Problem 20 · 2019 Math Kangaroo
Hard
Counting & Probability
sum-constraintcasework
The intersection points of the network of bars shown are labelled with the numbers 1 to 10. The sum \(S\) of the four numbers at the vertices of each square is the same for all three squares. What is the minimum possible value of \(S\)?
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Answer: C — 20
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Hint 1 of 2
Two of the ten points are shared between neighbouring squares, so they each count in two of the square sums.
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Hint 2 of 2
Adding all three square sums gives \(3S = 55 + (\text{the two shared vertices})\); minimise \(S\) by choosing those two values well.
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Approach: minimise the common square sum using the shared vertices
- The labels 1 to 10 add to 55. Summing the three equal square sums counts the two shared (middle) vertices twice, so \(3S = 55 + (\text{sum of the two shared vertices})\).
- We need that total divisible by 3; since \(55 \equiv 1 \pmod 3\), the two shared vertices must sum to \(2 \pmod 3\).
- The smallest such sum of two distinct labels is \(1 + 4 = 5\) (or \(2 + 3\)), giving \(3S = 60\), so \(S = 20\), and the remaining labels can be placed to make it work.
- Answer (C) 20.
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