.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