The problem goes like this.
Three fihserman go fishing in the middle of a stormy noght. They catch some fish, then land on a desert island and go to sleep. Later one of them wakes up and he says to himself, " I'll take my one third of the fish and I'll go away. " So he divides them into three parts. He finds one extra fish. He throws the extra fish into the sea. Then he takes his share and leaves that place. Next fisherman does the samehting...divides into three parts, one extra fish is left. Throws that into the sea. ( Without knowing that the first fisherman departed. )
Finally the third one also wakes up. Then he too does the samething. Divides it into three parts and throws the extra fish into the sea.
Now the question is, " What is the minimum number of fish? "
WHy Dirac comes into picture? Because it was given quickie problem to him at St. John's college. Quick as a flash he gives the answer as minus two.
The reasoning for the minus two is : Minus one + minus one + minus one + plus one = minus two. Now divide into three the extra one is plus one. But you have to throw the plus one into the sea. again minus two.
Clearly you must add a definite positive no. to get a sensible no fish.( of course positive.)
Answer : 25 ( Solved by me! )
[ The above problem is taken from the book " Unification of fundamental forces " by Sir. Abdus Salam, Nobel prize winner along with Sir. Weinberg and Sir. Glashow for Unification of wek forces with Electromagnetism. Hmmmm....I'm on the track. ]
This blog is for refining my ideas and post the things which I know/read.
Monday, August 22, 2005
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5 comments:
when I solved I got 13
1st fisher 4+4+4+1
2nd fisher 1+1+1+1
la fisher 0
can u gimme ur explination in a clear state without those minus into minus plus divded by n stuff?
ragz...
u see...(13-1)/3 = 4 and now the remaining ones are 8 and this no is not in the form of 3x+1
-2, is mathematically correct answer; the smallest possible number.
But it is not logically correct: as "-2" is not a realistic number of fish in "THAT SCENARIO".
It is "A PROBLEM WITH NO REALISTIC SOLUTION"
Explanation:-
Let first fisherman be: Mrs. A
Let second fisherman be: Mr. B
Let third fisherman be:Miss. C
Minimum number of fish that one can take is "1" [one]
Hence no. of fish took by our little girl, C = 1.
If our little kid, C took one after dividing the total fish, it is from three fishes.
Also our naughty girl C threw away one fish to make it as a whole number to divide.
That means, our darling saw only four fishes when she woke up.
( 4=3+1, 3 which she share for them and the extra one she threw)
Sooooo....., our great Mr.B leave four fishes to C.
Mr.B have thrown one fish, took 1/3rd of rest of the fishes and left 2/3rd portion of fishes.
If the left part ie., 2/3 of total fish that C saw, is four fishes, then, 1/3 will be two fishes.Also there is one fish thrown by this great man.
So number of fish undergone through the custody of Mr.B: 4+2+1=7
Similar way Mrs.A, that busy lady, go early after taking her share, by leaving 7 fishes, that 2/3rd portion for Mr.B and Miss.C.
IT WILL BE ALWAYS NOT POSSIBLE FOR 'Mrs.A' THE 'THIRD FOLD PART OF CALCULATION' TO EXIST "WITH" AN EVEN NUMBER BY IMPOSING THE CONDITION OF CARRY OVER OF THAT ONE THROWN FISH IN THE SECOND FOLD PART "ie., from Mr.B"
{This statement is valid as far as C is even, else B will be locked.}
Definitely she should be a mysterious mathematician.
Because: 2/3=(remaining fish)/(total fish)
Remaining fishes's count should be perfectly divisible by "2".
Our dear lady, who only can have an odd number of fishes as 2/3rd portion....we are unfortunate to have her with us, because she go away with what she took.
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