My book makes a demonstration using the concept of congruence to show the divisibility by 11. But I don't understand the last part.

  1. He stated with: \begin{align} 10\equiv 10 \mod11 \\ 1 \equiv 1 \mod11\\ 10^2\equiv1 \mod11 \to 10^{2n}\equiv1 \mod11 \end{align}

and then, $10^{2n}*10 \equiv 0 \mod11$

  1. Now for a number $n= n_{r}...n_5n_4n_3n_2n_1n_0$ wrote in base 10, we have that:

\begin{align} n_0 \equiv n_0\mod11\\ 10n_1+n_1 \equiv 0\mod11\\ 10^2n_2\equiv n_2\mod11\\ 10^3n_3+n_3 \equiv 0\mod11\\ ... \end{align}

  1. Adding term by term we have this, the part that I don't understand:

$n_1+n_3... \equiv n_0+n_2... \mod11$.

Because, When I sum the congruence I have this: $n_0+ (10a_1+a_1)+ 10^2a_2+ (10^3a_3+a_3)...\equiv a_0+a_2... \mod11$

How I get $n_1+n_3... \equiv n_0+n_2... \mod11$.??


The relations $\;10^{2k+1}a_{2k+1}+a_{2k+1}\equiv 0\mod 11$ simply means that $$10^{2k+1}a_{2k+1}\equiv-a_{2k+1}\mod 11,$$ so the relation becomes $$a_0-a_1+a_2-a_3+\dotsm\equiv 0\iff a_0 +a_2 +\dotsm\equiv a_1+ a_3+\dotsm\mod 11.$$


Some of your $n$'s changed into $a$'s. The think to note is that $10 \equiv -1 \pmod{11}.$ So the coefficients on the odd-numbered $n$'s are all really $-1$ and you can move them to the other side, where the $-1$ changes into a $+1$.

  • 1
    $\begingroup$ Right. We’re interested in the alternating sum of the digits. $\endgroup$ – Lubin Jun 29 '18 at 16:26

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