123456789 x 8 + 9 = 987654321
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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:
- $\ds \paren {b - 2} \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \paren {b - j} b^{n - j}$
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 $\map P n$ be the proposition:
- $\ds \paren {b - 2} \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \paren {b - j} b^{n - j}$
Basis for the Induction
$\map P 1$ is the case:
\(\ds \paren {b - 2} \sum_{j \mathop = 1}^1 j b^{1 - j} + 1\) | \(=\) | \(\ds \paren {b - 2} b^0 + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^1 \paren {b - j} b^{1 - j}\) |
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $\ds \paren {b - 2} \sum_{j \mathop = 1}^k j b^{k - j} + k = \sum_{j \mathop = 1}^k \paren {b - j} b^{k - j}$
from which it is to be shown that:
- $\ds \paren {b - 2} \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \paren {k + 1} = \sum_{j \mathop = 1}^{k + 1} \paren {b - j} b^{k + 1 - j}$
Induction Step
This is the induction step:
\(\ds \paren {b - 2} \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \paren {k + 1}\) | \(=\) | \(\ds \paren {b - 2} \sum_{j \mathop = 1}^k j b^{k + 1 - j} + k + 1 + \paren {b - 2} \paren {k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {b - 2} b \sum_{j \mathop = 1}^k j b^{k - j} + \paren {b k - b k} + k + 1 + b k - 2 k + b - 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b \paren {\paren {b - 2} \sum_{j \mathop = 1}^k j b^{k - j} + k} + b - k - 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b \paren {\sum_{j \mathop = 1}^k \paren {b - j} b^{k - j} } + b - k - 1\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^k \paren {b - j} b^{k + 1 - j} + b - \paren {k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^{k + 1} \paren {b - j} b^{k + 1 - j}\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \paren {b - 2} \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \paren {b - j} b^{n - j}$
$\blacksquare$