Solve the following system of equations:

$$
\begin{aligned}
& x_{1}+2 x_{2}+3 x_{3}+\ldots+9 x_{9}+10 x_{10}=55, \\
& x_{2}+2 x_{3}+3 x_{4}+\ldots+9 x_{10}+10 x_{1}=55, \\
& \vdots \\
& x_{10}+2 x_{1}+3 x_{2}+\ldots+9 x_{8}+10 x_{9}=55 .
\end{aligned}
$$
Adding the ten equations,

$$
\begin{gathered}
55 x_{1}+55 x_{2}+\ldots+55 x_{10}=550, \quad \text { hence } \\
x_{1}+x_{2}+\ldots+x_{10}=10
\end{gathered}
$$

Subtract the second equation from the first:

$$
9 x_{1}-\left(x_{2}+x_{3}+\cdots+x_{10}\right)=0
$$

On the left side, the subtrahend is $10-x_{1}$ according to (1), thus

$$
9 x_{1}-\left(10-x_{1}\right)=10 x_{1}-10=0, \quad \text { hence } \quad x_{1}=1
$$

Similarly, by forming the difference between the $(i+1)$-th and the $i$-th equation,

$$
9 x_{i}-\left(x_{i+1}+x_{i+2}+\cdots+x_{i-2}+x_{i-1}\right)=10 x_{i}-10=0
$$

results (for $i=10$ let $i+1=1$), that is,

$$
x_{i}=1
$$

The solution of the system of equations is thus

$$
x_{1}=x_{2}=\cdots=x_{10}=1
$$

Szakács Árpád (Árpád Gymn., Budapest II. o. t.)