Maths Olympiad Prep

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Combinatorics Difficulty 6.4 National olympiad Find the answer

Twelve cards lie in a row. The cards are of three kinds: with both sides white, both sides black, or with a white and a black side. Initially, nine of the twelve cards have a black side up. The cards 1-6 are turned, and subsequently four of the twelve cards have a black side up. Now cards 4-9 are turned, and six cards have a black side up. Finally, the cards 1-3 and 10-12 are turned, after which five cards have a black side up. How many cards of each kind are there?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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.

TypeInitially upInitially down
AAblackwhite
BBwhiteblack
CCwhitewhite
DDblackblack

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 AA and one of type CC or DD. The result of all three moves is that cards 7127-12 have been turned over, hence among these cards there must be four of type AA, and the combination of the other two must be one of the following:

(a) one of type AA and one of type BB;

(b) one of type CC and one of type DD;

(c) both of type CC;

(d) both of type DD.

Hence the unknown card among the cards 1-6 cannot be of type DD, 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 AA and three cards of type CC.

Alternative solution. Denote by a1,a2,,a12a_{1}, a_{2}, \ldots, a_{12} the sides of each card that are initially visible, and by b1,b2,,b12b_{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 a1,a2,a3,a4,a5,a6a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a7,a8,a9,a10,a11,a12a_{7}, a_{8}, a_{9}, a_{10}, a_{11}, a_{12}

(b) there are 4 black and 8 white sides among b1,b2,b3,b4,b5,b6,a7b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, a_{7}, a8,a9,a10,a11,a12a_{8}, a_{9}, a_{10}, a_{11}, a_{12};
(c) there are 6 black and 6 white sides among b1,b2,b3,a4,a5,a6,b7b_{1}, b_{2}, b_{3}, a_{4}, a_{5}, a_{6}, b_{7}, b8,b9,a10,a11,a12b_{8}, b_{9}, a_{10}, a_{11}, a_{12};

(d) there are 5 black and 7 white sides among a1,a2,a3,a4,a5,a6,b7a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, b_{7}, b8,b9,b10,b11,b12b_{8}, b_{9}, b_{10}, b_{11}, b_{12}.

Cases (b) and (d) together enumerate each of the sides aia_{i} and bib_{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.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.