# Why is $9 \times 11{\dots}12 = 100{\dots}08$?

While I was working on Luhn algorithm implementation, I discovered something unusual.

$$9 \times 2 = 18$$ $$9 \times 12 = 108$$ $$9 \times 112 = 1008$$ $$9 \times 1112 = 10008$$

Hope you can observe the pattern here.

What to prove this? What is it's significance?

• It's so good to see when people get interested in these beautiful things! and they start looking for explanation :) – Daniel Jan 21 '17 at 14:07
• Try doing the multiplications with pencil and paper rather than relying on a calculator ... – hmakholm left over Monica Jan 21 '17 at 14:45
• Hint $\ 9 = 10\!-\!1\$ divides $\ 10^N\!+\!8 = 10^N\!-\!1 + (10-1)\ \$ See Casting Out Nines. – Bill Dubuque Jan 21 '17 at 15:26

The repunit, $R_k = \overbrace{111\ldots 111}^{k \text{ ones}}$ , can be written as $R_k = \dfrac{10^k-1}{9}$

Your nice pattern corresponds to $9\times (R_k+1) = (10^k-1)+9 = 10^k+8$

Here's an idea.

This pattern is easier to understand: $$9 \times 11 = 99$$

$$9 \times 111 = 999$$

$$9 \times 1111 = 9999$$

Then see what happens when you add 9 to both sides.

One logic behind.

$9 \times (0 + 2) = 9 \times 0 + 9 \times 2 = 18$

$9 \times (10 + 2) = 9 \times 10 + 9 \times 2 = 108$

$9 \times (110 + 2) = 9 \times 110 + 9 \times 2 = 1008$

So on.

$$9 \times 11\cdots12 = 9 \times (11\cdots 11+1) = 99\cdots99 + 9.$$

Write $1...12$ as $$\overline{1...(\times\ n)...1} +1=\frac{10^n-1}{9}+1$$

Hence $$9\times (\frac{10^n-1}{9}+1)=10^n-1+9=10^n+8$$