# Cut the chain | 12 Sep 2009

Discussion in '\$1 Daily Competition' started by shabbir, Sep 12, 2009.

1. ### shabbirAdministratorStaff Member

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A Man has 23-link chain. He lives in a house whose rent is worth one link for one day. What is the fewest number of cuts he can make in the chain for paying rent for each day.

2. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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2 cuts.

He would cut link no. 4 and 11. ( counting from the same end )

3. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Explanation ::

After cutting links 4 and 11, he would be left with :
(A) Two chains of 1-link length each : link 4 and link 11.
(B) One chain of 3-link length : link 1~3
(C) One chain of 6-link length : link 5~10
(D) One chain of 12-link length : link 12~23

He then pays the 2 chains in (A) for the first 2 days.
On day 3, he gets back the 2 chains and gives (B) chain as rent.
Then pays 2 (A) chains for next two days.
And gets back 2 (A) chains + the (B) chain on day 6 and gives (C) chain as rent.
Then pays the 2 (A) chains and the (B) chain as rent for next 5 days.
And gets back all chains and pays the (D) chain as rent.

Then, he repeats the above steps till day 23.

4. ### shabbirAdministratorStaff Member

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Why. He need to Pay Daily Or else he could pay all 23 at one go.

5. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Of course he could have, and so there is no point of all this.
So, obviously I meant that only (one link/day), just I am lazy to type :p

OK. Let's have it your way :

Explanation ::

After cutting links 4 and 11, he would be left with :
(A) Two chains of 1-link length each : link 4 and link 11.
(B) One chain of 3-link length : link 1~3
(C) One chain of 6-link length : link 5~10
(D) One chain of 12-link length : link 12~23

He then pays the 2 chains in (A) for the first 2 days (one chain on each day).
On day 3, he gets back the 2 chains and gives (B) chain as rent.
Then pays 2 (A) chains for next two days (one chain on each day).
And gets back 2 (A) chains + the (B) chain on day 6 and gives (C) chain as rent.
Then pays the 2 (A) chains (one chain on each day) for day 7 and 8.
On day 9, he pays (B) chain as rent and gets back the 2 (A) chains.
He then pays the 2 (A) chains (one chain on each day) for next 2 days (till day 11).
And gets back all chains and pays the (D) chain as rent on day 12.

Then, he repeats the above steps till day 23.

Phew... > 100 chars added.

6. ### shabbirAdministratorStaff Member

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OK Got it.

7. ### naimishNew Member

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Congrs SP

8. ### SaswatPadhi~ Б0ЯИ Τ0 С0δЭ ~

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Thanx naimish :happy:

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