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 Another Chain in Payment (Posted on 2003-11-17)
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?

 See The Solution Submitted by Gamer Rating: 3.0000 (1 votes)

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 Most of the rest of the solution | Comment 2 of 5 |
Following on from Sam's correct solution to the first part of the problem...

With 3 breaks we have enough pieces to pay the person in days 1-3 using the broken links. Then we need a 4-link piece to pay him on day 4 (taking back the broken links).
On days 5-7 we again pay him using the broken links until day 8 when we need an 8-link piece to pay him (taking back the 3 broken links and the 4-link piece).
This is repeated which takes us to day 16 when we need a 16-link piece (taking back the 3 broken links, the 8-link and the 4-link).
Repeating again takes us through to day 32 when we need a 32-link piece.
Repeating again takes us through to day 63 when we run out of chain.
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Taking this to a general case with x broken links we would have a chain made up of:

Unfortunately this is as far as I go since my sum theory is sufficiently distant that I can't remember how to simplify the equation to give a neat solution as to the total number of links.

 Posted by fwaff on 2003-11-17 09:57:00

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