- #1

Math100

- 783

- 220

- 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 ##.

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 ##.