would it be equivalent to find this remainder? I was given this exercise: Suppose that  $N$ is written in the decimal system in the form $N=a_m a_{m-1} \dots a_1 a_0 $.

Prove that $11 \mid N \Leftrightarrow 11|a_0-a_1+ \dots+(-1)^{m-1}a_{m-1}+(-1)^m a_{m}$.

So,if I would have to find the remainder of the division of $N$ by $11$,could I find the remainder of the division of $a_0-a_1+ \dots+(-1)^{m-1}a_{m-1}+(-1)^m a_{m}$ by $11$ ??
The key of your problem is $11|(10^{k}-1)$ for all even $k$ and $11|(10^k+1)$ for all odd $k$. Since 
$$
N=a_m a_{m-1} \dots a_1 a_0=\sum_{i=0}^ma_i10^i,
$$
then
\begin{align}
11|N&amp;\iff 11|\sum_{i=0}^ma_i10^i\\
&amp;\iff 11|\left(\sum_{i\textrm{ odd}}^ma_i10^i+\sum_{i\textrm{ even}}^ma_i10^i\right)\\
&amp;\iff 11|\left(\sum_{i\textrm{ odd}}^ma_i(10^i+1)+\sum_{i\textrm{ even}}^ma_i(10^i-1)-\sum_{i\textrm{ odd}}^ma_i+\sum_{i\textrm{ even}}^ma_i\right).\\
\end{align}
Now using the two facts above, you will lead to what you wanted.
Mo you can not find it. If you have seen its proof this strategy works because because 9999...uptil even number of terms is removed as out will always be facor and in odd digits 10000...1 is divisible so we remove that factor
  sorry I mis understood your question I gave the explanation  for divisor . We can find the remainder it will be either number yor are getting let say n so it will be n or 11-n for n>0  and -n or n+11