Numbers divisible by $11$

Question

A number is divisible by $11$, when the difference between the sum of the digits in the odd positions counting from the left (the first, third, ....) and the sum of the remaining digits is either 0 or divisible by 11. Why is that?

Answer 1

\begin{align} abcde_{10} &= 10000a + 1000b + 100c + 10d + e \\ &= (9999a + a) + (1001b - b) + (99c + c) + (11d - d) + e \\ &= (9999a + 1001b + 99c + 11d) + (a-b+c-d+e) \\ &= 11(101a + 91b + 11c + d) + (a-b+c-d+e) \\ &\equiv a-b+c-d+e \pmod{11} \end{align}

Another way to look at this problem is to note that $10^{2n} \equiv (-1)^{2n} \equiv 1 \pmod{11}$

So $10^{2n}(10m + n) \equiv 1(-m + n) \equiv -m + n \pmod{11}$

So \begin{align} abcdef_{10} &\equiv 10^4(10a + b) + 10^2(10c + d) + 10^0(10e + f) \\ &\equiv (-a+b) + (-c + d) + (-e + f) \\ &\equiv -a+b-c+d-e+f \end{align}

Answer 2

Let $a$ be a whole number. We'll write $a$ as $a=a_0+a_1\cdot10+_{\cdots}+a_k\cdot 10^k$.

Let's check when $a (mod 11)=\overline{a}=0$ ($\overline a$ is just to make the writing more comfortable.

$\overline a=\overline{a_0+a_1\cdot10+_{\cdots}+a_k\cdot 10^k}=\overline{a_0+a_1\cdot(-1)+_{\cdots}+a_k\cdot (-1)^k}=\overline{a_0-a_1+_\cdots+a_k\cdot(-1)^k}$

and that equals zero only when $a_0-a_1+_\cdots+a_k\cdot(-1)^k$ equals to zero or something that is divisible by 11.