123456789 x 8 + 9 = 987654321

Theorem

\(\ds 1 \times 8 + 1\)

\(=\)

\(\ds 9\)

\(\ds 12 \times 8 + 2\)

\(=\)

\(\ds 98\)

\(\ds 123 \times 8 + 3\)

\(=\)

\(\ds 987\)

\(\ds 1234 \times 8 + 4\)

\(=\)

\(\ds 9876\)

\(\ds 12345 \times 8 + 5\)

\(=\)

\(\ds 98765\)

\(\ds 123456 \times 8 + 6\)

\(=\)

\(\ds 987654\)

\(\ds 1234567 \times 8 + 7\)

\(=\)

\(\ds 9876543\)

\(\ds 12345678 \times 8 + 8\)

\(=\)

\(\ds 98765432\)

\(\ds 123456789 \times 8 + 9\)

\(=\)

\(\ds 987654321\)

The above pattern is an instance of the identity:

$\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \left({b - j}\right) b^{n - j}$

where $b = 10$ and $n$ goes from $1$ to $9$.

Proof

The proof proceeds by induction.

Let $n, b \in \Z_{>0}$, where $b \ge 3$.

For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \left({b - j}\right) b^{n - j}$

Basis for the Induction

$P \left({1}\right)$ is the case:

\(\ds \left({b - 2}\right) \sum_{j \mathop = 1}^1 j b^{1 - j} + 1\)

\(=\)

\(\ds \left({b - 2}\right) b^0 + 1\)

\(\ds \)

\(=\)

\(\ds b - 1\)

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 1}^1 \left({b - j}\right) b^{1 - j}\)

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis

Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:

$\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^k j b^{k - j} + k = \sum_{j \mathop = 1}^k \left({b - j}\right) b^{k - j}$

from which it is to be shown that:

$\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \left({k + 1}\right) = \sum_{j \mathop = 1}^{k + 1} \left({b - j}\right) b^{k + 1 - j}$

Induction Step

This is the induction step:

\(\ds \left({b - 2}\right) \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \left({k + 1}\right)\)

\(=\)

\(\ds \left({b - 2}\right) \sum_{j \mathop = 1}^k j b^{k + 1 - j} + k + 1 + \left({b - 2}\right) \left({k + 1}\right)\)

\(\ds \)

\(=\)

\(\ds \left({b - 2}\right) b \sum_{j \mathop = 1}^k j b^{k - j} + \left({b k - b k}\right) + k + 1 + b k - 2 k + b - 2\)

\(\ds \)

\(=\)

\(\ds b \left({\left({b - 2}\right) \sum_{j \mathop = 1}^k j b^{k - j} + k}\right) + b - k - 1\)

\(\ds \)

\(=\)

\(\ds b \left({\sum_{j \mathop = 1}^k \left({b - j}\right) b^{k - j} }\right) + b - k - 1\)

Induction Hypothesis

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 1}^k \left({b - j}\right) b^{k + 1 - j} + b - \left({k + 1}\right)\)

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 1}^{k + 1} \left({b - j}\right) b^{k + 1 - j}\)

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \left({b - j}\right) b^{n - j}$

$\blacksquare$