from claim 1 shows that $\mathrm{O}$ can force a win.

If the initial configuration is even, odd, even, then E can move an outer even pebble to an odd number adjacent to the other even one. This forces $\mathrm{O}$ to move the even pebble, so the configuration after Os turn will have both outer pebbles odd.

Now the situation is identical to that one studied earlier but with the words even and odd interchanged, so E can arrange that in all subsequent configurations, the outer pebbles are both odd. The game must finish with adjacent places odd, even, odd and so Evie will win.

MARKERs' COMMENTS
This problem was found very hard. Only a minority of students found time to attempt it, and only a minority of attempts purported to be full solutions. A number of students had very good ideas, but wrote up only special cases: it is important to remember that a winning strategy has to explain what the winning player should do to win given any possible choice of moves by the other player.

In practice, those students frequently did well who observed (and who made clear that they observed) that moving two pebbles to adjacent positions forces the other player to move the outermost of the two on their next move.