5. In the set of real numbers, solve the system of equations

$$
\begin{aligned}
& x^{2006}+y^{2006}+z^{2006}=2 \\
& x^{2007}+y^{2007}+z^{2007}=2 \\
& x^{2008}+y^{2008}+z^{2008}=2
\end{aligned}
$$

## County Competition 2006, 4th grade, B variant - solutions to problems

Each correctly solved problem is worth 20 points.1. The difference between the second-to-last and the first term of a finite arithmetic sequence is 24. The sum of all terms is 102. The difference of the sequence is three times the first term. Determine that sequence.
Solution.

Let the first term of the desired sequence be $a$, the difference of the sequence $d$, and the number of terms in the sequence $n$. The desired sequence is

$$
a, a+d, a+2 d, \ldots, a+(n-1) d
$$

and $n>2$. According to the conditions of the problem, we have:

$$
\begin{aligned}
(a+(n-2) d)-a & =24 \\
\frac{n}{2}(2 a+(n-1) d) & =102 \\
d & =3 a
\end{aligned}
$$

Rearranging the first equation, we get $(n-2) d=24$, and due to (3),

$$
\text { (4) } \quad(n-2) a=8
$$

From (2) it follows that

$$
\text { (5) } \quad \frac{n}{2}(2 a+3(n-1))=102, \quad \text { or } \quad n a(3 n-1)=204
$$

(5 points)

From (4) it follows that $a=\frac{8}{n-2}$, and substituting into (5) we get the quadratic equation $6 n^{2}-53 n+102=0$.

Its solutions are $n_{1,2}=\frac{53 \pm \sqrt{53^{2}-4 \cdot 6 \cdot 102}}{12}$, i.e., $n_{1}=6, n_{2}=\frac{17}{6}$, and since $n$ is a natural number, $n=6$,

(3 points)

and further $a=2, d=6$, (2 points) so the desired sequence is $2,8,14,20,26,32$.