Use modular arithmetic to show that a number is divisible by 11 iff the sum of its alternating digits is divisible by 11

Question

We can expand the number $n = n_0 + 10n_1 + ... + (10^s)n_s$

Then we have $10^k ≡ (-1)^k \mod11$.

How do we go from here to here:

$n ≡ n_0 - n_1 + ... + (-1)^s n_s \mod 11$

I do not understand how the right hand side is congruent to (≡) n.

Furthermore, how do we jump from here to

"This shows if $11|n$ , then $n ≡ 0\mod 11$ (understand this) and so $n_0 - n_1 + ... + (-1)^s n_s = 0 \mod 11$ (do NOT understand where this jump in logic came from). Please explain.

Answer 1

We know that $10^k \equiv (-1)^k \pmod{11}$. If $k$ is even, then $(-1)^k \equiv 1 \pmod{11}$. If $k$ is odd, then $(-1)^k \equiv -1 \pmod{11}$. Hence,

\begin{align*} n & \equiv n_0 + 10n_1 + 10^2n_2 + 10^3n_3 + \cdots + 10^sn_s \pmod{11}\\ & \equiv n_0 + (-1)^1n_1 + (-1)^2n_2 + (-1)^3n_3 + \ldots + (-1)^sn_s \pmod{11}\\ & \equiv n_0 - n_1 + n_2 - n_3 + \cdots + (-1)^sn_s \pmod{11} \end{align*}

If $11|n$, then $n \equiv 0 \pmod{11}$, so

$$n_0 - n_1 + n_2 - n_3 + \cdots + (-1)^sn_s \equiv 0 \pmod{11}$$

Answer 2

HINT

Consider the number $n=n_1n_0=n_1\cdot 10 + n_0.$ You have that

$$(10n_1+n_0)\: (\mathrm{mod}\: 11)= 10n_1 \: (\mathrm{mod}\: 11) + n_0 \: (\mathrm{mod}\: 11)=-n_1+n_0 \: (\mathrm{mod}\: 11),$$

since as you have said $10\equiv -1\pmod{11}.$

Consider the number $n=n_2n_1n_0=n_2\cdot 10^2+n_1\cdot 10 + n_0.$ You have that

$$(100n_2+10n_1+n_0)\: (\mathrm{mod}\: 11)= 100n_2 \:(\mathrm{mod}\: 11)+10n_1 \: (\mathrm{mod}\: 11)+ n_0 \: (\mathrm{mod}\: 11)\\=(n_2-n_1+n_0) \:(\mathrm{mod}\: 11),$$

since as you have said $100\equiv 1 \pmod{11}$ and $10\equiv -1\pmod{11}.$

Can you get the general case from this particular cases?