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Math Help - Congruence Proof

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    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.
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    Hello,
    Quote Originally Posted by jzellt View Post
    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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