- #1
Hernaner28
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- 0
Homework Statement
Solve the following system:
[tex] \displaystyle \left\{ \begin{array}{*{35}{l}}
x\equiv 1\left( \bmod 2 \right) \\
x\equiv 2\left( \bmod 3 \right) \\
x\equiv 3\left( \bmod 4 \right) \\
x\equiv 4\left( \bmod 5 \right) \\
x\equiv 5\left( \bmod 6 \right) \\
x\equiv 6\left( \bmod 7 \right) \\
x\equiv 7\left( \bmod 8 \right) \\
x\equiv 8\left( \bmod 9 \right) \\
\end{array} \right.[/tex]
Homework Equations
Chinese remainder theorem.
The Attempt at a Solution
Well, what I did was to substract the module in each equation, so now I have:
[tex] \displaystyle \left\{ \begin{array}{*{35}{l}}
x\equiv -1\left( \bmod 2 \right) \\
x\equiv -1\left( \bmod 3 \right) \\
x\equiv -1\left( \bmod 4 \right) \\
x\equiv -1\left( \bmod 5 \right) \\
x\equiv -1\left( \bmod 6 \right) \\
x\equiv -1\left( \bmod 7 \right) \\
x\equiv -1\left( \bmod 8 \right) \\
x\equiv -1\left( \bmod 9 \right) \\
\end{array} \right.[/tex]
Now, if I try to solve one equation and then replace in the next it would be a tedious work to do. I've seen chinese remainder theorem but it doesn't help me, or at least I don't realize. What can I do next to simplify this system?
Thanks!