Four boys, $A, B, C$, and $D$, made 3 statements each about the same number $x$. We know that each of them has at least one correct statement, but also that at least one of their statements is false. Can $x$ be determined? The statements are:

A: (1) The reciprocal of $x$ is not less than 1.

(2) The decimal representation of $x$ does not contain the digit 6.

(3) The cube of $x$ is less than 221.

$B$: (4) $x$ is an even number.

(5) $x$ is a prime number.

(6) $x$ is an integer multiple of 5.

$C$: (7) $x$ cannot be expressed as the quotient of two integers,

(8) $x$ is less than 6.

(9) $x$ is the square of a natural number.

$D$: (10) $x$ is greater than 20.

(11) $x$ is positive and its base-10 logarithm is at least 2.

(12) $x$ is not less than 10.
According to all three statements of $B$, $x$ is an integer, and since at least one of these is true, $x$ is an integer and (7) is false.

All three statements of $D$ provide a lower bound for $x$: 20, 100, and 10. If $x \geq 100$ were true, then $D$ would have no false statements, so (11) is false. If $x<10$ were true, then $D$ would have no true statements, so (12) is true:

$$
10 \leq x<100
$$

As a result, of the statements of $A$, (1) and (3) are false, so (2) is true, and for $C$, (8) is also false, so (9) is true.

According to the above, $x$ can only be a number that satisfies the (*) inequality, is a perfect square, and does not contain the digit 6. Thus, $x$ cannot be even, as stated in (4), because all even perfect squares between 10 and 100 (16, 36, 64) contain the digit 6. This means that two statements of $B$, (4) and (5), are false, and only (6) can be true: $x$ is a perfect square, divisible by 5, and satisfies the $\left(^{*}\right)$ inequality. There is only one such number, 25, so according to our conditions, $x$ is uniquely determined.