Pirates Dividing Loot Question- NEW

  • Thread starter BR24
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In summary, the conversation discusses a mathematical problem given by a teacher involving four pirates who find a treasure and must divide it equally. They wake up in the morning to find that the pile of coins has been diminished and must figure out the smallest number of coins in the original pile. The conversation also includes a potential solution using a Diophantine equation and an algorithm for solving similar problems.
  • #1
BR24
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This is a long question so bear with me.

Four pirates find a treasure consisting of gold coins on a tiny island. They gather all the coins in a pile under a palm tree. Exhausted, they agree to wait until morning to split the coins.

At 1 in the morning, the 1st pirate wakes. He realizes the others can't be trusted, and decides to take his share now. He divides the coins into four equal piles, but there is one left over. He throws the extra one in the ocean, hides his coins, and put the rest back undr the palm tree.

At 2, Pirate 2 wakes up. Not realizing pirate 1 has already taken his share, he also divides up the remaining coins into 4 piles, with 1 left over. He throws the extra one in the ocean, hides his share and pus the rest of the coins back under the tree.

At 3 and 4 in the morning, the third and fourth pirates each wake up and carrry out the same actions.

In the morning, the pirates wake up, trying to look innocent. Noone says anything about the diminished coin pile. They divide the remaining pile into four piles for the fifth time, but this time there is no coin left over to throw in the ocean.

Find the smallest number of coins in the original pile.

I can't get this, my math teacher gave it to us. If you find the answer let me know, but if you have a method to this besides guess and check i'd like to know how.
 
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  • #2
This was more tedious than it sounded.

I get an equation that says

[tex] T = \frac{4}{3} \left( \frac{4}{3} \left[ \frac{4}{3} \left( 16P+1 \right) +1 \right] +1 \right) +1 [/tex]

T = total coins. P = how much each pirate gets at the division in the morning.

Expanding, and barring errors.

[tex]27T = 1024P + 148 +27[/tex]

At a guess, the answer might be the smallest positive P such that T is an integer.

Edit: This seems to work, oddly enough.
 
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  • #3
Thanks a lot man, just to make sure i worked it out right, did you get 765
 
  • #4
BR24 said:
Thanks a lot man, just to make sure i worked it out right, did you get 765

yes. In the last step I needed to find integer solutions for 27A = 1024B - 14. Is there an easier way to do this than making 27(?) tests?
 
  • #5
i don't know if there is an easier way, i'd like to know too. i used the equation you gave me
27t=1024p+175, started from zero and increased p which finally gave me a whole number when i got to 20.
 
  • #6
Phrak said:
yes. In the last step I needed to find integer solutions for 27A = 1024B - 14. Is there an easier way to do this than making 27(?) tests?
Yes, this is a Diophantine equation. See http://en.wikipedia.org/wiki/Diophantine_equation for an explanation.
You have
A = 1024B/27 - 14/27 = 37B + (25B - 14)/27
Since A and B are integers, (25B - 14)/27 must be an integer too.
(25B - 14)/27 = C
25B - 14 = 27C
25B = 27C + 14
B = 27C/25 - 14/25 = C + (2C + 14)/25
Now, you make
2C + 14 = 25D
2C = 25D - 14
C = 12D + D/2 - 7
D/2 must be an integer
D/2 = 1 is the minimum integer
Now you have D = 2
Working backwards I got A = 758 and not 765. This is not the right answer, because 758/4 gives a remainder of 2, not 1.

Working with the equation
27T = 1024P + 175
I got the good answer: T = 765.
 
  • #7
  • #8
CEL said:
Working backwards I got A = 758 and not 765. This is not the right answer, because 758/4 gives a remainder of 2, not 1.

No, that wasn't it. I defined A = t - (175 mod 27) + 1 to simplify calculator punching, or so I'd thought, thus the difference of 7.

Cool algorithm, by the way;

For

[tex] \Phi_1 A = \Phi_2 B + \Gamma [/tex]

each step in the algorithm takes

[tex] \Phi_{large}, \Phi_{small} [/tex]

to

[tex] Rem(\Phi_{large} / \Phi_{small} ), \Phi_{small} [/tex]

in alternating fashion.
 
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1.

What is the "Pirates Dividing Loot Question- NEW"?

The "Pirates Dividing Loot Question- NEW" is a hypothetical scenario used in game theory to understand how rational decision-making can be affected by varying incentives and outcomes.

2.

What is the basic premise of the "Pirates Dividing Loot Question- NEW"?

In the scenario, a group of pirates must decide how to divide a large amount of loot. Each pirate has a rank in the group, with the highest rank having the power to propose a division of the loot. The pirates must then vote on the proposal, and if more than half approve, the loot is divided as proposed. However, if less than half approve, the proposing pirate is thrown overboard and the process repeats until a proposal is approved.

3.

What makes the "Pirates Dividing Loot Question- NEW" a challenging problem?

The challenging aspect of this scenario is that each pirate must balance their own self-interest with the overall group dynamic. They must also consider the potential consequences of their decisions, as they may be thrown overboard if their proposal is rejected.

4.

What are some common strategies used by pirates in this scenario?

Some common strategies include the "greedy" approach, where the highest rank proposes a division that heavily favors themselves, or the "equal share" approach, where the highest rank proposes an equal split among all pirates. Other strategies involve forming alliances and bargaining with other pirates to secure a favorable outcome.

5.

What can the "Pirates Dividing Loot Question- NEW" teach us about decision-making?

This scenario highlights the complexity of decision-making when multiple parties are involved, and the importance of considering both self-interest and group dynamics. It also demonstrates how incentives can influence decision-making and the potential pitfalls of short-term thinking.

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