In a pile, there are 11 coins: 10 coins of common weight and one coin of different weight (lighter or heavier). They all look similar.
Using only a balance beam for only three times, show how you can determine the 'odd' coin.
Open problem (i cannot solve this myself): how many more coins (with the same weight as the ten) can we add to that pile so that three weighing still suffices? My conjecture is zero, though my friend guessed that adding one is possible. The best bound we can agree upon is < 2.
15 is the max number to achieve goal in 3 weighings. See example below ("0" represent coins)
1st weighing:
Lt side: 000 000 0
Rt side: 000 000 0
Not weighed: 0
2nd weighing:
Lt side: 000
Rt side: 000
Not weighed: 0
3rd weighing:
Lt side: 0
Rt side: 0
Not weighed: 0
Correct me if I'm wrong
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Posted by I
on 2002-10-23 09:16:37 |
answer
