Divisors of One More than Power of 10

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Theorem

Prime Factors

\(\ds 11\) \(=\) \(\ds 11\)
\(\ds 101\) \(=\) \(\ds 101\)
\(\ds 1001\) \(=\) \(\ds 7 \times 11 \times 13\)
\(\ds 10 \, 001\) \(=\) \(\ds 73 \times 137\)
\(\ds 100 \, 001\) \(=\) \(\ds 11 \times 9091\)
\(\ds 1 \, 000 \, 001\) \(=\) \(\ds 101 \times 9901\)
\(\ds 10 \, 000 \, 001\) \(=\) \(\ds 11 \times 909 \, 091\)
\(\ds 100 \, 000 \, 001\) \(=\) \(\ds 17 \times 5 \, 882 \, 353\)
\(\ds 1 \, 000 \, 000 \, 001\) \(=\) \(\ds 7 \times 11 \times 13 \times 19 \times 52 \, 579\)
\(\ds 10 \, 000 \, 000 \, 001\) \(=\) \(\ds 101 \times 3541 \times 27961\)
\(\ds 100 \, 000 \, 000 \, 001\) \(=\) \(\ds 11^2 \times 23 \times 4093 \times 8779\)
\(\ds 1 \, 000 \, 000 \, 000 \, 001\) \(=\) \(\ds 73 \times 137 \times 99 \, 990 \, 001\)


Number of Zero Digits Even

Let $N$ be a natural number of the form:

$N = 1 \underbrace {000 \ldots 0}_{\text {$2 k$ $0$'s} } 1$

that is, where the number of zero digits between the two $1$ digits is even.

Then $N$ can be expressed as:

$N = 11 \times \underbrace {9090 \ldots 90}_{\text {$k - 1$ $90$'s} } 91$


Number of Zero Digits Congruent to 2 Modulo 3

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$
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