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.

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,
$$
x=5a+1=10b+1=30c+21
$$
