1. ## Congruence Proof

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.

2. Hello,
Originally Posted by jzellt
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.
Assume that in base B, $n=n_kn_{k-1}\dots n_2n_1m$, where $n_i$ are its digits.

Then, $n=n_kB^k+n_{k-1}B^{k-1}+\dots+n_2B^2+n_1B+m$

$n=B \underbrace{\left(n_kB^{k-1}+n_{k-1}B^{k-2}+\dots+n_2B+n_1\right)}_{N}+m$

Hence $n=BN+m \equiv m \bmod B$