1. (2 points) For a natural number $x$, six statements are made:

$$
\begin{aligned}
& 3 x>91 \\
& x<37 \\
& 2 x \geq 21 \\
& x>7
\end{aligned}
$$

It is known that only three of them are true, and three are false. Find $x$.
Solution. Let's transform the original system of inequalities

$$
\begin{aligned}
& x>\frac{91}{3} \\
& x \geq \frac{21}{2} \\
& x>\frac{37}{4} \\
& x>7 \\
& x<27 \\
& x<120
\end{aligned}
$$

The first inequality does not hold, because otherwise four inequalities would be satisfied immediately. Therefore, $x \leq \frac{91}{3}$, and the sixth inequality is satisfied.

The second inequality also does not hold, because otherwise four inequalities would be satisfied: the second, third, fourth, and sixth. Therefore, $x<\frac{21}{2}$, and the fifth inequality is also satisfied.

The third inequality does not hold, because otherwise four inequalities would be satisfied: the third, fourth, fifth, and sixth. Therefore, $x \leq \frac{37}{4}$.

Thus, the fourth, fifth, and sixth inequalities are satisfied, while the first three are not.

Now it is immediately clear that the desired number $x \in\{8,9\}$.

Answer: $x \in\{8,9\}$