Solve congruence system $\ x\equiv m_i-1 \pmod{m_i}\,$ for $\,i = 1,\ldots, k$ Find natural number $x$ so that
$$\begin{align}x&\equiv 9\pmod{10}\\ x&\equiv8\pmod9\\ &\ \ \vdots\\ x&\equiv 1\pmod2\end{align}$$
 A: Hint: The unnatural number $-1$ works. 
A: Since  $\,m_i-1\equiv \color{#c00}{-1}\pmod{\!m_i}\,$ we can apply $ $ CCRT = $\rm\color{#c00}{constant}$ case optimization of CRT 
$$\begin{align} x\equiv \color{#c00}{-1}\!\!\pmod{\!m_i}&\iff x\equiv -1\!\!\pmod{{\rm lcm}\{m_i\}}\\[.4em]
\text{or, without using $\rm{\small CRT\!:}$}\ \ \ {\rm  all}\ \  m_i \mid x+1 &\iff {\rm lcm}\{m_i\}\mid x+1
\end{align}\qquad\qquad$$
The latter equivalence is by the Universal Property of LCM (= definition of LCM  in general)
Remark $ $ more generally this idea works for linearly related values & moduli:  $ $  if $\,(a,b) = 1\,$ then 
$$\left\{\,x\equiv d\!-\!ck\!\!\!\pmod{b\!-\!ak}\,\right\}_{k=0}^{n}\!\!\iff\! x\equiv \dfrac{ad\!-\!bc}a\!\!\!\pmod{{\rm lcm}\{b\!-\!ak\}_{k=0}^n}\quad  \   $$
$ $ e.g.  here $\,\ \underbrace{\left\{\,x \equiv 3-k\pmod{7-k}\,\right\}_{k=0}^2}_{\textstyle{x\equiv 3,2,1\pmod{\!7,6,5}}}\!\!\iff\!  x\equiv \dfrac{1(3)-7(1)}1\equiv -4\pmod{210}$
A: \begin{align}
   x &\equiv 9 \pmod{10} \\
   x &\equiv 8 \pmod 9 \\
   x &\equiv 7 \pmod 8 \\
   x &\equiv 6 \pmod 7 \\
   x &\equiv 5 \pmod 6 \\
   x &\equiv 4 \pmod 5 \\
   x &\equiv 3 \pmod 4 \\
   x &\equiv 2 \pmod 3 \\
   x &\equiv 1 \pmod 2 \\
\end{align}
Is equivalent to
\begin{align}
   x &\equiv -1 \pmod{10} \\
   x &\equiv -1 \pmod 9 \\
   x &\equiv -1 \pmod 8 \\
   x &\equiv -1 \pmod 7 \\
   x &\equiv -1 \pmod 6 \\
   x &\equiv -1 \pmod 5 \\
   x &\equiv -1 \pmod 4 \\
   x &\equiv -1 \pmod 3 \\
   x &\equiv -1 \pmod 2 \\
\end{align}
which is equivalent to
$$x \equiv -1 \mod{\operatorname{lcm}\{2,3,4,5,6,7,8,9,10\}}$$
$$x \equiv -1 \pmod{2520}$$
$$x \equiv 2519 \pmod{2520}$$
