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

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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:

\(\ds N\) \(=\) \(\ds a^3 + 1\)
\(\ds \) \(=\) \(\ds \paren {a + 1} \paren {a^2 - a + 1}\) Sum of Two Cubes

where it is noted that:

\(\ds \underbrace {99 \ldots 9}_{\text {$k$ $9$'s} } \underbrace {00 \ldots 0}_{\text {$k - 1$ $0$'s} }1\) \(=\) \(\ds \underbrace {99 \ldots 9}_{\text {$k$ $9$'s} } \underbrace {00 \ldots 0}_{\text {$k$ $0$'s} } + 1\)
\(\ds \) \(=\) \(\ds 10^k \paren {\underbrace {99 \ldots 9}_{\text {$k$ $9$'s} } } + 1\)
\(\ds \) \(=\) \(\ds 10^k \paren {10^k - 1} + 1\)
\(\ds \) \(=\) \(\ds a^2 - a + 1\)

Hence the result.

$\blacksquare$


Examples

\(\ds 1001\) \(=\) \(\ds 11 \times 91\)
\(\ds 1 \, 000 \, 001\) \(=\) \(\ds 101 \times 9901\)
\(\ds 1 \, 000 \, 000 \, 001\) \(=\) \(\ds 1001 \times 999 \, 001\)
\(\ds \) \(=\) \(\ds 7 \times 11 \times 13 \times 19 \times 52 \, 579\)
\(\ds 1 \, 000 \, 000 \, 000 \, 001\) \(=\) \(\ds 10 \, 001 \times 99 \, 990 \, 001\)
\(\ds \) \(=\) \(\ds 73 \times 137 \times 99 \, 990 \, 001\)


Also see


Sources

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