Quotient and Remainder to Number Base

Theorem
Let $$n \in \Z: n > 0$$ be an integer.

Let $$n$$ be expressed in base $b$:
 * $$n = \sum_{j = 0}^m {r_j b^j}$$

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

Then: where:
 * $$\left \lfloor {\frac n b} \right \rfloor = \left[{r_m r_{m-1} \ldots r_2 r_1}\right]_b$$
 * $$n \,\bmod\, b = r_0$$
 * $$\left \lfloor {.} \right \rfloor$$ denotes the floor function;
 * $$n \,\bmod\, b$$ denotes the modulo operation.

General Result

 * $$\left \lfloor {\frac n {b^s}} \right \rfloor = \left[{r_m r_{m-1} \ldots r_{s+1} r_s}\right]_b$$
 * $$n \,\bmod\, {b^s} = \sum_{j = 0}^{s-1} {r_j b^j} = \left[{r_{s-1} r_{s-2} \ldots r_1 r_0}\right]_b$$

where $$s \in \Z: 0 \le s \le m$$.

Proof
From the Quotient-Remainder Theorem, we have:


 * $$\exists q, r \in \Z: n = q b + r$$

where $$0 \le b < r$$.

We have that:

$$ $$ $$

Hence we can express $$n = q b + r$$ where:
 * $$q = \sum_{j = 1}^m {r_{j} b^{j-1}}$$
 * $$r = r_0$$

where $$\sum_{j = 1}^m {r_{j} b^{j-1}} = \left[{r_m r_{m-1} \ldots r_2 r_1}\right]_b$$.

The result follows from the definition of the modulo operation.

Proof of General Result
Follows directly by induction on $$s$$.

Example
This result is often used in computer algorithms for converting a date (in $$yyyymmdd$$ format) into a date object with separate day, month and year.

Performing the above "mod and div" operations on $$20100209$$, we get:

$$ $$ $$ $$