That digit is missing!

The number is $\underbrace{111....1}_{×24}m\underbrace{111....111}_{×25}$

$1001=7×11×13$

So, $111111=1001×111$ is also divisible by $13$

This means that $6$ consecutive $1$ are divisible by $13$

So, $24(6×4)$ $1$ 's are also divisible by $13$

So, both left and right of middle $2$ digits are divisible by $13$

This means that middle two digits also must be divisible by $13$ to make the given number divisible by $13$

So, $\overline{m1}$ is divisible by $13$

So, $\boxed{m=9}$

My this problem is similar

Let's look at the value of a 1 in each position. We work (mod 13) all the time here. $10^0=1,10^1=10,10^2=9,10^3=10×9=12,10^4=10×12=3,10^5=10×3=4,10^6=10×4=1$ . This pattern repeats every 6 positions, so the value of the digit at position 26 has the same value as the digit at position 2, which is 10.

A number of consisting of 50 1's has value $8×(1+10+9+12+3+4) +1+10=11$ , so with a 0 at position 26 we have $11-10=1$ .We need to add 12 to that to get a 0.

Since $9×10=90=12$ , we need a $\boxed{9}$ at that position.

$111111$ is the least all $1$ containing number which is divisible by $13$ . So the left part of missing number consists $24$ $1s$ . As $24$ is a multiple of 6 ,the left part or precisely $24$ 1s are divisible by $13$ . So, same as this the $24 1s$ from right is also divisible by $13$ . Only left one $1$ in $25 th$ position from right and the missing digit in $26th$ position from right. So, the missing number is $9$ ,because $91$ is the only two–digit number consisting $1$ in its unit position. So, answer is $9$ .