Solution
Assume the colors are A, B, C and the balls labeled A1, A2, B1, B2, C1, C2. We have to replace the labels with more informative AL, AH, BL, BH, CL, CH, where "L" stands for "light" and "H" stands for "heavy".
Weigh A1 + B1 against B2 + C1.
If A1 + B1 < B2 + C1 then
B1 < B2 because even if A1 < C1, B1 > B2 would lead to A1 + B1 = B2 + C1, at best. Thus B1 = BL, B2 = BH.
If A1 = AH then C1 = CH and
if C1 = CL then A1 = AL.
With this in mind, weigh A1 + C1 against B1 + B2.
A1 + C1 = B1 + B2 implies A1 = AL and C1 = CH.
A1 + C1 < B1 + B2 implies A1 = AL and C1 = CL.
A1 + C1 > B1 + B2 implies A1 = AH and C1 = CH.
If A1 + B1 = B2 + C1 then weigh B1 against B2.
The case where A1 + B1 > B2 + C1 is similar to the first one.
Weighing Coins, Balls, What Not ...
- The Oddball Problem, B. Bundy
- Weighing 12 coins, Dyson and Lyness' solution
- Weighing 12 coins, W. McWorter
- Thought Less Mathematics, D. Newman
- Weighing with counterbalances
- Odd Coin Problems, J. Wert
- Six Balls, Two Weighings
- 12 Coins in Verse
- Six Misnamed Coins, Two Weighings
- A Fake Among Eight Coins
- A Stack of Fake Coins
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Copyright © 1996-2012 Alexander Bogomolny