G4.1 Let $x$ be the number of candies satisfies the inequalities $120 \leq x \leq 150$. 2 candies will be remained if they are divided into groups of 5 candies each; 5 candies will be remained if they are divided into groups of 6 candies each. Find the value of $x$.
$x=5 m+2=6 n+5$, where $m$ and $n$ are integers.
$$
\begin{array}{l}
5 m-6 n=3 \\
5 \times 3-6 \times 2=15-12=3
\end{array}
$$
$\therefore m=3, n=2$ is a pair of solution
The general solution is $m=3+6 t, n=2+5 t$, where $t$ is any integer.
$$
\begin{array}{l}
x=5 m+2=5(3+6 t)+2=30 t+17 \\
120 \leq x \leq 150 \Rightarrow 120 \leq 30 t+17 \leq 150 \\
103 \leq 30 t \leq 133 \\
3.43<t<4.43 \Rightarrow t=4 \\
x=30 \times 4+17=137
\end{array}
$$