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I am studying number representations in base $b$, and in order to understand the method of "casting out 9s", "casting out 11s", and "casting out 99s" for a generic base, the following observations have been made: $$ \sum_{k=0}^p a_k \cdot b^k \equiv \sum_{k=0}^p a_k \mbox{ (mod }b-1), $$ $$ \sum_{k=0}^p a_k \cdot b^k \equiv \sum_{k=0}^p (-1)^k \cdot a_k \mbox{ (mod }b+1), $$ and $$ \sum_{k=0}^p a_k \cdot b^k \equiv \sum_{k=0}^{p/2} a_{2k+1}a_{2k} \mbox{ (mod }b^2-1), $$

(later edited) where the terms on the right-side $a_{2k+1}a_{2k}$ are just 2-digits numbers, not products, which I pointed out as "$\cdot$".

I can easily derive the first two formulas, but I am stuck with the third one. Of course, I know that $b^2-1=(b+1)(b-1)$, but I may be missing some properties properties of modular arithmetic. Can you help me?

EDIT After checking the 'casting out 99s' method, I realized that $a_{2k+1}a_{2k}$​ is not the product of the two coefficients, but the number obtained by juxtaposing them. So we just sum pairs of digits. But anyway, why is that the case?

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See that $b^k \equiv b^{2\lfloor k/2 \rfloor}b^{k-2\lfloor k/2 \rfloor} \equiv b^{k'} \pmod{b^2-1}$, where $k' \in {0,1}$ and $k\equiv k' \pmod{2}$. Thus, when you look at the sum $a_{2j}b^{2j} + a_{2j+1}b^{2j+1}$, you get:

$a_{2j}b^{2j} + a_{2j+1}b^{2j+1} \equiv a_{2j+1}b^1+a_{2j} = a_{2j+1}a_{2j}$, the desired two-digit number in base $b$

by24
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