Chinese Remainder Theorem with 0 mod n

$$ x+1 \equiv 1 (_{mod} 7) \\ x+1 \equiv 0 (_{mod} 6) \\ x+1 \equiv 0 (_{mod} 5) \\ x+1 \equiv 0 (_{mod} 4) \\ x+1 \equiv 0 (_{mod} 3) \\ x+1 \equiv 0 (_{mod} 2) $$ Hense $$ x+1 \equiv 1 (_{mod} 7) \\ x+1 \equiv 0 (_{mod} 60) $$ Then from 60, 120, 180, 240, 300 and 360 you can find $120 \equiv 1 (_{mod} 7)$ so $x=119+420k, k\in Z$ is a solution of system.

The first condition tells you that $x=7y$, so the system becomes \begin{align} 7y&\equiv 5\pmod{6} && (\textit{redundant})\\ 7y&\equiv 4\pmod{5}\\ 7y&\equiv 3\pmod{4}\\ 7y&\equiv 2\pmod{3}\\ 7y&\equiv 1\pmod{2} && (\textit{redundant}) \end{align}

There is no modular inverse of $0$. On the other hand, $7$ has a modular inverse modulo $k$, for $2\le k<7$.

I left the redundant equations just for completeness.

The three relations become then \begin{align} y&\equiv2\pmod{5}\\ y&\equiv1\pmod{3}\\ y&\equiv1\pmod{2} \end{align} that yields $y\equiv17\pmod{30}$.