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Ceremonial Stones (Posted on 2006-11-14) Difficulty: 5 of 5
A certain tribe had an interesting kind of annual festival, in which every male member of the tribe (regardless of age) had to contribute a levy of grain into the tribal store. Their unit of weight was roughly the same as our pound avoirdupois, and each tribesman had to contribute one pound of grain for every year of his age.

The contributions were weighed on the tribe's ceremonial scales, using a set of seven ceremonial stones. Each of these weighed an integral number of pounds, and it was an essential part of the ritual that not more than three of them should be used for each weighing, though they need not all be in the same pan.

If ever a tribesman lived to such an age that his contribution could no longer be weighed by using three or fewer stones, the levy of grain would terminate forever. And in the previous year, one old man had died only a few months short of attaining this critical age, greatly to the relief of the headman of the tribe.

It has been determined that the stones can measure the maximum age. What is this age and what were the weights of the seven ceremonial stones?

See The Solution Submitted by Bractals    
Rating: 3.0000 (7 votes)

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Some Thoughts upper limit | Comment 1 of 25

For a given set of 7 stones there are 7 weighings that are possible with one stone.  There are C(7,2)=21 weighings that can combine two stones on the same side of the scale, and the same number, 21, that have two stones on opposite pans.

There are C(7,3)=35 weights that can be made with three stones together, and 7*C(6,2)=105 that can oppose one stone against two in the opposite pan.

Altogether that's 7+2*21+35+105=189, so no more than 189 different weights can be measured.

But undoubtedly there'll be duplications, as well as totals above 189, each of which will prevent some number under 189 from being weighable.  So the answer will be somewhat under 189.


  Posted by Charlie on 2006-11-14 13:45:30
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