Problem 29 · 2021 Math Kangaroo
Stretch
Logic & Word Problems
work-backwardsum-constraint
Christina has eight coins whose weights in grams are different positive integers. Whenever she puts any two coins on one side of a balance scale and any two on the other side, the side containing the heaviest of those four coins is always the heavier side. What is the smallest possible weight of the heaviest coin?
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Answer: C — 34 g
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Hint 1 of 2
The hardest case is the heaviest coin with the lightest partner versus the next two heaviest.
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Hint 2 of 2
That forces each coin to exceed the sum of the previous two minus the lightest — a Fibonacci-style growth from 1, 2.
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Approach: push the worst pairing to a Fibonacci-type bound
- The binding condition is: heaviest + lightest > (2nd heaviest) + (3rd heaviest), with the analogous rule for every coin acting as the maximum.
- Taking the lightest as 1 forces each weight to be at least the sum of the two preceding ones.
- Minimal distinct weights are 1, 2, 3, 5, 8, 13, 21, 34.
- The smallest possible heaviest coin is 34 g, choice (C).
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