66th Belarusian Mathematical Olympiad

Answer: there are no such numbers. Suppose, contrary to our claim, that there are such numbers: $\overline{ab}$, $\overline{xy}$ and $\overline{pq}$. The problem condition is equivalent to the following statement: any two-digit number with tens digit $a$, $x$ or $p$, and with unit digit $b$, $y$ or $q$ is a prime number. So $b,y,q\in \{1,3,7,9\}$, hence, at least one of the digits 3 and 9 belongs to $\{b,y,q\}$. Therefore, none of the digits $a,x,p$ is divisible by 3, so $a,x,p\in M=\{1,2,4,5,7,8\}$. Thus, we can consider two-digit numbers with tens digits from the set $M$ only. Among numbers 11, 21, 41, 51, 71, 81 there are exactly two prime numbers with distinct tens and unit digits: 41 and 71. Therefore, none of the digits $b,y,q$ is equal to 1. It follows that $\{b,y,q\}=\{3,7,9\}$, which is impossible. Indeed, exactly two numbers 17 and 47 are prime among two-digit numbers 17, 27, 47, 57, 77, 87, while there should be at least three prime numbers among these numbers.