1. There are 12 carriages in the train, and they contain 384 passengers. In the second carriage, there are more than two passengers more than in the first, in the third carriage, there are more than two passengers more than in the second, and so on until the last carriage. The number of passengers in the last carriage does not exceed three times the number of passengers in the first carriage. How many passengers are in the first carriage?
Answer: There are no solutions.

Solution. Let $y_{k}$ be the number of passengers in the carriage with number $k, k=1,2,3, \ldots, 12$. According to the problem, $\sum_{k=1}^{12} y_{k}=384$. Additionally, it is stated that $y_{2} \geq y_{1}+3, y_{3} \geq y_{1}+6, \ldots, y_{11} \geq y_{1}+30, y_{12} \geq y_{1}+33$. Adding these inequalities,

$$
y_{2}+y_{3}+\ldots+y_{11}+y_{12} \geq y_{1}+3+y_{1}+6+\ldots+y_{1}+30+y_{1}+33
$$

or

$$
384-y_{1} \geq 11 y_{1}+198
$$

From this, we get $y_{1} \leq \frac{186}{12}=\frac{31}{2}=15.5$. Since $y_{1}$ is a natural number, then $y_{1} \leq 15$. From the conditions $y_{12} \geq y_{1}+33$ and $y_{12} \leq 3 y_{1}$, it follows that $3 y_{1} \geq y_{1}+33$ or $y_{1} \geq \frac{33}{2}=16.5$. Since $y_{1}$ is a natural number, then $y_{1} \geq 17$. The system $\left\{\begin{array}{l}y_{1} \leq 15, \\ y_{1} \geq 17,\end{array}\right.$ has no solutions.