Problem 21 · 2023 Math Kangaroo
Stretch
Counting & Probability
caseworkdivisibility
The numbers from 1 to 9 should be distributed among the 9 squares in the diagram according to the following rules: there should be one number in each square, and the sum of three adjacent numbers is always a multiple of 3. The numbers 3 and 1 are already placed. How many ways are there to place the remaining numbers?
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Answer: E — 24
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Hint 1 of 2
Three numbers in a row summing to a multiple of 3 forces a repeating pattern of residues mod 3.
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Hint 2 of 2
Group the nine positions by position mod 3; each group must hold one residue class, and the fixed 3 and 1 pin down two of the groups.
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Approach: use residues mod 3 across the strip
- Three adjacent numbers summing to a multiple of 3 forces position \(i\) and \(i+3\) to have the same residue mod 3, so the nine positions split into three groups of three by index mod 3.
- The numbers 1–9 also split by residue into \(\{3,6,9\}\equiv0\), \(\{1,4,7\}\equiv1\), \(\{2,5,8\}\equiv2\); each position-group must get one whole residue-group.
- The placed 3 (residue 0) and 1 (residue 1) pin which residue-group goes to their two position-groups, so the third assignment is forced too.
- Now arrange within groups: the group with no fixed number can be ordered \(3!=6\) ways, and each of the two groups with a fixed number has its other two numbers free (\(2!=2\) ways each), giving \(6\cdot2\cdot2=\) 24 ways (choice E).
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