What was the least number of coins that could have been stolen?

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Homework Help Overview

The problem involves determining the least number of coins stolen by a group of pirates, based on a series of modular conditions derived from their attempts to divide the coins. The subject area relates to modular arithmetic and the Chinese Remainder Theorem.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the Chinese Remainder Theorem to find a solution, with specific modular equations established for the number of coins. There is also a focus on verifying steps in the reasoning process, particularly regarding the calculations of modular inverses.

Discussion Status

The discussion includes attempts to clarify and correct specific steps in the reasoning. Some participants have pointed out potential errors in the calculations, leading to further examination of the modular relationships involved. There is an ongoing exploration of the implications of these corrections.

Contextual Notes

Participants are working within the constraints of the problem as presented, which includes specific modular conditions that must be satisfied. The discussion reflects a collaborative effort to ensure accuracy in the mathematical reasoning without reaching a definitive conclusion.

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Homework Statement
(Ancient Chinese Problem.) A band of ## 17 ## pirates stole a sack of gold coins. When they tried to divide the fortune into equal portions, ## 3 ## coins remained. In the ensuing brawl over who should get the extra coins, one pirate was killed. The wealth was redistributed, but this time an equal division left ## 10 ## coins. Again an argument developed in which another pirate was killed. But now the total fortune was evenly distributed among the survivors. What was the least number of coins that could have been stolen?
Relevant Equations
None.
Let ## x ## be the least number of coins that could have been stolen.
Then ## x\equiv 3\pmod {17}, x\equiv 10\pmod {16} ##, and ## x\equiv 0\pmod {15} ##.
Applying the Chinese Remainder Theorem produces:
## n=17\cdot 16\cdot 15=4080 ##.
This means ## N_{1}=\frac{4080}{17}=240, N_{2}=\frac{4080}{16}=255 ## and ## N_{3}=\frac{4080}{15}=272 ##.
Observe that
\begin{align*}
&240x_{1}\equiv 1\pmod {17}\implies 2x_{1}\equiv 1\pmod {17}\\
&\implies 18x_{1}\equiv 9\pmod {17}\implies x_{1}\equiv 9\pmod {17},\\
&255x_{2}\equiv 1\pmod {16}\implies -x_{2}\equiv 1\pmod {16}\\
&\implies x_{2}\equiv -15\pmod {16}\implies x_{2}\equiv 15\pmod {16},\\
&272x_{3}\equiv 1\pmod {15}\implies 2x_{3}\equiv 1\pmod {15}\\
&\implies 16x_{3}\equiv 8\pmod {15}\implies x_{3}\equiv 8\pmod {15}.\\
\end{align*}
Now we have ## x_{1}=9, x_{2}=15 ## and ## x_{3}=8 ##.
Thus ## x\equiv (9\cdot 3\cdot 240+15\cdot 10\cdot 255+0)\pmod {4080}\equiv 44730\pmod {4080}\equiv 3930\pmod {4080} ##.
Therefore, the least number of coins that could have been stolen is ## 3930 ##.
 
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Math100 said:
Homework Statement:: (Ancient Chinese Problem.) A band of ## 17 ## pirates stole a sack of gold coins. When they tried to divide the fortune into equal portions, ## 3 ## coins remained. In the ensuing brawl over who should get the extra coins, one pirate was killed. The wealth was redistributed, but this time an equal division left ## 10 ## coins. Again an argument developed in which another pirate was killed. But now the total fortune was evenly distributed among the survivors. What was the least number of coins that could have been stolen?
Relevant Equations:: None.

Let ## x ## be the least number of coins that could have been stolen.
Then ## x\equiv 3\pmod {17}, x\equiv 10\pmod {16} ##, and ## x\equiv 0\pmod {15} ##.
Applying the Chinese Remainder Theorem produces:
## n=17\cdot 16\cdot 15=4080 ##.
This means ## N_{1}=\frac{4080}{17}=240, N_{2}=\frac{4080}{16}=255 ## and ## N_{3}=\frac{4080}{15}=272 ##.
Observe that
\begin{align*}
&240x_{1}\equiv 1\pmod {17}\implies 2x_{1}\equiv 1\pmod {17}\\
&\implies 18x_{1}\equiv 9\pmod {17}\implies x_{1}\equiv 9\pmod {17},\\
&255x_{2}\equiv 1\pmod {16}\implies -x_{2}\equiv 1\pmod {16}\\
&\implies x_{2}\equiv -15\pmod {16}\implies x_{2}\equiv 15\pmod {16},\\
&272x_{3}\equiv 1\pmod {15}\implies 2x_{3}\equiv 1\pmod {15}\\
&\implies 16x_{3}\equiv 8\pmod {15}\implies x_{3}\equiv 8\pmod {15}.\\
\end{align*}
Now we have ## x_{1}=9, x_{2}=15 ## and ## x_{3}=8 ##.
Thus ## x\equiv (9\cdot 3\cdot 240+15\cdot 10\cdot 255+0)\pmod {4080}\equiv 44730\pmod {4080}\equiv 3930\pmod {4080} ##.
Therefore, the least number of coins that could have been stolen is ## 3930 ##.
Correct, except for one step. You have ##255x_{2}\equiv 1\pmod {16}\implies -x_{2}\equiv 1\pmod {16}##. Then multiplication with ##-1## yields ##x_2\equiv -1 \equiv 15 \pmod{16}.## This would be correct, but the step in between ##\implies x_{2}\equiv -15\pmod {16}## is not. If we had a single number with remainder ##-1## and ##-15## modulo ##16##, then ##-1-(-15)=14 \equiv 0 \pmod{16}## which is a contradiction.
 
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fresh_42 said:
Correct, except for one step. You have ##255x_{2}\equiv 1\pmod {16}\implies -x_{2}\equiv 1\pmod {16}##. Then multiplication with ##-1## yields ##x_2\equiv -1 \equiv 15 \pmod{16}.## This would be correct, but the step in between ##\implies x_{2}\equiv -15\pmod {16}## is not. If we had a single number with remainder ##-1## and ##-15## modulo ##16##, then ##-1-(-15)=14 \equiv 0 \pmod{16}## which is a contradiction.
Yes, thank you for pointing that out. I was wrong about that.
 
Math100 said:
Yes, thank you for pointing that out. I was wrong about that.
I only wanted to ensure you that I actually checked your proof. :biggrin:
 
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