Balti Tee 1997 · Ülesanne 20

Solution:

Answer: there are $9$ cards with one black and one white side and $3$ cards with both sides white.

Divide the cards into four types according to the table below.

Type	Initially up	Initially down
$A$	black	white
$B$	white	black
$C$	white	white
$D$	black	black

When the cards $1$-$6$ were turned, the number of cards with a black side up decreased by $5$. Hence among the cards $1$-$6$ there are five of type $A$ and one of type $C$ or $D$. The result of all three moves is that cards $7$-$12$ have been turned over, hence among these cards there must be four of type $A$, and the combination of the other two must be one of the following:

(a) one of type $A$ and one of type $B$; (b) one of type $C$ and one of type $D$; (c) both of type $C$; (d) both of type $D$.

Hence the unknown card among the cards $1$-$6$ cannot be of type $D$, since this would make too many cards having a black side up initially. For the same reason, the alternatives (a), (b) and (d) are impossible. Hence there were nine cards of type $A$ and three cards of type $C$.

Alternative solution.

Denote by $a_{1}, a_{2}, \ldots, a_{12}$ the sides of each card that are initially visible, and by $b_{1}, b_{2}, \ldots, b_{12}$ the initially invisible sides; each of these is either white or black. The conditions of the problem imply the following:

(a) there are $9$ black and $3$ white sides among $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}, a_{8}, a_{9}, a_{10}, a_{11}, a_{12}$;

(b) there are $4$ black and $8$ white sides among $b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, a_{7}, a_{8}, a_{9}, a_{10}, a_{11}, a_{12}$;

(c) there are $6$ black and $6$ white sides among $b_{1}, b_{2}, b_{3}, a_{4}, a_{5}, a_{6}, b_{7}, b_{8}, b_{9}, a_{10}, a_{11}, a_{12}$;

(d) there are $5$ black and $7$ white sides among $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{7}, b_{8}, b_{9}, b_{10}, b_{11}, b_{12}$.

Cases (b) and (d) together enumerate each of the sides $a_{i}$ and $b_{i}$ exactly once - hence there are $9$ black and $15$ white sides altogether. Therefore, all existing black sides are enumerated in (a), implying that we have $9$ cards with one black and one white side, and the remaining $3$ cards have both sides white.