Irish Mathematical Olympiad

(a) Denote the configuration of cards after round $k \ge 0$ by a binary sequence $a_1^{(k)}, a_2^{(k)}, a_3^{(k)}, \dots, a_n^{(k)}$, where the subscripts are interpreted modulo $n$ (here $k=0$ corresponds to the initial cards dealt to the players). The outcome of each round may then be computed recursively by
$$a_i^{(k+1)} = a_i^{(k)} + a_{i-1}^{(k)} + a_{i-2}^{(k)} \quad \forall i = 1, 2, \dots, n \quad (1)$$
for $k \ge 0$, where addition is modulo 2. Next, by induction on $r$ we show that for all $r \ge 0$ and all $k \ge 0$,
$$a_i^{(k+2r)} = a_i^{(k)} + a_{i-2r}^{(k)} + a_{i-2^{r+1}}^{(k)} \quad \forall i = 1, 2, \dots, n.$$
This is obviously true for $r=0$, as it matches (1). Assuming this equation holds for a fixed $r \ge 0$, we have, for all $i=1, 2, \dots, n$ and all $m \ge 0$,
$$a_i^{(m+2r)} = a_i^{(m)} + a_{i-2r}^{(m)} + a_{i-2^{r+1}}^{(m)}.$$
With $m = k + 2^r$ it follows that for all $i = 1, 2, \dots, n$ and all $k \ge 0$,
$$\begin{align*} a_i^{(k+2^{r+1})} &= a_i^{(k+2^r+2^r)} = a_i^{(k+2^r)} + a_{i-2^r}^{(k+2^r)} + a_{i-2^{r+1}}^{(k+2^r)} \\ &= (a_i^{(k)} + a_{i-2^r}^{(k)} + a_{i-2^{r+1}}^{(k)}) \\ &\quad + (a_{i-2^r}^{(k)} + a_{i-2^{r+1}}^{(k)} + a_{i-2^{r-2^{r+1}}}^{(k)}) \\ &\quad + (a_{i-2^{r+1}}^{(k)} + a_{i-2^{r-2^{r+1}}}^{(k)} + a_{i-2^{r+2}}^{(k)}) \\ &= a_i^{(k)} + a_{i-2^{r+1}}^{(k)} + a_{i-2^{r+2}}^{(k)} \end{align*}$$
and the results follows by the principle of induction.
Therefore, if $n = 2^r$ for any $r \ge 2$, after $n$ rounds we have $a_i^{(n)} = a_i^{(0)} + a_{i-n}^{(0)}$ and $a_{i-2n}^{(0)} = a_i^{(0)}$ for all $i$; we conclude that each such $n$ is playable.

(b) For any $n$ divisible by 3, we may set the initial configuration of cards to
$$a_i^{(0)} = \begin{cases} 1 & \text{if } i \equiv 0 \pmod 3 \\ 0 & \text{otherwise.} \end{cases}$$
Then it is easily shown that the next round yields the all-ones sequence, and this configuration will not change thereafter, i.e. $a_i^{(k)} = 1$ for all $i = 1, 2, \dots, n$ and for all $k \ge 1$. Therefore every $n$ divisible by 3 is not playable.