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Proof that a natural number multiplied by some integer results in a number with only one and zero as digits

The question is as stated. I'm really stumped on it. It seems intuitively true, but I don't really know what direction to look in. I've calculated some examples for some values, but there doesn't seem to be any particular rhyme or reason to it that would suggest a constructive proof. I've thought about the possible remainders of $9\cdot\cdot\cdot9/n$, and wondered if I could show that the remainder must be $0$ for some $n$, but that hasn't taken me far. I've been poking and prodding at it from several directions, but I've gotten nowhere really. This is a real brain-teaser. Can anyone give me some helpful hints?

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    $\begin{align} 99 & = 3\cdot3\cdot11 \\ 999 & = 3\cdot3\cdot3\cdot37 \\ 9999 & = 3\cdot3\cdot101 \\ 99999 & = 3\cdot3\cdot41\cdot271 \\ 999999 & 3\cdot3\cdot7\cdot11\cdot13\cdot37 \\ 9999999 & = 3\cdot3\cdot11\cdot73\cdot101\cdot137 \end{align}$ The claim seems to be that every positive integer whose last digit is not $0,2,4,5,6$, or $8$ will appear in one of these factorizations. (Not as one of the prime factors, but as the product of some of them.)2012-11-04
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    If I can believe the software that gave me these results, then $9999999999$ is divisible by $9091$, and $999999999999$ is divisible by $9901$, and $99999999999999$ is divisible by $909091$. I'm too lazy right now to explain what pattern I'm looking at, let alone find the obvious reason why it happens. But probably there is one.2012-11-04

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