Number of form 28000...0007 is Divisible by 7

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Theorem

Let $x$ be a natural number in the form:

$\sqbrk {28 \underbrace {000 \cdots 0}_n 7}_{10}$

Then $x$ is divisible by $7$.


Proof 1

We have:

\(\ds \sqbrk {28 \underbrace {000 \cdots 0}_n 7}_{10}\) \(=\) \(\ds 2 \times 10^{n + 2} + 8 \times 10^{n + 1} + 0 \times 10^n + \cdots + 0 \times 10^1 + 7\)
\(\ds \) \(=\) \(\ds 28 \times 10^{n + 1} + 7\)


Then:

\(\ds 28 \times 10^{n + 1} + 7\) \(\equiv\) \(\ds 0 \times 3^{n + 1} + 0\) \(\ds \pmod 7\) Fermat's Little Theorem and Congruence of Powers
\(\ds \) \(\equiv\) \(\ds 0 + 0\) \(\ds \pmod 7\) Congruence of Product
\(\ds \) \(\equiv\) \(\ds 0\) \(\ds \pmod 7\)

$\blacksquare$


Proof 2

We have:

\(\ds \sqbrk {28 \underbrace {000 \cdots 0}_n 7}_{10}\) \(=\) \(\ds 7 \times \sqbrk {4 \underbrace {000 \cdots 0}_n 1}_{10}\)

and the result follows by definition of divisibility.

$\blacksquare$