Proving divisibility tests using congruence relations

Question

For a positive integer $N$ which has the decimal representation $$N=\sum_{k=0}^n a_k\cdot10^k $$

Prove that $$11\mid N \Longleftrightarrow 11\mid \sum_{k=0}^n(-1)^k a_k $$ using congruence relations on $N$

Answer 1

Hint: We have $10\equiv -1\pmod{11}$ and therefore $10^k\equiv (-1)^k\pmod{11}$.

Answer 2

Hint $\ $ Note that the decimal (radix $10)$ representation writes $N$ as a polynomial function $\,f(10)\,$ of the radix, with integer coefficients (digits) $\,a_i,\,$ i.e.

$$\begin{eqnarray} N = f(10) &&= a_n 10^n +\,\cdots+a_1 10 + a_0 \\ {\rm where}\ \ f(x) &&= a_n\, x^n\,+\,\cdots\,+a_1\, x\, + a_0\end{eqnarray}$$

Thus $\ {\rm mod}\ 11\!:\,\ \color{#c00}{10\equiv -1}\,\Rightarrow\, f(\color{#c00}{10})\equiv f(\color{#c00}{-1})\,$ follows by the Polynomial Congruence Rule below.

Therefore $\ 11\mid N=f(10)\iff 11\mid f(-1),\,$ precisely what is claimed.

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Congruence Sum Rule $\rm\qquad\quad A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#c0f}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$

Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#c0f}{A+B - (a+b)} $

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

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

Congruence Power Rule $\rm\qquad \color{}{A\equiv a}\ \Rightarrow\ \color{#c00}{A^n\equiv a^n}\ \ (mod\ m)$

Proof $\ $ It is true for $\rm\,n=1\,$ and $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, \color{#c00}{A^{n+1}\equiv a^{n+1}},\,$ by the Product Rule, so the result follows by induction on $\,n.$

Polynomial Congruence Rule $\ $ If $\,f(x)\,$ is polynomial with integer coefficients then $\ A\equiv a\ \Rightarrow\ f(A)\equiv f(a)\,\pmod m.$

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