I want a proof of this conjecture:
Given a number $n$ with more than two figures written in a positional base-$b$ numeral system and let $n'$ be a base-$b$ number obtained by removing a consecutive pair of equal $b$-figures in $n$. Then $$\,n\!\!\mod (b+1)=\,n'\!\!\mod (b+1)$$
Example: $123345667 \mod 11 = 1245667 \mod 11$ for $b\ge 8$.
This is essentially related to a general form of the divisibility test for $11$ in our normal base numbering system.
Since $b\equiv -1 \bmod b{+}1$, $b^k\equiv -1^k \bmod b{+}1$, which alternates between positive and negative as $k$ increments. In particular $b^{k+2}\equiv b^k\bmod b{+}1$, so shifting a number by two places in the number repesentation does not change its residue $\bmod b{+}1$.
So when we have two adjacent equal digits $\ldots dd\ldots$, their net contribution to the residue $\bmod b{+}1$ is $(+d-d)=0$, and when we shift the higher digits down two places folowing their removal this also has no effect on the residue $\bmod b{+}1$, as required.