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Problem 1002

AMC 12 late, AIME early

Combinatorics Difficulty 4.5 Multiple choice CEMC Fermat · Canada · 2023

Twenty-five cards are randomly arranged in a grid, as shown.

Hide/Reveal Description of the Grid

The cards are arranged into five rows and five columns. The visible
side of each card has either a 0 or a 1. The number on each card is
given in the following 5 by 5 array. $\begin{array}{|c|c|c|c|c|} \hline 0&0&1&1&0\\\hline 1&0&0&1&0\\\hline 1&1&0&1&0\\\hline 0&0&1&0&0\\\hline 1&0&1&0&1\\\hline \end{array}$

Five of these cards have a 0 on one side and a 1 on the other side.
The remaining twenty cards either have a 0 on both sides or a 1 on both
sides. Loron chooses one row or one column and flips over each of the
five cards in that row or column, leaving the rest of the cards
untouched. After this operation, Loron determines the ratio of 0s to 1s
facing upwards. No matter which row or column Loron chooses, it is
not possible for this ratio to be

Pick one

Official solution

The given arrangement has 14 zeroes and 11 ones showing.

Loron can pick any row or column in which to flip the 5 cards over.
Furthermore, the row or column that Loron chooses can contain between 0
and 5 of the cards with different numbers on their two sides.

Of the 5 rows and 5 columns, 3 have 4 zeroes and 1 one, 2 have 3 zeroes
and 2 ones, and 5 have 2 zeroes and 3 ones.

This means that the number of zeroes cannot decrease by more than 4 when
the cards in a row or column are flipped, since the only way that the
zeroes could decrease by 5 is if all five cards in the row or column had
0 on the top face and 1 on the bottom face.

Therefore, there cannot be as few as $14 - 5 = 9$ zeroes after Loron flips the cards, which means that the
ratio cannot be $9:16$ , or (C). This means that the answer to the given problem is (C). For completeness, we will show that the other ratios are indeed achievable. If Loron chooses the first column and if this column includes 3 cards with ones on both sides, and 2 cards with zeroes on one side (facing up) and ones on the reverse side, then flipping the cards in this column yields $14 - 2 = 12$ zeroes and $11 + 2 = 13$ ones. Thus, the ratio $12:13$ (choice (A)) is possible. If Loron chooses the fifth column and if this column includes 1 card with a one on both sides and 4 cards with zeroes on one side (facing up) and ones on the reverse side, then flipping the cards in this column yields $14 - 4 = 10$ zeroes and $11 + 4 = 15$ ones. Thus, the ratio $10 : 15 = 2:3$ (choice (B)) is possible. If Loron chooses the first column and if the top 4 cards in this column have the same numbers on both sides and the bottom card has a one on the top side and a zero on the reverse side, then flipping the cards in this column yields $14 + 1 = 15$ zeroes and $11 - 1 = 10$ ones. Thus, the ratio $15 : 10 = 3:2$ (choice (D)) is possible. If Loron chooses the first column and if the first, fourth and fifth cards in this column have the same numbers on both sides and the second and third cards each has a one on the top side and a zero on the reverse side, then flipping the cards in this column yields $14 + 2 = 16$ zeroes and $11 - 2 = 9$ ones. Thus, the ratio $16 : 9$ (choice (E)) is possible. Therefore, the only ratio of the five that are given that is not possible is $9:16$ , or (C).

Figure for this problem

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.