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 Two sets of Coins (Posted on 2005-10-07)
You have two sets of five coins. Each set has four real 30g coins and one 31g fake coin. It is easy to find the fake coins using a balance scale a total of four times, two weighings for each set.

Can you find a strategy which uses only three weighings?

 See The Solution Submitted by Brian Smith Rating: 3.5000 (2 votes)

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 re: 3 simple steps | Comment 2 of 7 |

If ABC are weighed against abc and the result is equality, the possible heavy pairs are Aa, Ab, Ac, Ba, Bb, Bc, Ca, Cb, Cc, Dd, De, Ed, or Ed. That's 13 possibilities--which is more than can be determined in the two weighings left.  Your proposed second weighing of AB vs de would result in respective heavy sides of: L, L, L, L, L, L, =, =, =, R, R, R, R.  So if the result from that second weighing were not equality, then there'd be either 6 or 4 possibilities remaining--more than can be told in the remaining one weighing.

The following chart shows how the numbers of coins from each set in each pan partitions the set of possibilities in the first weighing.

The first two columns show the number from set 1 and from set 2 are in the left pan; the next two columns show the number from set 1 and from set 2 that are in the right pan.  The next two columns show how equality could be achieved by having no fake in either pan and by having one fake in each pain.  The columns after the colon column show the partitioning: number of ways to get equality, number of ways for the right pan to be heavy and number of ways for the left pan to be heavy.  None of these can be greater than 9, so as to assure the remaining two weighings will have at least a theoretic possibility of resolving the question.

`1 0; 0 1     16  1 :  17   4   41 0; 1 0     15  0 :  15   5   51 1; 0 2      8  2 :  10   8   71 1; 1 1      9  2 :  11   7   71 1; 2 0      8  2 :  10   8   72 0; 0 2      9  4 :  13   6   62 0; 1 1      8  2 :  10   7   82 0; 2 0      5  0 :   5  10  101 2; 0 3      0  3 :   3  12  101 2; 1 2      3  4 :   7   9   9 *1 2; 2 1      4  5 :   9   8   8 *1 2; 3 0      3  6 :   9   9   7 *2 1; 0 3      3  6 :   9   9   7 *2 1; 1 2      4  5 :   9   8   8 *2 1; 2 1      3  4 :   7   9   9 *2 1; 3 0      0  3 :   3  12  103 0; 0 3      4  9 :  13   6   63 0; 1 2      3  6 :   9   7   9 *3 0; 2 1      0  3 :   3  10  121 3; 2 2      0  8 :   8   8   9 *1 3; 3 1      1 10 :  11   7   71 3; 4 0      0 12 :  12   8   52 2; 1 3      0  8 :   8   9   8 *2 2; 2 2      1  8 :   9   8   8 *2 2; 3 1      0  8 :   8   9   8 *3 1; 0 4      0 12 :  12   8   5  3 1; 1 3      1 10 :  11   7   73 1; 2 2      0  8 :   8   8   9 *4 0; 0 4      1 16 :  17   4   44 0; 1 3      0 12 :  12   5   8`

Only those lines marked with an asterisk satisfy this criterion.

Due to symmetries (that is, it doesn't matter which set is which or which pan is which), the following are the only ones worth investigating as the first weighing:

1 2; 1 2      3  4 :   7   9   9 *
1 2; 2 1      4  5 :   9   8   8 *
1 2; 3 0      3  6 :   9   9   7 *
1 3; 2 2      0  8 :   8   8   9 *
2 2; 2 2      1  8 :   9   8   8 *

Note they all involve at least one of the pans having coins from both sets.

Edited on October 7, 2005, 7:15 pm
 Posted by Charlie on 2005-10-07 18:57:24

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