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.

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 $x\equiv21\pmod{30}$.
