Numbers Reversed when Multiplying by 9
Theorem
Numbers of the form $\sqbrk {10 (9) 89}_{10}$ are reversed when they are multiplied by $9$:
| \(\ds 1089 \times 9\) | \(=\) | \(\ds 9801\) | ||||||||||||
| \(\ds 10 \, 989 \times 9\) | \(=\) | \(\ds 98 \, 901\) | ||||||||||||
| \(\ds 109 \, 989 \times 9\) | \(=\) | \(\ds 989 \, 901\) |
and so on.
Proof
Let k represent the number of $9$s in the middle of the number.
For $k > 0$ We can rewrite the number as follows:
| \(\ds \sqbrk {10 (9) 89}_{10}\) | \(=\) | \(\ds 10 \times 10^{k + 2 } + 900 \sum_{i \mathop = 0}^{k - 1} 10^i + 89\) | Definition of Geometric Series |
Taking numbers of this form and multiplying by $9$ produces:
| \(\ds 9 \times \paren {10 \times 10^{k + 2 } + 900 \sum_{i \mathop = 0}^{k - 1} 10^i + 89 }\) | \(=\) | \(\ds 90 \times 10^{k + 2 } + 8100 \sum_{i \mathop = 0}^{k - 1} 10^i + 801\) |
The first part is composed of $k + 4$ digits. The first digit will be $9$ followed by $k + 3$ digits of $0$
| \(\ds 90 \times 10^{k + 2 }\) | \(=\) | \(\ds 9 \times 10^{k + 3 }\) |
The sum in the middle is composed of $k + 3$ digits. The first digit will be $8$ followed by $k - 1$ digits of $9$ and then the remaining three digits at the end are $100$
| \(\ds 8100 \sum_{i \mathop = 0}^{k - 1} 10^i\) | \(=\) | \(\ds 899 \cdots 99100\) |
Summing the three pieces, the final answer will have $k + 4$ digits.
The first digit is $9$
followed by $8$ which is the first digit of the middle part
followed by $k$ digits of $9$ where the last $9$ is the sum of the $1$ from the middle part and the $8$ of the last part
and then ending in $01$:
| \(\ds \sqbrk {10 (9) 89}_{10} \times 9\) | \(=\) | \(\ds \sqbrk {98 (9) 01}_{10}\) |
$\blacksquare$
Also see
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $1089$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1089$