Croatian Mathematical Society Competitions

Let us analyse how the moves affect certain sums. Denote by $a_k$ the numbers written at a certain moment, so that $a_1$ is the number written in the place where the one was written before any move was made, and $a_2,\dots ,a_{300}$ are the numbers written clockwise from $a_1$. After applying the first type of move, the numbers written are $$\begin{array}{rl}{b}_1& =a_1-a_{300}-a_2,\\ b_k& =a_k-a_{k-1}-a_{k+1},k=2,\dots ,299,\\ b_{300}& =a_{300}-a_{299}-a_1.\end{array}$$

a) Observe how the sum of all written numbers changes after each move. If the total sum of numbers $a_k$ is $$S=a_1+a_2+\cdots +a_{300},$$ then the sum of numbers $b_k$ is $$a_1-a_{300}-a_2+a_2-a_1-a_3+\cdots +a_{300}-a_{299}-a_1=-S$$ because each $a_k$ is added once and subtracted twice.

After the second type of move is applied, the sum of all numbers is either $S+2$ or $S-2$. We can conclude that none of the allowed moves changes the parity of the sum of all numbers. Since the sum is 1 (an odd number) in the beginning, we cannot obtain two ones and 298 zeros since their sum is 2 (an even number).

b) Observe how the moves affect the alternating sum. If the alternating sum of numbers $a_k$ is $$S^{\prime }=a_1-a_2+a_3-\cdots +a_{299}-a_{300},$$ then after applying the first type of move, the alternating sum of numbers $b_k$ is $$a_1-a_{300}-a_2-a_2+a_1+a_3+\cdots -a_{300}+a_{299}+a_1=3S^{\prime },$$ while the alternating sum remains $S^{\prime }$ after applying the second type of move. Assume that three consecutive ones and 297 zeros can be obtained. The original alternating sum before any move is made equals 1. When three consecutive ones and 297 zeros are written circular, the alternating sum is either 1 or -1, depending on where the ones are located. Since the allowed moves either triple the alternating sum or they do not affect it, we conclude that the three ones are located so that the alternating sum is 1, and no moves of the first type were allowed.

Observe the sum of numbers in places 3, 6, ..., 300, i.e. $$S_0=a_3+a_6+\cdots +a_{300}.$$ Originally, this sum is 0, and when three consecutive ones and 297 zeros are written, that sum is 1. However, applying only the second type of move does not change the parity of that sum. Namely, either none of the numbers $a_3,a_6,\dots ,a_{300}$ are changed, or we change exactly two of them by exactly 1, changing $S_0$ by 2 or -2. Since 0 and 1 are of different parity, this shows that we cannot obtain three consecutive ones and 297 zeros.