Binary counting

Suppose the first answer is $x$ , having $a = \lfloor \log_2 x \rfloor + 1$ digits, and the second answer is $y$ , having $b = \lfloor \log_2 y \rfloor + 1$ digits. Note that $x \ge 2^{a-1}, y \ge 2^{b-1}$ .

Note that $x+y$ counts all the digits. On the other hand, there are $2+a+b$ digits. Thus $2+a+b = x+y \ge 2^{a-1} + 2^{b-1}$ , or $(2^{a-1}-a) + (2^{b-1}-b) \le 2$ . But $2^{t-1} - t \ge 0$ for all natural number $t$ , so we have $2^{a-1}-a, 2^{b-1}-b \le 2$ .

By finding the derivative of $t \mapsto 2^{t-1}-t$ , we can see that it is increasing for $t \ge 2$ , and for $t = 4$ we have $2^{t-1} - t = 4 > 2$ , so $2^{t-1} - t > 2$ for all $t \ge 4$ . Thus $a,b \le 3$ , or $x,y \le 7$ . Also, $x,y \ge 1$ since they count at least the existing 0 and 1. There are just $49$ cases to search, which should be doable (if tiring) by hand.

The unique solution is $\boxed{(11_2, 100_2)}$ , where there are $3 = 11_2$ zeroes and $4 = 100_2$ ones.

binary 11 , 100

number of digis 3 , 4