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?
(In reply to
re: That's peculiar by Bractals)
Warm greetings to you Bractals!
I am working on your seven stones.
Thanks. I am clear on that now.
How do you measure the weight 14?
Easy. 1 + 3 + 9 vs 27
Can you check Charlie's first post "upper limit".
OK. Let's see.
From Charlie:
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.
I'm struggling with the English, especially in the first sentence. Obviously, there is only one weighing using a single stone. What is Charlie trying to say?
I will take a stab at the kind of counting argument I think may be going on here: Each stone may appear on the left, on the right or unused. That makes 3^N arrangements. But this overcounts the range of cases, since it counts identical arrangements with the left and right sides of the scale switched. The number of unique cases encompassable with N stones is thus (3^N1)/2. The 1 comes in when we exclude the null stone case.
How do we stand?
Note: One idea would be to for Perplexus to support a notification function. As it is, I could easily have overseen your message.
Edited on February 8, 2008, 10:19 am

Posted by FrankM
on 20080208 10:06:37 