Hamel wrote: Alice and Bob play the following game: They start with non-empty piles of coins. Taking turns, with Alice playing first, each player choose a pile with an even number of coins and moves half of the coins of this pile to the other pile. The game ends if a player cannot move, in which case the other player wins. Determine all pairs $(a,b)$ of positive integers such that if initially the two piles have $a$ and $b$ coins respectively, then Bob has a winning strategy. Very similar with http://emc.mnm.hr/wp-content/uploads/2016/12/EMC_2016_Seniors_ENG.pdf

I wouldn't state that. It probably has a whole different idea behind it.

We focus on the greatest power of $2$ dividing each pile. Let these be $x,y$. We begin by focussing on how these can change. If we choose pile $v_2(a)=x$ then: $$(x,y) \rightarrow \begin{cases} (x-1,y) & x \geq y+2 \\ (x-1,z) \, , \, z \geq x & x=y+1 \\ (x-1,x-1) & x \leq y \end{cases}$$Note in the first two moves the $min(x,y)$ is left unchaged. If at the beginning $v_2(a)=v_2(b)$ then only moves of the third type can be carried out so it is clear if $2 \vert v_2(a)$ then Bob wins otherwise Alice wins. Otherwise WLOG $v_2(b)>v_2(a)$. If $2 \not \vert v_2(a)$ then Alice first carries out the third type of move and then as above Alice will go on to win. If $2 \vert v_2(a)$ then if at any point either player reduces $min(x,y)$ (i.e. uses move type 3) then it decreases by exactly $1$ and hence we then have $2 \not \vert min(x,y)$ so the opponent will go on to win (using move type 3 repeatedly). Hence both players will only use move types $1,2$. This can continue as long as $x \neq y$ but both moves $1,2$ preserve $x,y$ being different so in the case the game is a draw. So Bob has a winning strategy when $\boxed{2 \vert v_2(a)=v_2(b)}$

This problem proposed by Dimitris Christofides (aka demetres) from Cyprus.

Isn't this problem trivial? If $a,b$ are both odd, the game doesn't start. If one is odd and the other even, it is easy to check that the game never ends and we always get an odd, even pair. So consider $a,b$ are both even. In the following solution, a number written as $2^uz$ for some $u$ will mean any number $n$ with $v_2(n)=u.$ Note that $6, 10$ are both $2z,$ while $3$ is $z.$ Also note that we use $z$ for this purpose and other variables will be used as usual, like $4=2k,$ where $k=2.$ Now $\left( 2x,2y \right) \mapsto (x, 2y+x).$ Hence, given a pair $(x,y),$ we can backtrace it to $(2x, y-x)$ or $(x-y, 2y),$ depending on whether $x>y$ or $x<y.$ Hence, in the end, we have a $(z,z)$ pair. This backtraces to a $(2z, 2z)$ pair. This backtraces to a $(4z,4z)$ pair and so on. Hence, Bob has a winning strategy iff the pair $(a,b)$ is a $\left( 2^{2k}z, 2^{2k}z \right)$ pair, i.e. $v_{2}(a)=v_2(b)$ is even.

Does anybody knows the cutoffs in 2018?

40 for gold, 29 for silver and 15 for bronze is what it says on the UK BMOS website.

If at any point $v_2(a) = v_2(b)$, then any move decreases both $v_2(a)$ and $v_2(b)$ by exactly $1$. Hence, a position where $v_2(a) = v_2(b)$ is winning if and only if $v_2(a)$ is odd. We have several cases. Case 1: $\text{min}(v_2(a), v_2(b))$ is odd. Then Alice can perform a move on $a$, leaving Bob with a losing position. Case 2: $\text{min}(v_2(a), v_2(b))$ is even and $v_2(a) = v_2(b)$. This is a losing position, so Bob is winning. Case 3: $\text{min}(v_2(a), v_2(b))$ and $v_2(a)\neq v_2(b)$. Call such a position $\emph{endless}$. Performing a move on the number with smaller $v_2$ in an endless position leaves the other player with a winning position. Hence, as long as the position remains endless, both players will only perform moves on the number with greater $v_2$. However, the position always remains endless after performing such a move. Hence, no player has a winning strategy in an endless position. In summary, Bob only has a winning strategy when $v_2(a) = v_2(b)$ and $v_2(a)$ is even.

If $a$ and $b$ are odd, then Bob wins. So, assume not. Claim 1: If one of a, b is odd, and the other is even, then the game is infinite. Let $a$ be even, $b$ odd. If $v_2(a) = 1$, then Alice gives half to $b$, and $b$ remains odd, while $v_2(a) \rightarrow v_2(a) - 1$. Since both the players are forced to act on $a$, we reach a point where $v_2(a) = 1$. Whoever's turn it is now, $a$ becomes odd regardless, and $b$ has to become even, which is yet again the same scenario. So, no winning strategy here. Assume both $a, b$ are even. If $v_2(a) = v_2(b)$, Alice's move will result in $v_2(a) \rightarrow v_2(a) - 1, v_2(b) \rightarrow v_2(a) - 1$. Now, it keeps on decreasing until both the $v_2$'s hit 1. Whoever makes the moves from $1 \rightarrow 0$, is the winner. If $v_2(a)$ is odd, Alice makes this move, and if it's even, Bob makes this move. Now, WLOG, $v_2(a) < v_2(b)$. Claim 2: $v_2(a)$ is odd means Alice wins. Alice acts on $a$ $\implies v_2(a) \rightarrow v_2(a) - 1$, $v_2(b) \implies v_2(a) \rightarrow v_2(a) - 1$. This is the same scenario as before, where $v_2(a) - 1$ is even, but Bob "starts" here, so Alice wins. Claim 3: $v_2(a)$ is even implies an infinite game. Alice is forced to act on $b$ here, else she loses. $v_2(b) \rightarrow v_2(b) - 1$, $v_2(a) \rightarrow v_2(a)$. Similarly for Bob. They keep acting on $b$, until we hit a point where $v_2(b) = v_2(a) + 1$. Now, regardless of whoever's move it is, $v_2(b) \rightarrow v_2(b) - 1$, while $v_2(a) \geq v_2(b)$ now, so it's the same scenario again, indicating an infinite game.