Divisibility by Power of 10

Theorem

Let $r \in \Z_{\ge 1}$ be a strictly positive integer.

An integer $N$ expressed in decimal notation is divisible by $10^r$ if and only if the last $r$ digits of $N$ are all $0$.

That is:

$N = \sqbrk {a_n \ldots a_2 a_1 a_0}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $10^r$

$a_0 + a_1 10 + a_2 10^2 + \cdots + a_r 10^r = 0$

Let $N$ be divisible by $10^r$.

Then:

$\ds N$

$\equiv$

$\ds 0 \pmod {10^r}$

$\ds \leadstoandfrom \ \ $

$\ds \sum_{k \mathop = 0}^n a_k 10^k$

$\equiv$

$\ds 0 \pmod {10^r}$

$\ds \leadstoandfrom \ \ $

$\ds \sum_{k \mathop = 0}^r a_k 10^r + \sum_{k \mathop = r + 1}^n a_k 10^k$

$\equiv$

$\ds 0 \pmod {10^r}$

$\ds \leadstoandfrom \ \ $

$\ds \sum_{k \mathop = 0}^r a_k 10^r + 10^r \sum_{k \mathop = r + 1}^n a_k 10^{k - r}$

$\equiv$

$\ds 0 \pmod {10^r}$

$\ds \leadstoandfrom \ \ $

$\ds \sum_{k \mathop = 0}^r a_k 10^r$

$\equiv$

$\ds 0 \pmod {10^r}$

as $10^r \equiv 0 \pmod {10^r}$

Hence the result.

$\blacksquare$