Simple exersice using CRT Calculate all solutions $0\leq x< 130$ for the following system of equations.
$x \equiv_2 1$
$x \equiv_5 2$
$x \equiv_{13} 3$
Solution:
$M=2\cdot 5\cdot 13=130$
So by CRT there exists a unique solution.
We get: 
$M_1=M/2=65$
$M_2 = M/5 = 26$
$M_3=M/13=10$
Now:
$65\cdot N_1 \equiv_2 1 \ \ \Rightarrow \ \ N_1 = 1$
$26\cdot N_2 \equiv_5 2 \ \ \Rightarrow \ \ N_2=2$
$10\cdot N_3 \equiv_{13} 3 \ \ \Rightarrow \ \ N_3=12$
$x = R_{130}(1\cdot 65\cdot 1 + 2\cdot 26 \cdot 2 + 3 \cdot 10 \cdot 12)=79$
This is not correct, as one can easily see with the second equation.
I can't find my mistake :/ what's wrong?

If you set
$$26\cdot N_2 \equiv_5 1 \ \ \Rightarrow \ \ N_2=1 \\
10\cdot N_3 \equiv_{13} 1 \ \ \Rightarrow \ \ N_3=4
$$
Then you'll get the right number
$$x = R_{130}(1\cdot M_1 \cdot N_1 + 2\cdot M_2 \cdot N_2 + 3 \cdot M_3 \cdot N_3)\\=R_{130}(1 \cdot 65\cdot 1 + 2\cdot 26 \cdot 2 + 3 \cdot 10 \cdot 12)\\
=R_{130}(65+52+120)=R_{130}(107)
$$
This satisfies the question.
$$107 \equiv_2 1, 107 \equiv_5 2, 107 \equiv_{13} 3$$
