Repeated decimal, part 3B

Comparing their decimal expansions, we see $\frac{100}{N}-\frac{1}{N}=0.1111\ldots=\frac{1}{9}$ .

In other words, $\frac{99}{N}=\frac{1}{9}$ and so $N=\boxed{891}$ .

$0.\underbrace{\overline{001122334455667789}}_{\text{18 digits}} = \dfrac {1122334455667789}{\underbrace{999999999999999999}_{\text{18 digits}}} = \dfrac 1{891}$

Therefore, $N = \boxed{891}$

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Proof:

$\begin{aligned} 0.\underbrace{\overline{001122334455667789}}_{\text{18 digits}} & = 0.001122334455667789 \sum_{k=0}^\infty 10^{-18k} \\ & = 0.001122334455667789 \times \frac 1{1-10^{-18}} \\ & = 0.001122334455667789 \times \frac {10^{18}}{10^{18}-1} \\ & = \frac {1122334455667789}{\underbrace{999999999999999999}_{\text{18 digits}}} \end{aligned}$