Reply to Re: Two interesting puzzles
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I know the 3 measurements way. I can get the heavier coin out of a total of 9 coins with 2 measurements, but I haven't figured out what happens when there's a total of 12 coins.
3 measurements:
1,2,3,4 VS 5,6,7,8
If left is heavier, continue with 1,2,3,4
If right is heavier, continue with 5,6,7,8
If they have the same weight, continue with 9,10,11,12
4 coins left. For simplicity, assume these are 1,2,3,4.
1,2 VS 3,4
Left heavier: continue with 1,2
Right heavier: continue with 3,4
2 coins left. For simplicity, assume these are 1,2
1 VS 2
And job done!
9 coins in 2 measurements:
1,2,3 VS 4,5,6
Left -> 1,2,3
Right -> 4,5,6
Neither -> 7,8,9
Assume 1,2,3
1 VS 2
Left -> 1
Right -> 2
Neither -> 3
12 anyone?
3 measurements:
1,2,3,4 VS 5,6,7,8
If left is heavier, continue with 1,2,3,4
If right is heavier, continue with 5,6,7,8
If they have the same weight, continue with 9,10,11,12
4 coins left. For simplicity, assume these are 1,2,3,4.
1,2 VS 3,4
Left heavier: continue with 1,2
Right heavier: continue with 3,4
2 coins left. For simplicity, assume these are 1,2
1 VS 2
And job done!
9 coins in 2 measurements:
1,2,3 VS 4,5,6
Left -> 1,2,3
Right -> 4,5,6
Neither -> 7,8,9
Assume 1,2,3
1 VS 2
Left -> 1
Right -> 2
Neither -> 3
12 anyone?
