**SOLUTION:**

**for question see:king and wine puzzle**

**Some of you might already had seen the solution of this puzzle in binary form like following:**

Number the bottles 1 to 1000 and write the number in binary format like bottle 1 = 0000000001 bottle 2=0000000010 ans so on.

**now take 10 prisoners and number them 1 to 10, now let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit. Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc.prisoner = 10 9 8 7 6 5 4 3 2 1bottle 845= 1 1 0 1 0 0 1 1 0 1**

For instance, bottle no. 845 would be sipped by 10,9,7,4,3 and 1. That way if bottle no. 845 was the poisoned one, only those prisoners would die.So after 4 weeks you have to check who had died and you can find that bottle.But this solution is somewhat artificial as some of my friends say.See the following

For instance, bottle no. 845 would be sipped by 10,9,7,4,3 and 1. That way if bottle no. 845 was the poisoned one, only those prisoners would die.So after 4 weeks you have to check who had died and you can find that bottle.But this solution is somewhat artificial as some of my friends say.See the following

first let us take 1 prisoner,he can drink one bottle if its poisoned he will die.so this way u will need 1000 prisoners

first let us take 1 prisoner,he can drink one bottle if its poisoned he will die.so this way u will need 1000 prisoners

**Take 2 prisoners and name them as a1,a2**

**their group**

**is like**

**a1**

**a2**

**a1,a2**

**none total:4=2^2**

**and make this four groups drink 4 bottles(named) if a2 alone dies u can identify that bottle**

Take 3 prisoners a1,a2,a3.group them as

Take 3 prisoners a1,a2,a3.group them as

**a1**

**a2**

**a3**

**a1,a2**

**a2,a3**

**a1,a3**

**a1,a2,a3**

**none total:8=2^3 if a1,a2 alone dead that wine bottle is poisoned**

so to test 1000 bottle

so to test 1000 bottle

**2^n>1000**

which gives the answer n=10

which gives the answer n=10

any comments and suggestions are welcome.

any comments and suggestions are welcome.

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