A math club with 6 members agree on a real number \( b \) that the incoming seventh member must guess. After they arrived, they received the following clues in sequence:
I. \( b \) is an even number.

II. Adding \( b \) to its cube results in a positive number less than 8000.

III. \( b \) is a real number, and its square is 13.

IV. \( b \) is an integer, divisible by 7.

V. \( b \) is a rational number.

VI. \( b \) is an integer, divisible by 14.

They also informed him that among the pairs I. and II., III. and IV., V. and VI., one is always true and the other is false. - How did the seventh member of the club figure out the number?
If the VI. enlightenment is true, then V. is also true, since the integers belong to the rational numbers. Thus, only the more specific VI. can be false, and V. is true: $b$ is a rational number.

Accordingly, III. cannot be true, because there is no rational number whose square is 13. Therefore, the IV. enlightenment is true: $b$ is an integer, divisible by 7.

Now, I. cannot be true, because numbers divisible by 7 and even are also divisible by 14, which VI. has excluded, so $b$ is odd, and II. is true. Thus, $b$ is positive, otherwise $b^3$ would not be positive, and the sum $b + b^3$ could not be positive; furthermore, $b$ is less than 20, because the cube of larger numbers is also larger, so the sum $b + b^3$ is larger if we start from a larger number, and $20 + 20^3 > 8000$.

According to all this, $b = 7$, because this is the only positive, less than 20, odd, and divisible by 7 integer.

Note. We will prove the statement used in the exclusion of III. enlightenment. Suppose that for the irreducible rational number $p / q$, $(p / q)^2 = p^2 / q^2 = 13$ (that is, $p$ and $q$ are integers, and they have no common divisor, and of course $p^2$ and $q^2$ are also integers). Thus, $p^2 = 13 q^2$, so $p^2$ is divisible by 13, but then $p$ is also, since 13 is a prime number, so $p = 13 r$, where $r$ is an integer. Substituting this, after simplification, $13 r^2 = q^2$, and repeating the reasoning, $q$ is also divisible by 13, which contradicts our assumption. Therefore, there is indeed no rational number whose square is equal to 13.