Finding Integer with Chinese Remainder Theorem

[Math100]

Homework Statement

A certain integer between $1$ and $1200$ leaves the remainders $1, 2, 6$ when divided by $9, 11, 13$, respectively. What is the integer?

Relevant Equations

None.

Consider a certain integer between $1$ and $1200$.
Then $x\equiv 1\pmod {9}, x\equiv 10\pmod {11}$ and $x\equiv 0\pmod {13}$.
Applying the Chinese Remainder Theorem produces:
$n=9\cdot 11\cdot 13=1287$.
This means $N_{1}=\frac{1287}{9}=143, N_{2}=\frac{1287}{11}=117$ and $N_{3}=\frac{1287}{13}=99$.
Observe that
\begin{align*}
&143x_{1}\equiv 1\pmod {9}\implies 8x_{1}\equiv 1\pmod {9}\implies x_{1}\equiv 8\pmod {9},\\
&117x_{2}\equiv 1\pmod {11}\implies -4x_{2}\equiv 1\pmod {11}\\
&\implies -12x_{2}\equiv 3\pmod {11}\implies -x_{2}\equiv 3\pmod {11}\implies x_{2}\equiv 8\pmod {11},\\
&99x_{3}\equiv 1\pmod {13}\implies -5x_{3}\equiv 1\pmod {13}\\
&-15x_{3}\equiv 3\pmod {13}\implies -2x_{3}\equiv 3\pmod {13}\\
&-12x_{3}\equiv 18\pmod {13}\implies x_{3}\equiv 5\pmod {13}.\\
\end{align*}
Now we have $x_{1}=8, x_{2}=8$ and $x_{3}=5$.
Thus $x\equiv (1\cdot 143\cdot 8+2\cdot 117\cdot 8+6\cdot 99\cdot 5)\pmod {1287}\equiv 5986\pmod {1287}\equiv 838\pmod {1287}$.
Therefore, the integer is $838$.

 

[robphy]

Your integer as 838 checks out.
https://www.desmos.com/calculator/8kbm63myjr

 

[fresh_42]

Everything is fine until ...

Homework Statement:: A certain integer between $1$ and $1200$ leaves the remainders $1, 2, 6$ when divided by $9, 11, 13$, respectively. What is the integer?
Relevant Equations:: None.

Consider a certain integer between $1$ and $1200$.
Then $x\equiv 1\pmod {9}, x\equiv 10\pmod {11}$ and $x\equiv 0\pmod {13}$.

... which I did not understand. Shouldn't it be $(1,2,6)$ instead of $(1,10,0)?$ Especially, because ...

Applying the Chinese Remainder Theorem produces:
$n=9\cdot 11\cdot 13=1287$.
This means $N_{1}=\frac{1287}{9}=143, N_{2}=\frac{1287}{11}=117$ and $N_{3}=\frac{1287}{13}=99$.
Observe that
\begin{align*}
&143x_{1}\equiv 1\pmod {9}\implies 8x_{1}\equiv 1\pmod {9}\implies x_{1}\equiv 8\pmod {9},\\
&117x_{2}\equiv 1\pmod {11}\implies -4x_{2}\equiv 1\pmod {11}\\
&\implies -12x_{2}\equiv 3\pmod {11}\implies -x_{2}\equiv 3\pmod {11}\implies x_{2}\equiv 8\pmod {11},\\
&99x_{3}\equiv 1\pmod {13}\implies -5x_{3}\equiv 1\pmod {13}\\
&-15x_{3}\equiv 3\pmod {13}\implies -2x_{3}\equiv 3\pmod {13}\\
&-12x_{3}\equiv 18\pmod {13}\implies x_{3}\equiv 5\pmod {13}.\\
\end{align*}
Now we have $x_{1}=8, x_{2}=8$ and $x_{3}=5$.
Thus $x\equiv (1\cdot 143\cdot 8+2\cdot 117\cdot 8+6\cdot 99\cdot 5)\pmod {1287}\equiv 5986\pmod {1287}\equiv 838\pmod {1287}$.
Therefore, the integer is $838$.

... everything after that is correct again.

 

[Math100]

Everything is fine until ...... which I did not understand. Shouldn't it be $(1,2,6)$ instead of $(1,10,0)?$ Especially, because ...

... everything after that is correct again.

Yes, I was wrong. It should definitely be (1, 2, 6).