.Open Car talk puzzler Mar 4th 2000 . Introduction I heard this one on the radio. It was in a program called "CarTalk": .See http://cartalk.cars.com/Radio/ which is kind of indescribable but fun. Each week they set the audience a little puzzle .See 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: .Box 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? .Close.Box . Note 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 PROBLEM::=following .Net n::\$Nat, the number of fish at the start. i::\$Nat, the number of fish taken away by the first fisherman. j::\$Nat, the number of fish taken away by the second fisherman. k::\$Nat, the number of fish taken away by the third fisherman. |-(1): \$n=3*\$i+1. |-(2):2*\$i=3*\$j+1. |-(3):2*\$j=3*\$k+1. .Close.Net . The Solution SOLUTION::=following .Net |-\$PROBLEM. (1)|-(4):2*\$n=3*2*\$i+2. (4,2)|-(5):2*\$n=3*(3*\$j+1)+2. (5)|-(6): 2*\$n=9*\$j+5. (6)|-(7): 4*\$n=9*2*\$j+10. (7,3)|-(8): 4*\$n=9*(3*\$k+1)+10. (8)|-(9): 4*\$n=27*\$k+19. (9)|-(10):if \$k=1 or 2 then no \$n. (9)|-(11):if \$k=3 then \$n=25 and \$i=8 and \$j=5. (9)|-(12):as \$k increases so does the \$n that satisfies \$PROBLEM. (10, 11, 12)|-(13): \$min(\$n. 4*\$n=27*\$k+19) = 25. .Close.Net . Proof of 10 27*1+19 and 27*2+19 are not divisible by 4. . Glossary Nat::=`the natural numbers 1,2,3, and so one`, .See http://www/dick/maths/math_42_Numbers.html#Nat min::=`the minimum value of a variable satisfying some property`, .See http://www/dick/maths/math_21_Order.html#MINMAX .Close Car talk puzzler Mar 4th 2000