Show that if n is a natural number and m is the last digit of n written in base B , then n and m are congruent mod B.
Hello,
Assume that in base B, $\displaystyle n=n_kn_{k-1}\dots n_2n_1m$, where $\displaystyle n_i$ are its digits.
Then, $\displaystyle n=n_kB^k+n_{k-1}B^{k-1}+\dots+n_2B^2+n_1B+m$
$\displaystyle n=B \underbrace{\left(n_kB^{k-1}+n_{k-1}B^{k-2}+\dots+n_2B+n_1\right)}_{N}+m$
Hence $\displaystyle n=BN+m \equiv m \bmod B$