4. Students were supposed to calculate the example $x+y \cdot z$ for a three-digit number $x$ and two-digit numbers $y$ and $z$. Martin knows how to multiply and add numbers written in the decimal system, but he forgot the rule about the precedence of multiplication over addition. Therefore, he got an interesting number that reads the same from left to right as from right to left, but the correct result was 2004 less. Determine the numbers $x, y$, and $z$.

II. round of category $\mathrm{C}$ takes place

on Tuesday, March 23, 2004

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. These details will be communicated to the students before the competition begins.
4. Martin calculated the value of $(x+y) z$ instead of $x+y z$, so according to the problem statement,

$$
(x+y) z-(x+y z)=2004, \quad \text { or } \quad x \cdot(z-1)=2004=12 \cdot 167
$$

where 167 is a prime number. The factors $x$ and $z-1$ can be determined by realizing that $z$ is a two-digit number, so $9 \leq z-1 \leq 98$. We see that necessarily $z-1=12$ and $x=167$, hence $z=13$. Martin therefore calculated the number $V=(167+y) \cdot 13$. The number $V$ is thus a four-digit number, and since it reads the same forwards and backwards, it has the form $\overline{a b b a}=1001 a+110 b$. Since $1001=13 \cdot 77$, the equation $(167+y) \cdot 13=13 \cdot 77 a+110 b$ must hold, from which it follows that the digit $b$ is divisible by thirteen, so $b=0$. Substituting this, we get (after dividing by thirteen) the equation $167+y=77 a$, which, considering the inequalities $10 \leq y \leq 99$, means that the digit $a$ equals 3, thus $y=64$.

In the second part of the solution, we could also proceed as follows. For the number $V=(167+y) \cdot 13$, the inequalities $10 \leq y \leq 99$ yield the estimates $2301 \leq V \leq 3458$. Therefore, we determine which of the numbers $\overline{2 b b 2}$, where $b \in\{3,4,5,6,7,8,9\}$, and the numbers $\overline{3 b b 3}$, where $b \in\{0,1,2,3,4\}$, are divisible by thirteen. Although these twelve numbers can be quickly tested on a calculator, let's do it generally by partially dividing them by thirteen:

$$
\begin{aligned}
& \overline{2 b b 2}=2002+110 b=13 \cdot(154+8 b)+6 b, \\
& \overline{3 b b 3}=3003+110 b=13 \cdot(231+8 b)+6 b .
\end{aligned}
$$

We see that only the number $\overline{3 b b 3}$ for $b=0$ satisfies the condition, when $167+y=231$, so $y=64$.

Answer: The students were supposed to calculate the example $167+64 \cdot 13$, i.e., $x=167, y=64$ and $z=13$.

For a complete solution, award 6 points, of which 1 point for setting up the equation $(x+y) z-(x+y z)=2004$, 1 point for transforming it into the product form $x(z-1)=2004$, 2 points for finding the factors $x$ and $z-1$, and 2 points for discussing the divisibility of the numbers $\overline{a b b a}$ by 13.