Croatian Mathematical Competitions

a. We consider three different cases of adding a red ball depending on the colours of its neighbours.
If we add the ball between two blue balls, then we have the case $PP \rightarrow CCC$ in which the number of blue balls decreases by two. If we add the ball between two red balls, we have $CC \rightarrow PCP$ and the number of blue balls increases by two. In the third case, we add the ball between one blue and one red ball ($CP \rightarrow PCC$) and the number of blue balls stays the same. We conclude that adding a red ball does not change the parity of the number of blue balls. Also removing a red ball does not change the parity of the number of blue balls.

In the beginning there is no blue balls, so after each operation the number of blue balls is even. Hence it is not possible that we obtain exactly $2013$ blue balls.

b. In a given configuration of balls, we label the blue balls clockwise $P_1, P_2, \dots, P_{2k}$ starting with an arbitrary blue ball. Let $m_i$ ($i \in \{1, 2, \dots, 2k\}$) be the number of red balls between balls $P_i$ and $P_{i+1}$ ($P_{n+1} = P_1$). For example, in the following picture we have $m_1 = 2, m_2 = 3, m_3 = 1, m_4 = 0, m_5 = 4, m_6 = 1$.

Furthermore, let us denote $S = m_1 - m_2 + m_3 - \dots + m_{2k-1} - m_{2k}$. If the configuration consists of $n$ red balls and no blue balls we denote $S = n$. Let us prove that either $S$ is divisible by three for all configurations or $S$ is not divisible by three for all configurations. We again consider three cases:

1. We add a red ball between two blue balls, $P_i$ and $P_{i+1}$:
$$P_{i-1} \underbrace{\dots P_i}_{m_{i-1}} \underbrace{\dots P_{i+1}}_{m_i=0} \rightarrow P_{i-1}' \underbrace{\dots \overbrace{CC}^{\downarrow} \dots P_i'}_{m_i'}$$

2. We add a red ball between two red balls:
$$P_i \underbrace{\dots CC \dots}_{m_i} P_{i+1} \rightarrow P_i \underbrace{\dots P_{i+1}}_{m_i'} \underbrace{\overbrace{C}^{\downarrow}}_{m_{i+1}'=1} P_{i+1} \underbrace{\dots P_{i+3}}_{m_{i+2}'}$$
Similiarly as before, $S = S_0 + m_i$, a $S' = S_0 + m_i' - 1 + m_{i+2}$. But now $m_i = m_i' + m_{i+2}' + 2$, so $S' = S - 3$.

(Analogous statement holds in the case if there is no blue balls on the circle.)

3. We add a red ball between one red and one blue ball (to the right of the blue ball $P_i$):
$$P_{i-1} \underbrace{\overbrace{P_i}^{m_{i-1}} \overbrace{C \dots P_{i+1}}^{m_i}} \rightarrow P_{i-1} \underbrace{\overbrace{\dots \overbrace{C}^{m'_{i-1}} C}^{\downarrow} P_i}_{m'_i} \dots P_i$$
In this case $S = S_0 + m_{i-1} - m_i$ and $S' = S_0 + m'_{i-1} - m'_i$. Since $m'_{i-1} = m_{i-1} + 2$ and $m'_i = m_i - 1$, we have $S' = S + 3$.

In any case, $S$ either decreases or increases by $3$. In the beginning number $S$ is $2$ and hence $S$ can never be divisible by three. We cannot obtain exactly two blue balls since $S$ would be $0$.