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### 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/
```
Associated Topics:
High School Basic Algebra
Middle School Algebra
Middle School Word Problems

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