The Sailor's Reward Problem and Prime Numbers

Date: 12/11/2004 at 15:07:55
From: Michele
Subject: The Sailor's Reward Problem and Prime Numbers

The problem is from Malba Tahan's book, The Man Who Counted.  Briefly, 
there were between 200-300 coins in a chest.  The first sailor went to 
the chest, divided the coins into three equal piles with one coin left 
over, which he threw into the sea.  He took one pile and left the rest 
in the chest.  The next sailor did the same thing, dividing by three, 
throwing away the one left over and leaving the rest.  The third 
sailor did exactly the same thing.  Finally, they got to port and the 
tax man divided the remaining coins into three piles.  Each sailor got 
a pile and the tax man took the one left over.  The question is to 
figure out how many coins were in the chest to begin with and how 
many each sailor got. 

This is easy enough to figure out by eliminating possibilities. 

The beginning value had to be 241 coins.  One of my students asked if 
the answer had to be prime.  241 is a prime number, but is there 
anything in the problem that would REQUIRE the number to be prime?  In 
other words, could we have known from reading the problem, before 
listing and eliminating possibilities, that the number would be 
prime?  This would certainly have cut down on the number of 
possibilities we sorted through.

I don't think that the problem itself requires a prime answer, but I 
wanted to ask an expert, so I can share the answer with my students.

Date: 01/06/2005 at 01:56:16
From: Doctor Jeremiah
Subject: Re: The Sailor's Reward Problem and Prime Numbers

Hi Michele,

Say the ending value each sailor gets when the tax man takes one 
coin is X.  So when they reach port there are 3X+1 coins.

After the third sailor took some he put 3X+1 coins back in the 
chest.  That 3X+1 is 2/3 of the amount that existed before he got 
there (minus the one that he threw away)  So before the third sailor 
got there there were 1+(3X+1)*3/2 coins.

After the second sailor took some he put 1+(3X+1)*3/2 coins back in 
the chest.  That 1+(3X+1)*3/2 is 2/3 of the amount that existed 
before he got there (minus the one that he threw away)  So before 
the second sailor got there there were 1+(1+(3X+1)*3/2)*3/2 coins.

After the first sailor took some he put 1+(1+(3X+1)*3/2)*3/2 coins 
back in the chest.  That 1+(1+(3X+1)*3/2)*3/2 is 2/3 of the amount 
that existed before he got there (minus the one that he threw away)  
So before the first sailor got there there were 1+(1+(1+(3X+1)*3/2)
*3/2)*3/2 coins.

 Total coins = 1+(1+(1+(3X+1)*3/2)*3/2)*3/2
 Total coins = (81X+65)/8

So the question is: is there any value of X that makes 81X+65 
divisible by 8?  Here is a list of all the ones under 100:

     X  (81X+65)/8
     7      79     (prime)
    15     160 
    23     241     (prime)
    31     322 
    39     403 
    47     484 
    55     565 
    63     646 
    71     727     (prime)
    79     808 
    87     889 
    95     970 

So really, most of them aren't prime--you were just lucky that they 
asked for 200-300 coins.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/