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.
$\blacksquare$