Puzzle:

A man needs to pay his rent and he was out of money. He found that his rent was worth about one gold link on his chain per day. What is the fewest number of cuts he can make in his 23-link chain to pay the rent for up to 23 days?

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Solution:

Solution:

Considering that the chain is not closed, it requires only two links to be cut. Cut link number 4 and link number 11 counting from the same beginning link. He then has 2 pieces of length 1 (the cut links), and one of 3, 6, and 12. He can then pay the rent as follows. One each of the first two days he can give a cut link. On the third day he gives the chain of 3 and gets his two cut links back. He uses them on days 4 and 5, and then trades all given so far and gives the 6-link chain on day 6. He then again repeats the first steps for days 7-11. On day 12 he gets all those links back and gives the 12-link chain. The then repeats the actions of the first 11 days to go all the way though day 23. For those knowing numbering systems, it will be noticed that this is basically a trinary numbering scheme.

## 9 comments:

the minimum cuts is 5.

first he makes one cut to break the chain after that he makes cuts in such a way that each set has 1 link, 2 links, 4 links, 7links, 9links, which solves his problem.

fist he he gives one ring after that

he gives two ring and take the one ring, third he give the one ring like that he he pays the 23 rings

YES.

I AGREE WITH MASHERLA.

ITS 5.

yah its good solution given by macherla keep it up macherla

yes nice answer man

yeah!its lyk that

yup, right...but there are other possibilities too...like, 1,2,4,8,8.

Can 0 be the answer? He can pay the rent on the last day. :P

Say, if I have to pay for 63:days at one link a day out of my chain which has 63 links provided I could cut only 3 links, cause I need to pay for the coppersmith who charges 1 ₹ for each link separated from my packet and I v only 3₹ .any one who can solve it

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