## Task 3 - 150933

For a number $x$, the following four pairs $\left(A_{1}, A_{2}\right),\left(B_{1}, B_{2}\right),\left(C_{1}, C_{2}\right),\left(D_{1}, D_{2}\right)$ of statements are made, of which exactly one is true and exactly one is false.

Investigate whether there is a number $x$ that satisfies this requirement! Determine, if this is the case, every such number $x$!

A1) There is no number other than $x$ that satisfies the requirement of this task.

A2) $x$ is a natural number, in whose (decimal) representation a digit appears twice.

B1) $x-5$ is an integer divisible by 6.

B2) $x+1$ is an integer divisible by 12.

C1) $x$ is a natural number whose (decimal) representation begins with the digit 3.

C2) $x$ is the number 389.

D1) $x$ is a three-digit prime number with $300<x<399$, i.e., one of the numbers $307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397$.

D2) $x$ is a natural number consisting of three identical digits.
Since $C_{2}$ directly implies $C_{1}$, but both statements cannot be true, $C_{2}$ must be false and $C_{1}$ must be true. Therefore, $x$, if it exists, is a natural number that begins with the digit 3 but is not 389.

If statement $D_{2}$ were true, it would have to be the number 333. However, this contradicts both statements $B_{1}$ and $B_{2}$, since neither 328 is divisible by 6 nor 334 by 12. Therefore, $D_{2}$ must be false and $D_{1}$ must be true, meaning that $x$ must be one of the prime numbers different from 389 in the given list.

If statement $B_{2}$ were true, then $x+1$ would also be divisible by 6, and thus $x+1-6 = x-5$ would also satisfy statement $B_{1}$. Therefore, $x-5$ must be divisible by 6, while $x+1 = (x-5) + 6$ must not be divisible by 12. This is only possible if $x-5$ is already divisible by 12. In the list, only 317 and 353 satisfy this, so $x$ must be one of these two numbers.

If $A_{2}$ were true, it would be uniquely identified as 353, since only in this remaining possible number do two identical digits occur. However, this would also make statement $A_{1}$ true due to the now demonstrated uniqueness, which contradicts the condition of the problem statement.

If, on the other hand, $A_{2}$ is false, then no digit may appear twice in the representation of $x$. This rules out 353, leaving 317 as the unique solution, making statement $A_{1}$ true.

Thus, $x=317$ satisfies exactly the first statements while the second ones are false; and it is also the only number that meets the problem statement.