Re: pirates of the caribbean
Here is the answer for your question...
Pirates1= 1
Pirates2= 0
Pirates3= 1
Pirates4= 0
Pirates5(me) = 98
we already discuss this problem in our brainteasers forum. Here is the reference link...
[url]http://www.geekinterview.com/talk/2198-thiefs.html[/url]
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suresh
Re: pirates of the caribbean
Actually i asked dis question!!
Re: pirates of the caribbean
[QUOTE=smart_coder;13575]Actually i asked dis question!![/QUOTE]
can anyone plz describe me the solution..
m not able to get the logic...
Re: pirates of the caribbean
My solution is:
Pirate 5 (Me): 25
Pirate 4 : 25
Pirate 3 : 25
Pirate 2 : 13
Pirate 1 : 12
Here Pirate 4 and 3 will vote for me. I have my own vote at my side. This will keep me in majority. Pirate 4 and 3 will think that if they do not accept my solution then they will murder me but 100 coins will be then distributed among them. Probably they will get 25 only which anyway they are getting. So why not accept my solution?
Pirate 1 and 2 will not accept. But they what will happen? Nothing. They constitute lesser than 50 % in the total quorum. So anyway I will get more than others and I will remain alive.
Correct me if I am wrong.
Re: pirates of the caribbean
pirate 5 = 22
pirate 4 = 21
pirate 3 = 20
pirate 2 = 19
pirate 1 = 18
Re: pirates of the caribbean
Suresh got the answer correct. But My question has two parts, one part is to answer this question and the second part is to derive a generalized solution for N number of pirates :)