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Tue Sep 18 15:25:42 PDT 2007


    Car talk puzzler Mar 4th 2000


      I heard this one on the radio. It was in a program called "CarTalk": [ http://cartalk.cars.com/Radio/ ] which is kind of indescribable but fun. Each week they set the audience a little puzzle [ http://cartalk.cars.com/Radio/Puzzler/ ]

      On February the 26th (or march the 4th, when the radio waves finally made it over the San Bernardino mountains...) the CarGuys set the following Puzler:

        Going Fishing

        Three guys finish a day of fishing and decide to divide the catch equally in the morning. In the middle of the night one of the guys has to take off for a drugstore. He goes to take his third of the fish, and notices that the number that they caught is not divisible by three. He throws one of the fish overboard, takes his third and leaves. A few hours later, another guy wakes up with horrible stomach pains, he goes to take his third, and notices that he, too, can't take a third unless he throws one fish away. He throws one fish overboard, takes his third, and goes home. Finally, the third guy gets up in the morning and goes to get his third and again, can't take a third. What is the smallest number of fish by which this little scenario could have taken place?


      It is a variation of the coconuts and the sailors (to say nothing of the monkey) problem publicized by Martin Gardner in an earlier millenium.

      The Problem

    1. PROBLEM::=following
      1. n::Nat, the number of fish at the start.
      2. i::Nat, the number of fish taken away by the first fisherman.
      3. j::Nat, the number of fish taken away by the second fisherman.
      4. k::Nat, the number of fish taken away by the third fisherman.

      5. |- (1): n=3*i+1.
      6. |- (2): 2*i=3*j+1.
      7. |- (3): 2*j=3*k+1.

      (End of Net)

      The Solution

    2. SOLUTION::=following

      1. |-PROBLEM.
      2. (1)|- (4): 2*n=3*2*i+2.
      3. (4, 2)|- (5): 2*n=3*(3*j+1)+2.
      4. (5)|- (6): 2*n=9*j+5.
      5. (6)|- (7): 4*n=9*2*j+10.
      6. (7, 3)|- (8): 4*n=9*(3*k+1)+10.
      7. (8)|- (9): 4*n=27*k+19.

      8. (9)|- (10): if k=1 or 2 then no n.
      9. (9)|- (11): if k=3 then n=25 and i=8 and j=5.
      10. (9)|- (12): as k increases so does the n that satisfies PROBLEM.
      11. (10, 11, 12)|- (13): min(n. 4*n=27*k+19) = 25.

      (End of Net)

      Proof of 10

    3. 27*1+19 and 27*2+19 are not divisible by 4.


    4. Nat::=the natural numbers 1,2,3, and so one, [ Nat in math_42_Numbers ]
    5. min::=the minimum value of a variable satisfying some property, [ MINMAX in math_21_Order ]

    . . . . . . . . . ( end of section Car talk puzzler Mar 4th 2000) <<Contents | End>>