Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$ 
Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$.

I tried
$$x\equiv5\pmod6\\x\equiv4\pmod5\\x\equiv3\pmod4\\x\equiv2\pmod3$$What $x$ value?

HINT: Notice that your congruences are equivalent to the following ones:
$$\left\{\begin{align*}
x\equiv-1\pmod6\\
x\equiv-1\pmod5\\
x\equiv-1\pmod4\\
x\equiv-1\pmod3
\end{align*}\right.$$
In other words, $x+1$ is divisible by $6,5,4$, and $3$. What’s the smallest positive integer with that property?
