Problem 11 · 2023 Math Kangaroo
Stretch
Counting & Probability
divisibilitycareful-counting
The numbers from 1 to 9 are to be distributed to the nine squares in the diagram according to the following rules: There is to be one number in each square. The sum of three adjacent numbers is always a multiple of 3. The numbers 7 and 9 are already written in. How many ways are there to insert the remaining numbers?
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Answer: E — 24
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Hint 1 of 3
Sort 1–9 by remainder mod 3: \(\{3,6,9\}\) give 0, \(\{1,4,7\}\) give 1, \(\{2,5,8\}\) give 2.
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Hint 2 of 3
Comparing two overlapping triples shows the remainders must repeat every third square.
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Hint 3 of 3
Use the fixed 7 and 9 to decide which remainder class goes in which of the three position-classes.
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Approach: the mod-3 remainders repeat with period 3, then count
- If squares \(i,i+1,i+2\) and \(i+1,i+2,i+3\) both sum to a multiple of 3, subtracting shows square \(i\) and square \(i+3\) have the same remainder, so the remainders repeat in a period-3 pattern across the nine squares.
- Thus one remainder class (three numbers) fills positions \(1,4,7\), another fills \(2,5,8\), the third fills \(3,6,9\); the placed 7 (remainder 1) and 9 (remainder 0) lock those two classes onto their position groups, leaving the remainder-2 class for the third group.
- Each of the three classes can be arranged inside its three positions in \(3! = 6\) ways, but the 7 and 9 are fixed, so the count is \(2 \cdot 2 \cdot 6 = \mathbf{24}\) ways.
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