 Alternating digit sum

The alternating digit sum of a nonnegative integer in radix $b\ge 2$ iswith a positive sign on the units digit. Starting the signs from the opposite end can change the overall sign, so the convention should be stated.

Because $b\equiv -1(\mathrm{mod}b+1)$, the positional expansion gives $n\equiv A_b(n)(\mathrm{mod}b+1)$. This proves a divisibility test: $b+1$ divides the number exactly when it divides its alternating digit sum. In decimal notation, $1023$ has alternating sum $3-2+0-1=0$ and equals $11\cdot 93$.

The rule differs from the ordinary digit sum test modulo $b-1$, which instead uses $b\equiv 1$. Both follow from the same modular arithmetic principle of replacing the radix by its residue before evaluating the digit polynomial.