problem based on LCM and HCF Find the least number which when divided by 7,9,11 produce 1,2,3 as reminders .
I tried a method which works only when the difference between the elements is same. But in this case 7-1=6 , 9-2=7, 11-3=8 . So that method is not working.
Please, provide the way by which I can solve the problem. Answer of the problem is 344.
 A: Here $\,m_i-x_i\,$ is not constant but, rather $\,m_i - \color{#c00}2x_i = 5,\, $ so $\,\color{#c00}2x_i\equiv -5\pmod{m_i},\,$ so
$\quad 2x \equiv -5\,$ mod $\,7,9,11\iff x \equiv \dfrac{-5}{2}\equiv \dfrac{7\cdot 9\cdot 11-5 }{2}\equiv 344 \pmod{7\cdot 9\cdot 11}$
A: This may be a bit longer, but it is a bit more mechanical of a solution.
To solve
$$
\begin{align}
x&\equiv1\pmod{7}\\
x&\equiv2\pmod{9}\\
x&\equiv3\pmod{11}
\end{align}
$$
we can solve three simpler problems. To solve the simpler equations, we can use the Extended Euclidean Algorithm as implemented in this answer.

$$
\begin{align}
x_7&\equiv1\pmod{7}\\
x_7&\equiv0\pmod{9}\\
x_7&\equiv0\pmod{11}
\end{align}
$$
$x\equiv0\pmod{9}$ and $x\equiv0\pmod{11}\iff x\equiv0\pmod{99}$.
$$
\begin{array}{r}
&&14&7\\\hline
1&0&1&-7\\
0&1&-14&99\\
99&7&1&0
\end{array}
$$
Therefore, $x_7=99$.

$$
\begin{align}
x_9&\equiv0\pmod{7}\\
x_9&\equiv1\pmod{9}\\
x_9&\equiv0\pmod{11}
\end{align}
$$
$x\equiv0\pmod{7}$ and $x\equiv0\pmod{11}\iff x\equiv0\pmod{77}$.
$$
\begin{array}{r}
&&8&1&1&4\\\hline
1&0&1&-1&2&-9\\
0&1&-8&9&-17&77\\
77&9&5&4&1&0
\end{array}
$$
Therefore, $x_9=154$.

$$
\begin{align}
x_{11}&\equiv0\pmod{7}\\
x_{11}&\equiv0\pmod{9}\\
x_{11}&\equiv1\pmod{11}
\end{align}
$$
$x\equiv0\pmod{7}$ and $x\equiv0\pmod{9}\iff x\equiv0\pmod{63}$.
$$
\begin{array}{r}
&&5&1&2&1&2\\\hline
1&0&1&-1&3&-4&11\\
0&1&-5&6&-17&23&-63\\
63&11&8&3&2&1&0
\end{array}
$$
Therefore, $x_{11}=-252$.

We can put the three simpler answers together by
$$
x\equiv x_7+2x_9+3x_{11}=-349\equiv344\pmod{693}
$$
That is
$$
\begin{bmatrix}
1\\0\\0
\end{bmatrix}
+
2\begin{bmatrix}
0\\1\\0
\end{bmatrix}
+
3\begin{bmatrix}
0\\0\\1
\end{bmatrix}
=
\begin{bmatrix}
1\\2\\3
\end{bmatrix}
\begin{array}{l}
\pmod{7}\\
\pmod{9}\\
\pmod{11}
\end{array}
$$
