A cool problem!

We will first find the number of $10$ digit binary strings. Let $a_n = \text{ number of n digit strings with 1 in the leftmost slot}$ $b_n = \text{ number of n digit strings with 0 in the leftmost slot}$

We know that $a_n = b_{n-1} + b_{n-2}$ since the leftmost part of a string that ends in $1$ can start in two ways. Either $10$ or $110$ . In the former, it is just $b_{n-1}$ and just appending one $1$ . In the later, it is just $b_{n-2}$ and just appending two $1$ 's. Similarly, we know that $b_n = a_{n-1} + a_{n-2}$ . We will let $X_n = a_n + b_n = \text{ total number of n digit strings}$ .

By adding the two recursions we have, we get that $X_n = X_{n-1} + X_{n-2}$

We know that $X_1 = 2$ because we have $0$ and $1$ .

We know that $X_2 = 4$ because we have $00$ , $01$ , $10$ , $11$ .

To check our recursion, we have $X_3 = 6$ , because there are $2^3$ ways without restriction, then we subtract two for the extraneous strings which are $111$ and $000$ . Therefore we have $X_3=6$ .

We continue from here.

$X_1 = 2$ $X_2 = 4$ $X_3 = 6$ $X_4 = 10$ $X_5 = 16$ $X_6 = 26$ $X_7 = 42$ $X_8 = 68$ $X_9 = 110$ $X_{10} = 178$ .

Therefore we know that there are $178$ valid 10-strings. Now notice that for the $k^{th}$ slot for any string, the number of strings with one in the k slot is equal to the number of strings with zero in the k slot.

This is because we can simply switch 1 to 0 and 0 to 1 to form a one - one correspondence.

Therefore, for each slot, there are $178/2 = 89$ ones. (Zeroes do not matter because they sum to zero.)

Therefore, our answer is just $1111111111 \cdot 89 = \boxed{98888888879}$ .