## Task 1 - 340731

Albrecht is supposed to determine a natural number between 1 and 1000000. Dirk, Evelyn, and Franziska each make exactly one true and one false statement (the order is not specified):

Dirk: (1) The number in question has fewer than three decimal places.

(2) If the number in question is decomposed into prime factors, exactly two different prime numbers appear in this decomposition, each (at least once, but) possibly also multiple times. Evelyn: (1) The number in question is divisible by 9.

(2) The number in question is not divisible by 27.

Franziska: (1) The number in question is 91809.

(2) The number in question is divisible by 101.

Show that the number in question can be uniquely determined in this way, and find this number!
If the conditions of the task are met by a natural number $n$, then:

If Evelyn's statement (2) were false, i.e., $n$ is divisible by 27, then $n$ would also be divisible by 9, meaning Evelyn's statement (1) would also be false. This contradicts the conditions of the task. Therefore, Evelyn's statement (2) is true and (according to the conditions of the task) her statement (1) is false.

This means: $n$ is divisible by 9, but not by 27.

If Franziska's statement (1) were true, then because 91809 : $101=909$, Franziska's statement (2) would also be true. Again, this cannot be according to the conditions of the task, so it follows: Statement (1) is false, statement (2) is true, thus $n$ is not 91809, but is divisible by 101.

Therefore, it is shown: If $n$ is factored into prime factors, the prime number 3 appears exactly twice, and the prime number 101 appears at least once. Thus, $n$ is at least a three-digit number, and Dirk's statement (1) is false. Therefore, Dirk's statement (2) is true.

Apart from the prime numbers 3 and 101, no other prime number appears in the factorization of $n$. If 101 appeared exactly twice, then $n=3^{2} \cdot 101^2=91809$, which has already been refuted; if 101 appeared more than twice, then $n \geq 3^{2} \cdot 101^3>1003$, so $n$ would not lie between 1 and 1000000. Therefore, 101 appears exactly once; it follows that $n=3^{2} \cdot 101=909$.

Thus, the number sought is uniquely determined by the conditions of the task, and it is 909.