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Possible Duplicate:
Proof that a natural number multiplied by some integer results in a number with only one and zero as digits
Why (directly!) does every number divide 9, 99, 999, … or 10, 100, 1000, …, or their product?

Let $n$ be a natural number co-prime with 10, and $m$, another natural number consisting entirely of $1$'s. How do you show that that for every $n$, there exists an $m$ such that $n$ divides $m$?

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    You should state your question as a question, not an order. You should also state what you have tried and what exactly is giving you trouble. Especially since this is most likely a homework.2012-11-01
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    Hint: A repunit is of the form $\frac{10^k-1}{10-1}$. First do the case of $\gcd(n,9) = 1$. How can you guarantee a $k$ exists such that $n|(10^k-1)$?2012-11-01
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    Related: http://math.stackexchange.com/q/4758/2012-11-01
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    Related: http://math.stackexchange.com/questions/165160/all-odd-primes-except-5-divide-a-number-made-up-of-all-1s2012-11-01
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    @tomasz I proceeded just like dinoboy said, i.e, wrote $9n+1=10^k$. Obviously this is easy to prove for the multiples of $3$ but what about the others. I'd really appreciate help. This problem's been bugging me for a while now.2012-11-01
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    Also related: http://math.stackexchange.com/questions/204645/why-directly-does-every-number-divide-9-99-999-or-10-100-1000/204653#2046532012-11-01
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    yet another: http://math.stackexchange.com/questions/83932/proof-that-a-natural-number-multiplied-by-some-integer-results-in-a-number-with/83936#839362012-11-01

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