Monkey and the Shipwreck

7/28/2019 Monkey and the Shipwreck

1/4

Statement of the Problem

Given m sailors and a pile of coconuts, each sailor in sequence takesm

1of the coconuts

left after the previous sailor removed his share and gives n (less than m) coconuts to a monkey.

Then in the morning, they divide the remaining coconuts in m ways and give the n coconuts

which are left over to the monkey. Ifn is the same at each division, then how many Ccoconuts

were there originally?

Solution:

Assuming Ciis the number of coconuts left after the previous the sailor removed his share

where . Since n coconuts were given to the monkey at each division, then the first

sailor took nCm

1

coconuts. That leaves C1 coconuts left after the first sailor took his share.

In equation, we write

nm

mC

m

mC

nCm

mC

nCm

nCC

11

1

1

1

1

1

.

Take note that m

m 1is the proportion

of the coconuts that was left after each sailor in

sequence took his share. So after the second sailor took his share, there were C2 coconuts left,

where,

n

m

mn

m

mC

m

mC

nnm

mC

m

m

m

mC

nCm

mC

111

111

1

22

2

2

12


2/4

After the third sailor took his share, there were C3 coconuts left, where,

nm

mn

m

mn

m

mC

m

mC

nnm

mnm

mCm

m

m

mC

nCm

mC

1111

1111

1

233

3

22

3

23

After the fourth sailor took his share, there were C4 coconuts left, where,

nm

mn

m

mn

m

mn

m

mC

m

mC

nnm

mn

m

mn

m

mC

m

m

m

mC

nCm

mC

11111

11111

1

2344

4

233

4

34

After the fifth sailor took his share, there were C5 coconuts left, where,

nm

mn

m

mn

m

mn

m

mn

m

mC

m

mC

nnm

mn

m

mn

m

mn

m

mC

m

m

m

mC

nCm

mC

111111

111111

1

23455

5

2344

5

45


3/4

Clearly, from the second term to the last term of the equations, they form a geometric

progression, where the common ratio r=

m

m 1. Thus, successive Cis can be generated.

terms

mmm

mn

m

mn

m

mn

m

mn

m

mC

m

mC

nm

mn

m

mn

m

mn

m

mn

m

mn

m

mC

m

mC

nm

m

nm

m

nm

m

nm

m

nm

m

Cm

m

C

m

21

234566

6

23455

5

11111

1111111

111111

In the morning, afterm sailors kept their share, there were Cm coconuts left. For the finaldivision, n coconuts were given to the monkey and each man got Ycoconuts, where,

termsm

mm

m

m

termsm

mmm

m

m

m

m

m

m

m

m

m

m

n

m

CmY

nnm

mn

m

mn

m

mn

m

mC

m

m

mY

nCm

Y

)1(

21

1

)1(

21

111111

111111

1

Notice that the terms in the bracket is a geometric progression. It can be simplified as,

m

m

m

m

n

m

CmY

m

m

m

m

m

n

m

CmY

m

m

m

m

m

m

1

11

1

11

11

1

1

1

1

1


4/4

Further algebraic manipulation and clearing out fractions will yield to simple equation.

11

11

11

11

11

1

1

11

1

1

1

1

1

1

1

mCmnYm

mCmnYm

m

m

m

CmnY

m

mnnm

CmY

m

mn

m

CmY

mm

mmm

m

m

m

m

m

m

m

m

m

m

The resulting equation looks a lot simpler, but it still may not be clear how to make

progress. Since m and n are whole number constants, Y and Cmust be whole numbers also.

Clearly, there are infinite number of solutions. Solving forCin terms ofm, notice that 1m

m

does

not divide mm 1 and vice versa. Thus, 1mm must divide 1mC . By definition ofcongruence, we have

1

1

mod1

mod01

m

m

mmC

mmC

Solving forY, we can apply the same principle. mm 1 must divide nY . Thus,

mm

mnY

mnY

1mod1mod0

In the original problem with m = 5 and n = 1, 15625mod4C and

1024mod1Y . The smallest positive solution to the problem is there are 15625 coconuts atthe beginning.

Documents

Monkey and the Shipwreck