## Pirate Portions

A handful of pirates distributed some golden coins among themselves, such that:
• the share of Pirate 1 plus half the total share of the rest (all pirates except #1)
• = the share of Pirate 2 plus 1/3 of the share of the rest (all pirates except #2)
• = ...
• = the share of Pirate N plus 1/(N + 1) of the share of the rest (all pirates except #N)
Captain Cutthroat, the puzzle-loving pirate, informed his captive Jolly Rogers of this strange method of distribution. Moreover, he promised to set Jolly free if he could correctly tell him the share of each pirate.
"The total number of coins is less than 1000, and more than 50% of us received an odd number of coins."
Jolly Rogers got to work immediately, and after some time, asked the Captain, "Can you give me a little hint?"
"The least a pirate got is this number."
Jolly Rogers was still not sure until the Captain boasted, "And I got more than 10 times that fellow."

Can you guess the number of pirates and the share of each?

Solutions were received from Dane Brooke, Joseph DeVincentis, Tim Edmonds, Denis Borris, Jimmy Chng Gim Hong, John Hewson, Radu Ionescu, Claudio Baiocchi, David Green, Jayavel Sounderpandian, Graeme McRae. Jayavel Sounderpandian had the most concise solution. Here it is [with my comments]:

The only proportions that satisfy the conditions bulleted in the problem statement are:
2 Pirates: 3:4
3 Pirates: 5:11:13
4 Pirates: 1:19:25:28

[Thus, any solution must be multiple of one of the above proportions.
The first statement "more than 50%..." rules out 2 pirates and restricts the other proportions to odd multiples.
The second statement "The least ..." forces the multiple for the 4-pirate proportion to be 5, 15, etc (so that the possible solutions both have the same minimum value). If the multiple is greater than 13, then the total number of coins is greater than 1000, so the multiple can only be 5.
The last statement "...more than 10 times..." forces the solution to be that for 4-pirates.]

Given the rest of the conditions, the answer is:
4 pirates with shares 5, 95, 125, 140.

Graeme McRae made his own website ( http://mcraefamily.com/MathHelp/PuzzlePirateAnswer.htm) to show the solution.
Tim Edmonds sent:
```I have to say I found the end a bit confusing in terms of getting rid of
the possible solutions. He's what I did, maybe I'll find my error on route :-)

First I found out possible solutions with 3 pirates (getting a,b,c coins)
since we know there must be at least three from the puzzle statement. The
two pirate solution is clearly a=3, b-=4 but the puzzle seems has three
equalities so I'm assuming 3 minimum.

a+(b+c)/2 = b+ (a+c)/3 = c+(a+b)/4 as per statement.

Resolving and re-arranging, I got 13a=5c. Therefore the smallest integer
solution is a=5, c=13 (and so b=11) and all multiples thereof.

Then I looked at four pirates :

a+(b+c+d)/2 = b+(a+c+d)/3 = c+ (a+b+d)/4 = d+ (a+b+c)/5

Resolving and re-arranging I got 28c=25d. Therefore the smallest integar
solution is c=25, d=28 giving a=1, b=19 and multiples thereof.

Moving onto five pirates .. I have not been able to get an answer out. I
have not tried higher pirates (this would be an algebraic nightmare!)
```

Joe DeVincentis sent:
```f we let X equal to the total amount of gold, and P1, P2, ... equal the
shares of the pirates, then we can write the main equation as:

1/2 X + 1/2 P1 =
1/3 X + 2/3 P2 =
...
1/(N+1) X + N/(N+1) PN

Taking consecutive pairs of terms from this, we get the following:
2/3 P2 - 1/2 P1 = 1/6 X  or  4 P2 - 3 P1 = X
3/4 P3 - 2/3 P2 = 1/12 X  or  9 P3 - 8 P2 = X
...

Alternatively, we can write these as:
P2 = (X + 3 P1)/4
P3 = (X + 8 P2)/9
...

If we substitute the expression for P2 into the expression for P3, we get:
P3 = 1/9 X + 8/9 P2
= 1/9 X + 2/9 X + 2/3 P1
= 1/3 X + 2/3 P1

Similarly, for other PN:
P4 = 3/8 X + 5/8 P1
P5 = 2/5 X + 3/5 P1
P6 = 5/12 X + 7/12 P1
...
PN = (N-1)/(2N) X + (N+1)/(2N) P1

Now, if we assume there are two pirates:
X = P1 + P2
= P1 + 1/4 X + 3/4 P1
= 1/4 X + 7/4 P1
3/4 X = 7/4 P1
P1 = 3/7 X
P2 = 4/7 X

If there are three pirates:
X = P1 + P2 + P3
= P1 + 1/4 X + 3/4 P1 + 1/3 X + 2/3 P1
= 7/12 X + 29/12 P1
5/12 X = 29/12 P1
P1 = 5/29 X
P2 = 11/29 X
P3 = 13/29 X

If there are four pirates:
X = P1 + P2 + P3 + P4
= P1 + 1/4 X + 3/4 P1 + 1/3 X + 2/3 P1 + 3/8 X + 5/8 P1
= 23/24 X + 73/24 P1
1/24 X = 73/24 P1
P1 = 1/73 X
P2 = 19/73 X
P3 = 25/73 X
P4 = 28/73 X

If there are more than four pirates, then the similar algebra
to the above gives X = (greater than 1) X + (positive) P1
which gives P1 is negative.  Thus there are no more than 4 pirates.

The clue "more than 50% of us received an odd number of coins"
eliminates the possibility of there being only 2 pirates, because
the one who received 4/7 of the coins would receive an even number
of coins.

At this point, Jolly Rogers can deduce that there are either 3 pirates, who
received numbers of coins 5, 11, and 13 times an odd number from 1 to 33,
or there are 4 pirates, who received numbers of coins 1, 19, 25, and 28
times an odd number of coins from 1 to 13.

When Jolly Roger is told the least number of coins a pirate got, if that
number was 15, 25, 35, ... then he'd know there were three pirates, and
the numbers of coins each pirate got.  If it was 1, 3, 7, 9, 11, or 13,
then he'd know there were four pirates.  Only if the fewest coins was 5
would he still not know.

In order for one pirate to get 10 times what another one got, there
must be 4 pirates, and the multiplier is 5.  The pirates therefore
got 5, 95, 125, and 140 coins.
```

Mail to Ken