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

• Math100
In summary, the least number of coins that could have been stolen in the Ancient Chinese Problem is ##3930##. This is determined by applying the Chinese Remainder Theorem to the given congruences and solving for the smallest possible value of ##x##.
Math100
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 ##.

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.

Math100
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.

Math100

## 1. What is the minimum number of coins that could have been stolen?

The minimum number of coins that could have been stolen is one. This is because even if only one coin is stolen, it still meets the criteria of being the least number of coins stolen.

## 2. Is it possible for no coins to have been stolen?

Yes, it is possible for no coins to have been stolen. This would mean that the person did not steal any coins at all.

## 3. How do you determine the least number of coins stolen?

The least number of coins stolen can be determined by looking at the total number of coins that were originally present and the number of coins that are missing. The difference between these two numbers would be the least number of coins stolen.

## 4. Can the least number of coins stolen be negative?

No, the least number of coins stolen cannot be negative. This is because negative numbers represent a quantity less than zero, and it is not possible to have a negative number of coins.

## 5. Does the value of the coins matter in determining the least number of coins stolen?

No, the value of the coins does not matter in determining the least number of coins stolen. The only factor that matters is the number of coins that are missing, regardless of their individual values.

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