Unknown Number of Digits

Note the following expansions to be used later:

$(\overbrace{11...11}^{k} = 10^{k-1} + 10^{k-2} + 10^{k-3}... + 10^{0}$

$1\overbrace{00...00}^{k}1 = 1 \times 10^{k} + 1$

When $k=0$ , the value of $11 \div 11 = 1$ is an integer. Then, for integers $k \geq 1$ :

$1\overbrace{00...00}^{k}1 \times \overbrace{11...11}^{k} = 10^{k} \times (10^{k-1} + ... + 10^{0}) + 1 \times (10^{k-1} + ... + 10^{0})$ $= 10^{2k-1} + 10^{2k-2} + 10^{2k-3} + ... + 10^{k} + 10^{k-1} + 10^{k-2} + 10^{k-3} + ... + 10^{0}$ $= 1\overbrace{11...11}^{2k}1$

$\therefore 1\overbrace{11...11}^{2k}1 \div 1\overbrace{00...00}^{k}1 = \overbrace{11...11}^{k}$

Together, this proves that the value is an integer for all integer values of $k \geq 0$. The answer is always an integer .

Note that $1\overbrace{11\cdots 11}^{2k}1=\overbrace{11\cdots 11}^{2k+2} = \frac{\overbrace{99\cdots 99}^{2k+2}}{9}=\frac{10^{2k+2}-1}{9}=\frac{(10^{k+1}+1)(10^{k+1}-1)}{9}$ and

$1\overbrace{00\cdots 00}^{k} 1 =1\overbrace{00\cdots 00}^{k+1}+ 1 = 10^{k+1}+1$

Thus the problem becomes

$\frac{(10^{k+1}+1)(10^{k+1}-1)}{9} \times \frac{1}{10^{k+1}+1} = \frac{10^{k+1}-1}{9}$

Noting the way the first term was decomposed, it can be easily deduced that

$\frac{10^{k+1}-1}{9}=\overbrace{11\cdots 11}^{k+1}$

Hence, the result is always an integer