Solve ##x\equiv 1\pmod {3},x\equiv 2\pmod {5},x\equiv 3\pmod 7 ##

[Math100]

Homework Statement

Solve the following set of simultaneous congruences:
$x\equiv 1\pmod {3}, x\equiv 2\pmod {5}, x\equiv 3\pmod {7}$.

Relevant Equations

None.

Consider the following set of simultaneous congruences:
$x\equiv 1\pmod {3}, x\equiv 2\pmod {5}, x\equiv 3\pmod {7}$.
Applying the Chinese Remainder Theorem produces:
$n=3\cdot 5\cdot 7=105$.
Now we define $N_{k}=\frac{n}{n_{k}}$ for $k=1, 2,..., r$.
Observe that $N_{1}=\frac{105}{3}=35, N_{2}=\frac{105}{5}=21$ and $N_{3}=\frac{105}{7}=15$.
Then
\begin{align*}
&35x_{1}\equiv 1\pmod {3}\\
&21x_{2}\equiv 1\pmod {5}\\
&15x_{3}\equiv 1\pmod {7}.\\
\end{align*}
This means $x_{1}=2, x_{2}=1$ and $x_{3}=1$.
Thus $x\equiv (1\cdot 35\cdot 2+2\cdot 21\cdot 1+3\cdot 15\cdot 1)\pmod {105}\equiv 157\pmod {105}\equiv 52\pmod {105}$.
Therefore, $x\equiv 52\pmod {105}$.

 

[fresh_42]

Homework Statement:: Solve the following set of simultaneous congruences:
$x\equiv 1\pmod {3}, x\equiv 2\pmod {5}, x\equiv 3\pmod {7}$.
Relevant Equations:: None.

Consider the following set of simultaneous congruences:
$x\equiv 1\pmod {3}, x\equiv 2\pmod {5}, x\equiv 3\pmod {7}$.
Applying the Chinese Remainder Theorem produces:
$n=3\cdot 5\cdot 7=105$.
Now we define $N_{k}=\frac{n}{n_{k}}$ for $k=1, 2,..., r$.
Observe that $N_{1}=\frac{105}{3}=35, N_{2}=\frac{105}{5}=21$ and $N_{3}=\frac{105}{7}=15$.
Then
\begin{align*}
&35x_{1}\equiv 1\pmod {3}\\
&21x_{2}\equiv 1\pmod {5}\\
&15x_{3}\equiv 1\pmod {7}.\\
\end{align*}
This means $x_{1}=2, x_{2}=1$ and $x_{3}=1$.
Thus $x\equiv (1\cdot 35\cdot 2+2\cdot 21\cdot 1+3\cdot 15\cdot 1)\pmod {105}\equiv 157\pmod {105}\equiv 52\pmod {105}$.
Therefore, $x\equiv 52\pmod {105}$.

Perfect!