Evdokimov M.A.

Wise men \$ will be given three cards each, and each will see only their own cards. After the distribution, the wise men said the following in turn.

\$A\$: "One of my cards has the number 8".

\$B\$: "All the numbers on my cards are prime".

\$C\$: "All the numbers on my cards are composite, and they have a common prime divisor".

\$D\$: "Then I know what cards each of you have".

What cards does \$A\$ have, if everyone spoke the truth?
Prime numbers can only be at $\$ \mathrm{~A}, \mathrm{~B}$ and $\$ \mathrm{D} \$$$. On the cards $\$ \mathrm{~B} \$$ - three out of five possible prime numbers $(2,3,5,7,11)$. The other two prime numbers are with \$D\$, otherwise he wouldn't know which of the prime numbers are with $\$ B \$$, and which are with \$A\$.

On the cards \$C\$ there can be triples $(4,6,10),(4,6,12),(4,10,12),(6,10,12)$ or $(6,9,12)$. Only if \$D has 6 or 12, he can determine which exact triple is with \$C\$.

So, \$D\$ has two prime numbers and one of the numbers 6 and 12, \$C$ respectively - 4, 6, 10, or 4, 10, 12, \$B$ three prime numbers, and \$A - numbers $1,8,9$.

## Answer

1, 8 and 9.