Numbers which Multiplied by 2 are the Reverse of when Added to 2

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Theorem

\(\ds 47 + 2\) \(=\) \(\ds 49\)
\(\ds 47 \times 2\) \(=\) \(\ds 94\)


\(\ds 497 + 2\) \(=\) \(\ds 499\)
\(\ds 497 \times 2\) \(=\) \(\ds 994\)


\(\ds 4997 + 2\) \(=\) \(\ds 4999\)
\(\ds 4997 \times 2\) \(=\) \(\ds 9994\)

... and so on:


Proof

We have that:

$\ds \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7} + 2 = 4 \times 10^n + \sum_{k \mathop = 0}^{n - 1} 9 \times 10^k$

using the Basis Representation Theorem.


It remains to be demonstrated that:

$\ds 2 \times \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7} = \sum_{k \mathop = 1}^n 9 \times 10^k + 4$

again using the Basis Representation Theorem.


Thus:

\(\ds 2 \times \paren {4 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 7}\) \(=\) \(\ds 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 2 \times 9 \times 10^k + 2 \times 7\)
\(\ds \) \(=\) \(\ds 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 18 \times 10^k + 14\)
\(\ds \) \(=\) \(\ds 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + \sum_{k \mathop = 1}^{n - 1} 10 \times 10^k + 10\)
\(\ds \) \(=\) \(\ds 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + \sum_{k \mathop = 0}^{n - 1} 10 \times 10^k\)
\(\ds \) \(=\) \(\ds 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + \sum_{k \mathop = 1}^n 10^k\)
\(\ds \) \(=\) \(\ds 8 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 8 \times 10^k + 4 + 10^n + \sum_{k \mathop = 1}^{n - 1} 10^k\)
\(\ds \) \(=\) \(\ds 9 \times 10^n + \sum_{k \mathop = 1}^{n - 1} 9 \times 10^k + 4\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n 9 \times 10^k + 4\)

$\blacksquare$


Sources

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