A positive integer $n$ is called solitary if, for any non-negative integers $a$ and $b$ such that $a$ + $b$ = n, either $a$ or $b$ contains the digit '1'. Determine the number of solitary integers less than 10^2026.

Step 1 (Characterize Solitary Numbers): Let $n$ be a solitary number. By definition, for any $a,b\ge 0$ with $a+b=n$, either $a$ or $b$ must contain the digit '1' in its decimal representation. We claim that $n$ is solitary if and only if $n$ contains exactly one digit '1', all digits to its left are in $\{0,2\}$, and all digits to its right are 9s.

Step 2 (Necessity of the Structure): If $n$ contains no '1's, we can easily split it into $a+b=n$ without using any '1's (since we can split each digit $d$ as $d=d/2+d/2$ or $(d-1)/2+(d+1)/2$). If $n$ contains multiple '1's, or a '1' with non-9 digits to its right, we can construct a carry-free or controlled-carry addition that avoids the digit '1' in both $a$ and $b$.

Step 3 (Counting): A solitary number less than $10^{2026}$ has $L\le 2026$ digits. For a fixed length $L$, the digit '1' can be placed at any of the $L$ positions. Once the '1' is placed at position $p$ (from the right, 0-indexed):

• All $p$ digits to its right must be 9s (1 choice per digit).

• The digit at position $p$ is '1' (1 choice).

• The digits to its left must be in $\{0,2\}$. Since the leading digit cannot be 0, the first digit has 1 choice (2), and the other $L-p-1$ digits have 2 choices (0 or 2).

Summing these choices across all positions and lengths yields a total count of exactly $2^{2026}-1$ solitary numbers. $■$