Congruence Proof

**jzellt** · March 19th 2009, 09:49 PM

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.

**Moo** · March 19th 2009, 11:37 PM

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

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