Math Problem

[Blopa]

Hey there forum, I have been reading this awesome book called The Man Who Calculated (really recommended, it's a good recreational book), in this book the main character faces different mathematical problems which he solves easily, in the book it appears one which he only answers but doesnt explains how he solved it, so I in my curiosity tried to solve the problem through math, but it became troublesome, I would really appreaciate if someone here helps me.

 

Here's the story of the problem:

 

3 sailors got some money for doing a very good work, the amount of golden coins they were going to get was between 200 and 300. They were going to split it in the morning.

 

During the night one of them woke up and decided he was going to take his share before the morning, so he went to where the money was kept and divided the coins in 3, there was one coin left, so he thought "we are going to fight over this coin in the morning", so he decided to throw it to the sea. He took his coins and left the rest in there.

 

Later in the night, another one of the sailors woke up with the same idea, but without knowing the other one had already taken his part, so he went and splited the coins in 3, and there was also one coin left, so after thinking the same as the other sailor, he threw it to the sea, he took his share and leaved the rest.

 

After that the story repeats with the third sailor.

 

And later, another sailor assigned by the captain in the morning went to help them divide it, finding out there was (again) one coin left, which he took for himself for the trouble.

 

 

That's the story, I tried to solve it, but I can't, I wonder if one of you guys can solve it, I would be thankful for the instruction.

[Bignose]

Do you want help to solve it yourself, or do you want an answer?

 

The first split must go like this. Let's call the original total of coins N.

 

N0 must be in the form 3*P+1 because N is split into 3 equal piles and the extra 1 tossed away.

 

So, N = 3P + 1.

 

Then, after the 1st man leaves, the pile remaining is 2P.

 

And repeat this procedure again for the next man:

 

2P = 3Q + 1

 

and the pile remaining after the 2nd man would be 2Q.

 

Repeat for the next 2 splits.

 

I think that the last clue is that the original pile was between 200 and 300 coins.

 

I think that the one flaw in the problem is that each of the 3 men should have known that the original amount was between 200 and 300 right? Well, after the 1st man takes his cut, the remaining pile would no longer be between 200 and 300, and the second and third men should have been upset that the amount wasn't right.

 

I.e. say the original amount was N=298. After the 1st man splits this pile and then tosses the extra 1, and then takes his third, the pile would be 198 = 2 * 99. Well, the second guy should have been upset that this pile was less than 200. Any number between 200 and 300 will have this same issue.

 

But, if you assume that this fact wouldn't bother people, I think that the problem is solvable. If the above hints aren't enough, I'll come back and try to solve it later.

[Blopa]

I understand your analysis I did the same, but still couldnt solve it...

how do you think you can solve it with that?

 

And yeah the key is in the 200<N<300 but I cant figure out how to use it...

Thanks by the way

 

And I really dont remember how he worked the thing about they noticing the missing coins, I just paraphrased the problem, I read the book a year ago and remember the problem today

Edited December 7, 2010 by Blopa

[Bignose]

 

 

I get the following equations describing all the splits:

 

1st split: N = 3P+1

2nd split: 2P = 3Q+1

3rd split: 2Q = 3R+1

4th split: 2R = 3S+1

 

5 unknowns, 4 equations. The last piece of knowledge is that N is an integer between 200 and 300.

 

There is probably a formal way to solve this, but I wrote a small program in Python:

 

def thisfunc(x):

... R2=3*x+1

... Q2=1.5*R2+1

... P2=1.5*Q2+1

... N=1.5*P2+1

... return N

 

then ran values from 19 to 28 (the range of integer inputs that yield results between 200 and 300). You actually only have to run the odd values, since even values will end up with fractional coins after the first step.

 

The only one that yields an integer value after each step is 23. All the others end up with fractions of coins.

 

 

 

[Blopa]

Yeah, I knew the answer

241

 

but... I was wondering of the formal way to do it, testing the numbers isnt really a good way to solve it, even though, I admire you for testing it with a program (I use Actionscript (kids programming) to program), and thanks for helping ^^

Edited December 7, 2010 by Blopa