Number of form 4666...6669 is Divisible by 7

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Theorem

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

$\sqbrk {4 \underbrace {666 \cdots 6}_n 9}_{10}$

Then $x$ is divisible by $7$.


Proof

We have:

\(\ds \sqbrk {4 \underbrace {666 \cdots 6}_n 9}_{10}\) \(=\) \(\ds 4 \times 10^{n + 1} + 6 \times 10^n + \cdots + 6 \times 10^1 + 9\)
\(\ds \) \(=\) \(\ds 4 \times 10^{n + 1} + \dfrac {6 \times \paren {10^n - 1} } {\paren {10 - 1} } \times 10 + 9\) Sum of Geometric Sequence
\(\ds \) \(=\) \(\ds 4 \times 10^{n + 1} \times \dfrac 3 3 + \dfrac {6 \times \paren {10^n - 1} } 9 \times 10 + 9 \times \dfrac 3 3\) multiplying by 1 to obtain Common Denominator
\(\ds \) \(=\) \(\ds \dfrac {12 \times 10^{n + 1} } 3 + \dfrac {2 \times \paren {10^{n + 1} - 10} } 3 + \dfrac {27} 3\)
\(\ds \) \(=\) \(\ds \dfrac {14 \times 10^{n + 1} + 7} 3\)


The sum of the digits in $14 \times 10^{n + 1} + 7$ is in fact $12 = 1 + 4 + 7$ which is a multiple of $3$.

Hence, by Divisibility by $3$, $\dfrac {14 \times 10^{n + 1} + 7} 3$ is indeed an integer.


Then:

\(\ds (14 \times 10^{n + 1} + 7) \div 3\) \(\equiv\) \(\ds (0 \times 3^{n + 1} + 0) \div 3\) \(\ds \pmod 7\) Congruence of Powers
\(\ds \) \(\equiv\) \(\ds (0 + 0) \div 3\) \(\ds \pmod 7\) Congruence of Product
\(\ds \) \(\equiv\) \(\ds 0 \div 3\) \(\ds \pmod 7\) Congruence of Quotient
\(\ds \) \(\equiv\) \(\ds 0\) \(\ds \pmod 7\)

$\blacksquare$