Digit goes to Recycle bin

We get $2^n - d \cdot 10^k = 2^m$ where $d$ is the first decimal digit of $2^n$ , and $k+1$ is the number of digits of $2^n$ . Rearranging, we get $2^m(2^{n-m}-1) = d \cdot 10^k$ So we can divide through by the $2^m$ on both sides to get $2^{n-m}-1 = d' \cdot 5^k$ where $d'$ is odd (since the left side is odd, dividing through by $2^m$ must strip out all the factors of $2$ on the right side).

Now the key observation is that $2^m$ has only one fewer digit than $2^n$ , so $2^{n-m} \le 100$ , so $1 \le n-m \le 6$ . Stepping through the cases gives $n-m=4, k = 1, d' = 3$ as the only solution, so $2^n$ has two digits, and so the only solutions are $2^n = 32,64$ . The answer is $5+6 = \fbox{11}$ .