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Another Chain in Payment (Posted on 2003-11-17) Difficulty: 2 of 5
You have 23 links in a gold chain (just like before), and will pay one link per day to someone. Which two chain pieces should you break in order to pay this person?

What would be the longest chain you could have in order to pay this person (one each day) by breaking 3 pieces in the chain?

Is there a rule for how many days you could pay someone under the same circumstances if you could break x pieces in the chain?

  Submitted by Gamer    
Rating: 3.0000 (1 votes)
Solution: (Hide)
The first chain could be split so that the fourth link from one end and sixth link from the other end was cut. This would make chains of 3, 1, 12, 1, 6.

1: Pay with the 1 chain
2: Pay with the 1 chain
3: Take back both 1 chains, pay with the 3 chain
4: Pay with the 1 chain
5: Pay with the 1 chain
6: Take back 1 and 3 chains, pay with the 6 chain
7: Pay with the 1 chain
8: Pay with the 1 chain
9: Take back both 1 chains , pay with the 3 chain
10: Pay with the 1 chain
11: Pay with the 1 chain
12: Take back 1, 3 and 6 chains, pay with the 12 chain
13: Pay with the 1 chain
14: Pay with the 1 chain
15: Take back both 1 chains, pay with the 3 chain
16: Pay with the 1 chain
17: Pay with the 1 chain
18: Take back 1 and 3 chains, pay with the 6 chain
19: Pay with the 1 chain
20: Pay with the 1 chain
21: Take back both 1 chains, pay with the 3 chain
22: Pay with the 1 chain
23: Pay with the 1 chain


4, 1, 16, 1, 32, 1, 8 is the next set.

1: Pay with a 1 link
2: Pay with a 1 link
3: Pay with a 1 link
4: Take back all 1 links, pay with the 4 link
5: Pay with a 1 link
6: Pay with a 1 link
7: Pay with a 1 link
8: Take back the 4 link and the 1 links, pay with the 8 link...

Paying by continuing on with powers of two would work for this situation.

Fwaff figured that (2^(x+1))(x+1) - 1 would work as a formula.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
not sufficient informationdonnmike2003-12-06 02:22:32
Comment to LevikS2003-11-17 10:45:11
SolutionThe last little bit...fwaff2003-11-17 10:10:52
SolutionMost of the rest of the solutionfwaff2003-11-17 09:57:00
Some ThoughtsPartial SolutionSam2003-11-17 09:07:31
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