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 Ceremonial Stones (Posted on 2006-11-14)
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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 Generalisation | Comment 20 of 26 |

What is wanted is a sequence aJ such that each integer <=NJ can be expressed uniquely as a some of sum of terms mJ*aJ where each multiplier mJ is either -1,0 or +1.

This is given by aJ = 3^(J-1) so that the tribal ancient was about to experience his 14th birthday (=one more than 1+3+9).

_________________________

Still confused? With weights 1,3,9 we can uniquely generate each age from 1 to 13:

1= 1

2= 3-1

3= 3

4= 3+1

5= 9-3-1

6=9-3

7=9-3+1

8=9-1

9=9

10=9+1

11=9+3-1

12=9+3

13=9+3+1

aJ is chosen such that putting aJ on the left and all smaller stones in the right gives the first new number, namely 1 more than the sum of the stones a1 thru aJ-1

 Posted by FrankM on 2008-02-03 11:46:36

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