Maths Olympiad Prep

Track / Stage 6 / 63 of 400 #1063 of 1964

Problem 1063

National olympiad, first round
Combinatorics Difficulty 6.1 Prove it

9.4. Is it possible to place one chip in some cells of an 8 by 8 chessboard so that the number of chips in any two adjacent rows differs by a factor of 3, and in any two adjacent columns by a factor of 4? There must be at least one chip on the board.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Answer: No.

Solution. A row or column of an 8x8 board cannot contain more than 8 chips, so the minimum number of chips in a row is 1 or 2; otherwise, one of the adjacent rows would contain at least 9 chips. From the condition, it easily follows that in the first case, the rows contain 1,3,1,3,1,3,1,31,3,1,3,1,3,1,3 or 3,1,3,1,3,1,3,13,1,3,1,3,1,3,1 chips, and in the second case, 2,6,2,6,2,6,2,62,6,2,6,2,6,2,6 or 6,2,6,2,6,2,6,26,2,6,2,6,2,6,2 chips, which means a total of 16 or 32 chips. If we divide the columns into pairs of adjacent columns, the number of chips in each pair of adjacent columns must be divisible by 5. Therefore, the total number of chips on the board must also be divisible by 5. However, 16 and 32 are not divisible by 5 - a contradiction.

Grading Criteria. Clear justification that the rows contain 1,3,1,3,1,3,1,31,3,1,3,1,3,1,3 or 2,6,2,6,2,6,2,62,6,2,6,2,6,2,6 chips: 3 points. Clear justification that the total number of chips on the board must be divisible by 5: 3 points. Obtaining a contradiction: 1 point.

Solutions where the phrase "three times" is interpreted as "at least three times" or "no more than three times": not considered.

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