Solve the simultaneous linear congruences

[Math100]

Homework Statement

Solve the simultaneous linear congruences $3x\equiv 2\pmod {5}, 3x\equiv 4\pmod {7}, 3x\equiv 6\pmod {11}$.

Relevant Equations

Let $p, q$ be coprime. Then the system of equations $x\equiv a\pmod {p}, x\equiv b\pmod {q}$ has a unique solution for $x$ modulo $pq$.

Consider the following set of simultaneous linear congruences:
$3x\equiv 2\pmod {5}, 3x\equiv 4\pmod {7}, 3x\equiv 6\pmod {11}$.
Observe that
\begin{align*}
&3x\equiv 2\pmod {5}\implies 6x\equiv 4\pmod {5}\implies x\equiv 4\pmod {5}\\
&3x\equiv 4\pmod {7}\implies 15x\equiv 20\pmod {7}\implies x\equiv 6\pmod {7}\\
&3x\equiv 6\pmod {11}\implies 12x\equiv 24\pmod {11}\implies x\equiv 2\pmod {11}.\\
\end{align*}
Applying the Chinese Remainder Theorem produces:
$n=5\cdot 7\cdot 11=385$.
Now we define $N_{k}=\frac{n}{n_{k}}$ for $k=1,2,...,r$.
Note that $N_{1}=\frac{385}{5}=77, N_{2}=\frac{385}{7}=55$ and $N_{3}=\frac{385}{11}=35$.
Then $77x_{1}\equiv 1\pmod {5}, 55x_{2}\equiv 1\pmod {7}$ and $35x_{3}\equiv 1\pmod {11}$.
This implies $x_{1}=3, x_{2}=6$ and $x_{3}=6$.
Thus $x\equiv (4\cdot 77\cdot 3+6\cdot 55\cdot 6+2\cdot 35\cdot 6)\pmod {385}\equiv 3324\pmod {385}\equiv 244\pmod {385}$.
Therefore, $x\equiv 244\pmod {385}$.

 

[fresh_42]

Homework Statement:: Solve the simultaneous linear congruences $3x\equiv 2\pmod {5}, 3x\equiv 4\pmod {7}, 3x\equiv 6\pmod {11}$.
Relevant Equations:: Let $p, q$ be coprime. Then the system of equations $x\equiv a\pmod {p}, x\equiv b\pmod {q}$ has a unique solution for $x$ modulo $pq$.

Consider the following set of simultaneous linear congruences:
$3x\equiv 2\pmod {5}, 3x\equiv 4\pmod {7}, 3x\equiv 6\pmod {11}$.
Observe that
\begin{align*}
&3x\equiv 2\pmod {5}\implies 6x\equiv 4\pmod {5}\implies x\equiv 4\pmod {5}\\
&3x\equiv 4\pmod {7}\implies 15x\equiv 20\pmod {7}\implies x\equiv 6\pmod {7}\\
&3x\equiv 6\pmod {11}\implies 12x\equiv 24\pmod {11}\implies x\equiv 2\pmod {11}.\\
\end{align*}
Applying the Chinese Remainder Theorem produces:
$n=5\cdot 7\cdot 11=385$.
Now we define $N_{k}=\frac{n}{n_{k}}$ for $k=1,2,...,r$.
Note that $N_{1}=\frac{385}{5}=77, N_{2}=\frac{385}{7}=55$ and $N_{3}=\frac{385}{11}=35$.
Then $77x_{1}\equiv 1\pmod {5}, 55x_{2}\equiv 1\pmod {7}$ and $35x_{3}\equiv 1\pmod {11}$.
This implies $x_{1}=3, x_{2}=6$ and $x_{3}=6$.
Thus $x\equiv (4\cdot 77\cdot 3+6\cdot 55\cdot 6+2\cdot 35\cdot 6)\pmod {385}\equiv 3324\pmod {385}\equiv 244\pmod {385}$.
Therefore, $x\equiv 244\pmod {385}$.

Correct, if I haven't overlooked a typo. The result is definitely correct.

 

[Math100]

Correct, if I haven't overlooked a typo. The result is definitely correct.

Is there a typo?

 

[fresh_42]

Is there a typo?

No.

 

[WWGD]

You can verify the solution yourself:
244(3)=732=2Mod5=4mod7=6Mod11