You have eight bags, each of them containing 48 coins.
Five of these bags contain only true coins, the rest of them contain fake coins. Fake coins weigh 1 gram less than the real coins.
You do not know what bags have fake coins and what bags have real coins. You do not know also, besides that it is an integer value, the weight of the real coins.
You can use a digital or analog reading scale with precision up to 1 gram.
Making only one weighing and using the minimum number of coins, how can you find the bags containing the fake coins?
The best that I can come up with requires 49 coins from one of the bags. The numbers from each bag are:
0 1 3 6 11 18 30 49
If all the bags had good coins the sum would weigh 118 grams. Depending on which three numbered bags contain bad coins, the following totals would appear:
1 2 3 114 2 4 7 81
1 2 4 111 2 4 8 62
1 2 5 106 2 5 6 88
1 2 6 99 2 5 7 76
1 2 7 87 2 5 8 57
1 2 8 68 2 6 7 69
1 3 4 109 2 6 8 50
1 3 5 104 2 7 8 38
1 3 6 97 3 4 5 98
1 3 7 85 3 4 6 91
1 3 8 66 3 4 7 79
1 4 5 101 3 4 8 60
1 4 6 94 3 5 6 86
1 4 7 82 3 5 7 74
1 4 8 63 3 5 8 55
1 5 6 89 3 6 7 67
1 5 7 77 3 6 8 48
1 5 8 58 3 7 8 36
1 6 7 70 4 5 6 83
1 6 8 51 4 5 7 71
1 7 8 39 4 5 8 52
2 3 4 108 4 6 7 64
2 3 5 103 4 6 8 45
2 3 6 96 4 7 8 33
2 3 7 84 5 6 7 59
2 3 8 65 5 6 8 40
2 4 5 100 5 7 8 28
2 4 6 93 6 7 8 21
These totals are all unique.
This is based on the good coins weighing 1 gram apiece. That of course is not possible, as it would mean the light coins would weigh zero. But the total can be reduced by a mulitple of 118 such that a weight of less than 118 is obtained, and then you use the above table.
The weights used are interesting in that each one is a prime number higher than the one before except that one increment is 1 and another is 12, rather than 11. But if we substitute 11 and use 0 1 3 6 11 18 29 48 as the numbers to use, we get ambiguous results:
1 2 3 112 2 4 7 80
1 2 4 109 2 4 8 61
1 2 5 104 2 5 6 86 *****
1 2 6 97 2 5 7 75
1 2 7 86 2 5 8 56
1 2 8 67 2 6 7 68
1 3 4 107 2 6 8 49
1 3 5 102 2 7 8 38
1 3 6 95 3 4 5 96
1 3 7 84 3 4 6 89
1 3 8 65 3 4 7 78
1 4 5 99 3 4 8 59
1 4 6 92 3 5 6 84 *****
1 4 7 81 3 5 7 73
1 4 8 62 3 5 8 54
1 5 6 87 3 6 7 66
1 5 7 76 3 6 8 47
1 5 8 57 3 7 8 36
1 6 7 69 4 5 6 81 *****
1 6 8 50 4 5 7 70
1 7 8 39 4 5 8 51
2 3 4 106 4 6 7 63
2 3 5 101 4 6 8 44
2 3 6 94 4 7 8 33
2 3 7 83 5 6 7 58
2 3 8 64 5 6 8 39 *****
2 4 5 98 5 7 8 28
2 4 6 91 6 7 8 21
where the *'s beside 5,6,8 39, for example indicate that the 39 duplicates that of 1,7,8.
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Posted by Charlie
on 2008-10-23 13:35:52 |
not quite -- my best
