Macedonian Mathematical Olympiad

Let us suppose that the sequence $1,x_2,\dots ,x_n,1$ satisfy the condition of the problem. Then $$|1-x_2|=|x_2-x_3|=\cdots =|x_n-1|=C.(1)$$ Also $$x_n-1=(x_n-x_{n-1})+(x_{n-1}-x_{n-2})+\cdots +(x_2-1)(2)$$ where the number of pairs of brackets of the right-hand side is $n-1$. We consider two cases: $n$ is odd and $n$ is even number.

Case 1. $n$ is odd, i.e. $n-1$ is even number. Then the equality (2) get the form ${\Delta }_{1}C=\underset{n-1}{\underbrace{{\Delta }_{2}C{\Delta }_{3}C\dots {\Delta }_{n}C}}$, where ${\Delta }_i\in \{-,+\}$, $1\le i\le n$. The choice of ${\Delta }_i\in \{-,+\}$, $2\le i\le n$ determine a sequence $1,x_2,\dots ,x_n,1$, satisfying the sequence (1). Let we

note that the sum of the right-hand side of the equality ${\Delta }_{1}C=\underset{n-1}{\underbrace{{\Delta }_{2}C{\Delta }_{3}C\dots {\Delta }_{n}C}}$ (for one choice of ${\Delta }_i\in \{-,+\},2\le i\le n$) is even number of times $C$, and the left-hand side is $\pm C$, i.e. odd number times $C$. Therefore, the right-hand side is equal to $\pm 2kC$, for some $k\in \mathbb{N}$, and the left-hand side is equal to $\pm C$. The last is possible only if $C=0$ and then we obtain the sequence $1=x_2=\cdots =x_n$, i.e. there exists only one sequence with the desired property.

Case 2. $n$ is even, i.e. $n-1$ is odd number. If $C=0$, then $1=x_2=\cdots =x_n$. Let $C>0$. Then again from the equality (2) we obtain $${\Delta }_{1}C=\underset{n-1}{\underbrace{{\Delta }_{2}C{\Delta }_{3}C\dots {\Delta }_{n}C}}.$$ Hence, the problem can be transformed to find the number of choices of ${\Delta }_i\in \{-,+\},2\le i\le n$ such that ${\Delta }_{1}C=\underset{n-1}{\underbrace{{\Delta }_{2}C{\Delta }_{3}C\dots {\Delta }_{n}C}}$. If on the left-hand side of the equality we have $+C$, then it must, on the right-hand side of the equality, the difference between the number of the positive and the number of the negative constants to be exactly 1, i.e. $+C$ must appear $\frac{n-2}{2}+1=\frac{n}{2}$ times, and $-C$ appears exactly $\frac{n-2}{2}$ ways. Such type of choices can be made on $\left(\genfrac{}{}{0ex}{}{n-1}{n/2}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{(n-2)/2}\right)$ ways (for a sequence with length $n-1$ we choose $n/2$ positions where we will put $+C$, and on the rest we will put $-C$). On the exact same number of ways can be written the sum of the right-hand side is the sum of the left-hand side is $-C$. So, the total number of such representations is $2\left(\genfrac{}{}{0ex}{}{n-1}{n/2}\right)$. Finally, if with $r=r(n,C)$ we denote the number of the sequences satisfying the condition of the problem, then $$r=\{\begin{array}{ll}1,& C=0\\ 0,& n\text{ is odd and }C>0.\\ 2\left(\genfrac{}{}{0ex}{}{n-1}{n/2}\right),& n\text{ is even and }C>0.\end{array}$$