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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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marked as duplicate by Micah, Arkamis, Phira, Chris Eagle, Gerry Myerson Nov 2 '12 at 11:55

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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. –  tomasz Nov 1 '12 at 17:02
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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 $n|(10^k-1)$? –  dinoboy Nov 1 '12 at 17:06
    
Related: math.stackexchange.com/q/4758 –  Jonas Meyer Nov 1 '12 at 17:09
    
    
@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. –  Bolt64 Nov 1 '12 at 17:35

1 Answer 1

Consider the fraction $\frac{1}{n}$. Since $n$ is coprime to $10$, the fraction is a purely repeating fraction with period $k$. Denote the repeating block by $r$. Then taking $9$ copies of $r$ appended with itself (i.e. $x=\underbrace{rr\cdots r}_{9\ \text{times}}$) we have that $x$ is necessarily divisible by $9$. Therefore $$\frac{1}{n} = 0.x\overline{x} \implies \frac{10^{9k}}{n} - x = \frac{1}{n}$$ This then gives $$10^{9k} - 1 = xn \implies \frac{10^{9k} - 1}{9} = \frac{x}{9}n$$

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Isn't this argument a bit circular? The only proof of repeating decimals I know uses the question basically. –  dinoboy Nov 1 '12 at 18:47
    
@dinoboy The proof of repeating decimals that I know is based on the order of $10$ modulo $n$. What is the proof you are thinking of? –  EuYu Nov 1 '12 at 18:50
    
Well, this problem is almost equivalent to showing 10 has an order modulo n. So that's why I say that this is slightly circular. –  dinoboy Nov 1 '12 at 19:01
    
@dinoboy I'm not sure I follow. The fact that $10$ has an order modulo $n$ comes simply from the fact that $n$ and $10$ are coprime. I've never even heard of repunits being used to prove that order exists. –  EuYu Nov 1 '12 at 19:06
    
Well yes. But note that proving a repunit exists that satisfies the conditions is basically equivalent to showing 10 has an order n. Both of them instantly imply each other. Basically what I'm saying is a "non-circular" proof of this problem requires showing your statement that a has an order modulo b when a,b are coprime. –  dinoboy Nov 1 '12 at 19:12

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