Divisors of One More than Power of 10/Number of Zero Digits Congruent to 2 Modulo 3

Theorem
Let $N$ be a natural number of the form:
 * $N = 1000 \ldots 01$

where the number of zero digits between the two $1$ digits is of the form $3 k - 1$.

Then $N$ has divisors:
 * $1 \underbrace {00 \ldots 0}_{\text {$k - 1$ $0$'s} } 1$
 * where the number of zero digits between the two $1$ digits is $k - 1$
 * $\underbrace {99 \ldots 9}_{\text {$k$ $9$'s} } \underbrace {00 \ldots 0}_{\text {$k - 1$ $0$'s} }1$

Proof
By definition, $N$ can be expressed as:
 * $N = 10^{3 k} + 1$

Let $a := 10^k$.

Then we have:

where it is noted that:

Hence the result.

Also see

 * Henry Ernest Dudeney: Modern Puzzles: $62$ -- Factorizing