Congruence of Sum of Digits to Base Less 1

Theorem
Let $$x \in \Z$$, and $$b \in \N, b > 1$$.

Let $$x$$ be written in base $b$:


 * $$x = \left[{r_m r_{m-1} \ldots r_2 r_1 r_0}\right]_b$$.

Then $$\sum_{j=0}^m r_j \equiv x \pmod {b-1}$$.

That is, the sum of the digits of any integer $$x$$ in base $b$ notation is congruent to $x$ modulo $b-1$.

Proof
Let $$x \in \Z, x > 0$$, and $$b \in \N, b > 1$$.

Then from the Basis Representation Theorem, $$x$$ can be expressed uniquely as:


 * $$x = \sum_{j = 0}^m r_j b^j, r_0, r_1, \ldots, r_m \in \left\{{0, 1, \ldots, b-1}\right\}$$.

Proof by induction:

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition $$\sum_{j = 0}^n r_j \equiv x \pmod {b-1}$$.

Basis for the Induction

 * $$P(1)$$ is trivially true: $$\forall x: 0 \le x \le b: x \equiv x \pmod {b-1}$$. This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:


 * $$\sum_{j = 0}^k r_j \equiv \sum_{j = 0}^k r_j b^j \pmod {b-1}$$.

Then we need to show:


 * $$\sum_{j = 0}^{k+1} r_j \equiv \sum_{j = 0}^{k+1} r_j b^j \pmod {b-1}$$.

Induction Step
This is our induction step:

Let $$y$$ be expressed as $$y = \sum_{j = 0}^{k+1} {r_j b^j}, r_0, r_1, \ldots, r_{k+1} \in \left\{{0, 1, \ldots, b}\right\}$$.

Then $$y = \sum_{j = 0}^k r_j b^j + r_{k+1} b^{k+1}$$.

Now from Congruence of Powers, $$b \equiv 1 \pmod {b-1} \implies b^n \equiv 1^n \pmod {b-1} \implies b^n \equiv 1 \pmod {b-1}$$.

So by modulo multiplication: $$r_{k+1} b^{k+1} \equiv r_{k+1} \pmod {b-1}$$.

From the induction hypothesis: $$\sum_{j = 0}^{k+1} r_j \equiv y \pmod {b-1}$$

Thus by modulo addition: $$\sum_{j = 0}^{k+1} r_j \equiv \sum_{j = 0}^k r_j + r_{k+1} \pmod {b-1}$$.

Hence $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction:


 * $$\sum_{j = 0}^n r_j \equiv \sum_{j = 0}^n r_j b^j \pmod {b-1}$$.