41. For which $n$ can $n$ numbers $a_{1}, a_{2}, \ldots, a_{n}$ (not all equal to zero) be placed on a circle so that for any $k \leqslant n$ the sum of $k$ consecutive numbers starting from $a_{k}$ is zero?
83.41. Answer. Such an arrangement is possible if and only if $n$ is not a power of two. Hint. Let $s_{k}=a_{1}+a_{2}+\ldots+a_{k-1}$. Then the equalities specified in the condition can be rewritten as follows: for any $k$ we have $s_{k}=s_{2 k}$ (and there is also the equality $s_{1}=0$). Since there are $n$ independent variables $s_{k}$ and $n$ equations of the specified form for them, the equality of all $a_{k}$ to zero is equivalent to the equality of all variables $s_{k}$ to each other, i.e., to the fact that for any indices $k$ and $m$ there exist powers of two $2^{a}$ and $2^{b}$ such that $2^{a} k \equiv 2^{b} m(\bmod n)$; or, equivalently, for any $k$ there exists a power of two $2^{c}$ such that $2^{c} k$ is divisible by $n$.