Solve $x\equiv 1\bmod2, x \equiv 1\bmod5$ and $x \equiv 0\bmod3$ $$x\equiv 1\mod2\\ x \equiv 1\mod5\\x \equiv 0\mod3$$
Somehow I got the wrong solution
Here's how I got them
$b_i$ |  $N_i$ | inverse| Product
2     |  20    | 4 |160
2 | 12 | 3 | 72
0 | 15 | 3 | 0
Sum of products is 232 then you modulus by 2*3*5=30 which is $x\equiv 232\bmod30\equiv 52\bmod30$, but this turned out to be wrong.
 A: Why modulo $60$? The least common multiple is $2\cdot3\cdot5=30$.
Indeed, $x=5a+1$; from $5a+1\equiv1\pmod{2}$ we deduce $a=2b$; then
$$
10b+1\equiv0\pmod{3}
$$
is the same as $b\equiv2\pmod{3}$, so $b=3c+2$. Recapitulating,
$$
x=5a+1=10b+1=30c+21
$$
A: In this case you don't even need CRT. The relations $x\equiv1\pmod2$ and $x\equiv0\pmod3$ suggest that $x$ must be an odd multiple of $3$.
Next, $x\equiv1\pmod5\implies x=1,6,11,16,21,26,31,\cdots$ from which $21$ is the only odd multiple of $3$. 
Since $\operatorname{lcm}(2,3,5)=30$, the solution is therefore $x\equiv21\pmod{30}$.
A: The scheme is as follows. We start with $N=2\cdot 3\cdot 5=30$, which is related to the following data:


*

*$a_1=1$, $n_1=2$, and build $b_1=N/n_1=15$, and the inverse of $b_1$ modulo $n_1$ is $c_1=1$, 

*$a_2=1$, $n_2=5$, and build $b_2=N/n_2=6$, and the inverse of $b_2$ modulo $n_2$ is $c_2=1$, 

*$a_3=0$, $n_3=3$, and build $b_3=N/n_3=10$, and the inverse of $b_3$ modulo $n_3$ doesn't matter (but it is $c_3=1$).


Now consider 
$$a_1b_1c_1+a_2b_2c_2+a_3b_3c_3=15+6+0=21\ .$$
It works.
