Solve congruence system $\ x\equiv m_i-1 \pmod{m_i}\,$ for $\,i = 1,\ldots, k$

Find natural number $x$ so that $$\begin{align}x&\equiv 9\pmod{10}\\ x&\equiv8\pmod9\\ &\ \ \vdots\\ x&\equiv 1\pmod2\end{align}$$

Solution 1:

Hint: The unnatural number $-1$ works.

Solution 2:

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 $\,\ \underbrace{\left\{\,x \equiv 3-k\pmod{7-k}\,\right\}_{k=0}^2}_{\textstyle{x\equiv 3,2,1\pmod{\!7,6,5}}}\!\!\iff\! x\equiv \dfrac{1(3)-7(1)}1\equiv -4\pmod{210}$