Solving $N\equiv1\pmod2,N\equiv2\pmod3,\dots,N\equiv n-1\pmod n$ If I have:
\begin{align*}
N &\equiv 1 &\pmod{2}\\
N &\equiv 2 &\pmod{3}\\
N &\equiv 3 &\pmod{4}\\
&\vdots\\ 
N &\equiv n - 1 &\pmod{n}
\end{align*}
How could I solve for $N$? Is there any property relates to this problem?  
Update
Base on Moron hint, we have:
\begin{align*}
N + 1 &\equiv 0 &\pmod{2}\\
N + 1 &\equiv 0 &\pmod{3}\\
N + 1 &\equiv 0 &\pmod{4}\\
\vdots\\
N + 1 &\equiv 0 &\pmod{n}
\end{align*}
Hence, 
$$N + 1 \equiv 0 \pmod{\mathrm{lcm}(2\cdot 3\cdots n}$$
$$\therefore N + 1 = lcm(2.3.4...n) * k \text{ for some integers k } $$
$$\implies N = lcm(2.3.4...n) * k - 1$$
Does it look right?  
Thanks,
Chan

Hint: Consider the possible values for $N+1$.
