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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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sayantankhan
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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 01 '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 01 '12 at 17:06
  • Related: http://math.stackexchange.com/q/4758/ – Jonas Meyer Nov 01 '12 at 17:09
  • Related: http://math.stackexchange.com/questions/165160/all-odd-primes-except-5-divide-a-number-made-up-of-all-1s – lab bhattacharjee Nov 01 '12 at 17:20
  • @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. – sayantankhan Nov 01 '12 at 17:35
  • Also related: http://math.stackexchange.com/questions/204645/why-directly-does-every-number-divide-9-99-999-or-10-100-1000/204653#204653 – Micah Nov 01 '12 at 18:13
  • yet another: http://math.stackexchange.com/questions/83932/proof-that-a-natural-number-multiplied-by-some-integer-results-in-a-number-with/83936#83936 – Chris Eagle Nov 01 '12 at 21:52

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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 $$10^{9k} - 1 = xn \implies \frac{10^{9k} - 1}{9} = \frac{x}{9}n$$

EuYu
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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 01 '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 01 '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 01 '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 01 '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 01 '12 at 19:12
  • @dinoboy The statements may be equivalent (I'm still not sure how "for every $n$, there exists a repunit $m$ such that $n$ divides $m$" implies $a$ has an order modulo $b$, I'd like to see the proof actually), but that does not mean they're circular. Nowhere in a proof that $(a,\ b)=1$ implies $a$ has order modulo $b$ is the statement in question assumed. It is standard for almost all classes in elementary number theory to prove these statements from the ground up. – EuYu Nov 01 '12 at 19:20
  • I've found a much more elegant solution in Arthur Engel's Problem Solving Strategies. It goes something like this: For any $n$ co-prime with 10, take $11$, $111$, $1111$, until $11...11$(n+1) digits. Take their remainders mod n. If any of them is $0$, we're done or else it lies between $1$ and $(n-1)$. By Pigeonhole Principle, two of the remainders must be same. Then the difference of of their corresponding repunits(of the form 111..1000..00) is a multiple of $n$. And since $n$ is co-prime with 10, removing the zeroes doesn't matter. Hence we get a repunit multiple of $n$. – sayantankhan Nov 03 '12 at 11:07