Maths Olympiad Prep

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Combinatorics Difficulty 6.2 National Olympiad Prove it Brazil

The cells of a 3×33 \times 3 table were numbered from 11 to 99, each number appearing exactly once. For each row the cell with the greatest number is colored red and the cell with the smallest number is colored green. Let AA be the smallest of the numbers in the red cells and BB the greatest of the numbers in the green cells.

a. Show a distribution of the numbers in the table such that AB=4A - B = 4.

b. Show a distribution of the numbers in the table such that AB=3A - B = -3.

c. Is it possible that A=4A = 4 and B=3B = 3?

Solution

a. For example,

741
852
963

Notice that A=min{7,8,9}=7A = \min\{7, 8, 9\} = 7 and B=max{1,2,3}=3B = \max\{1, 2, 3\} = 3, so AB=4A - B = 4.

b. For example,

123
457
689

Notice that A=min{3,7,9}=3A = \min\{3, 7, 9\} = 3 and B=max{1,4,6}=6B = \max\{1, 4, 6\} = 6, so AB=3A - B = -3.

c. No, it's not possible. Since B=3B = 3 is the greatest of the numbers in green cells, which are the smallest ones in their respective rows, the numbers 11, 22 and 33 must be on different rows. Now A=4A = 4 means that 44 is the greatest number of a row, but it could only happen if two of the numbers 11, 22, 33 are on the same row as 44, which cannot happen.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.