\section*{Task 1 - 041231}

In a mathematical circle, six participants agree on a real number \(a\), which the seventh participant, who had left the room beforehand, is supposed to determine. Upon his return, he receives the following information:
1. \(a\) is a rational number.
2. \(a\) is an integer that is divisible by 14.
3. \(a\) is a real number whose square is equal to 13.
4. \(a\) is an integer that is divisible by 7.
5. \(a\) is a real number that satisfies the following inequality. \(0 < a^3 + a < 8000\).
6. \(a\) is an even number.

He learns that of the statements 1 and 2, 3 and 4, as well as 5 and 6, one is true and one is false in each pair.

What is the number \(a\)? How did the seventh participant determine the number?
}
Assuming that statement 2 is true, then statement 1 is also true. However, this is impossible according to the above condition. Therefore, statement 1 is true and statement 2 is false.

Since statement 1 is true, statement 3 is false and statement 4 is true. \(a\) is thus one of the numbers

\[
\ldots,-35,-21,-7,7,21,35, \ldots, 7(2 n+1), \ldots
\]

(n is an integer). Now, since statement 2 is false, statement 6 is also false and statement 5 is true. Therefore, the number \(a\) is positive.

Now, \(7^{3}+720^{3}+20>8000\). Thus, \(a=7\).

\section*{Adapted from \([5]\)}

\section*{2.Solution:}

The number is 7. The number can be uniquely determined with the following considerations:

Assume the number is not an integer (i.e., not a whole number).

Then statements 2, 4, and 6 are false, and statements 1, 3, and 5 must be true. However, there is no rational number whose square is 13, so statements 1 and 3 cannot both be true.

Therefore, the assumption is false.

Thus, the number \(a\) is an integer. Statement 3 is therefore false, and statement 4 is true. Additionally, statement 1 is true and statement 2 is false. A number divisible by 7 but not by 14 must be odd. Therefore, statement 6 is false and statement 5 is true.

Thus, the number \(a\) is an integer and hence rational, odd, and it holds that

\[
0<a^{3}+a<8000
\]

If \(a\) were negative, then \(a^{3}+a\) would also be negative, contradicting (1).

For \(a \geq 0\), it follows from (1) that \(a^{3} \leq a^{3}+a<8000\), so \(a \leq 20\). In the range from 0 to 20, there is only one odd number divisible by 7, namely \(a=7\).

Indeed, for \(a=7\), statements 1, 4, and 5 are true \(\left(0<7^{3}+7=350<8000\right)\) and statements 2, 3, and 6 are false.