Note $n = c_0\! + c_1 1000 + \cdots\! + c_k 1000^k\! = f(1000)$ is a *polynomial* in $1000$ with *integer* coef's $\,c_i\,$ thus $\,{\rm mod}\ 7\!:\ \color{#c00}{1000}\equiv 10^3\equiv 3^3\equiv \color{#c00}{-1}\,\Rightarrow\, n = f(\color{#c00}{1000})\equiv f(\color{#c00}{-1}) \equiv c_0 - c_1 + \cdots + (-1)^k c_k$ follows by applying the **Polynomial Congruence Rule** below.

Similarly $\bmod 7\!:\ \color{#c00}{100\equiv 2}\Rightarrow n = f(\color{#c00}{100})\equiv f(\color{#c00}2)\,$ where $f$ is its radix $100$ polynomial as above.

[**Note** $ $ If congruences are unfamiliar then instead see the rules in *divisibility* form]

Below are the basic congruence rules. $ $ Let $\rm\ A,B,a,b\,$ be any integers.

**Congruence Sum Rule** $\rm\qquad\quad A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#90f}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$

**Proof** $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#90f}{A+B - (a+b)} $

**Congruence Product Rule** $\rm\quad\ A\equiv a,\ \ and \ \ B\equiv b\ \Rightarrow\ \color{#0a0}{AB\equiv ab}\ \ \ (mod\ m)$

**Proof** $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{#0a0}{AB - ab} $

**Congruence Power Rule** $\rm\qquad \color{}{A\equiv a}\ \Rightarrow\ \color{#c00}{A^n\equiv a^n}\ \ (mod\ m)\ \ $ for all naturals $\rm\,n.$

**Proof** $\ $ For $\rm\,n=0\,$ it's $\,1\equiv 1\,$ so true. $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, \color{#c00}{A^{n+1}\equiv a^{n+1}},\,$ by the Product Rule, so it follows by induction on $\rm\,n.\ $ **Warning** this does not remain true more generally if we analogously also replace the power $\,\rm n\,$ by any $\rm\,N\equiv n\pmod{\! m},\,$ see "Beware" below.

**Congruence Inverse Rule** $\rm \quad\ \color{#c00}{A\equiv a}\ \Rightarrow\ A^{-1}\equiv a^{-1},\ $ if $\rm\,A^{-1}\,$ exists.

**Proof** $\rm\,\ A^{-1} \equiv A^{-1} \color{#c00}a a^{-1}\equiv A^{-1} \color{#c00}A a^{-1} \equiv a^{-1}$ by PR = Product Rule $ $ (note $\rm\, a^{-1}$ exists by $\,\rm \color{#c00}aA^{-1}\equiv \color{#c00}AA^{-1}\equiv 1\,$ by PR). $ $ *Alternatively:* $ $ if $\rm\, A^{-1}\equiv b\,$ then PR
$\rm\Rightarrow 1\equiv \color{#c00}Ab\equiv \color{#c00}ab\,$ so $\rm\, {a}^{-1} \equiv b\equiv A^{-1}\,$ by uniqueness of inverses.

**Congruence Quotient Rule** $ $ If $\rm\,(B,n)= 1\,$ then $\rm\!\bmod n\!:\, {\begin{align}\rm A\equiv a\\ \rm B\equiv b\end{align}\,\Rightarrow\, \dfrac{A}B\equiv \dfrac{a}b\:\! \overset{\rm def}\equiv\, ab^{-1}}$

**Proof** $\ $ See this answer, and see here for modular **fractions**.

**Polynomial Congruence Rule** $\ $ If $\rm\,f(x)\,$ is *polynomial* with *integer* coefficients then $\rm\ A\equiv a\ \Rightarrow\ f(A)\equiv f(a)\,\pmod m.\ $ Note: this is equivalent to the Factor Theorem.

**Proof** $\ $ By induction on $\rm\, n = $ degree $\rm f.\,$ Clear if $\rm\, n = 0.\,$ Else $\rm\,f(x) = f(0) + x\,g(x)\,$ for $\rm\,g(x)\,$ a polynomial with integer coefficients of degree $\rm < n.\,$ By induction $\rm g(A)\equiv g(a)$ so $\rm \color{#0a0}{A g(A)\equiv\! a g(a)}\,$ by the Product Rule. Hence $\rm \,f(A) = f(0)+\color{#0a0}{Ag(A)}\equiv f(0)+\color{#0a0}{ag(a)} = f(a)\,$ by the Sum Rule.

**Beware** $ $ that such rules need not hold true for other operations, e.g.
the exponential analog of above $\rm A^B\equiv\, a^b$ is not generally true (unless $\rm B = b,\,$ so it follows by the Power Rule, or the Polynomial Rule with $\rm\,f(x) = x^{\rm b}),$ e.g. $\rm\!\bmod {\rm prime}\ p\!:\ \color{#c00}{p\equiv 0}\,$ but $\rm\,a^{\large\color{#c00} p}\equiv a^{ \color{#c00}0}\!\!\iff\! a\equiv 1,\,$ by little Fermat. But there is a more limited rule for integer powers - see modular order reduction,