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How can I prove that any number coprime with 10 (either prime or not) can divide at least one number of the form 111...111?

Maximilian
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    What number theory do you know? These numbers can be written as $$\frac{10^m-1}{9}$$ with $m$ being the number of $1$s. – Thomas Andrews Feb 16 '16 at 03:45
  • Welcome to Math.SE! It is useful when asking questions to provide some context: for example, what your thoughts are on the problem, what you have tried, where you are stuck, and things you know that might be relevant to the problem (i.e. a little of your background). This enables other users to give you better feedback and answers. – Ben Sheller Feb 16 '16 at 03:47
  • Thanks! I'm taking Discrete Math for CS and even though I tried working this exercise out, I couldn't even find a way to start. I found out that I can prove this but for prime numbers by using fermat's Little Theorem, but I can't figure out how to show that this works for non prime numbers (such as 9, 63 or 93) – Maximilian Feb 16 '16 at 06:00
  • Suppose it doesn't. Then consider the sequence $1,11,111,1111,.. \pmod n $ where $n$ is coprime to $10$. Can you apply pigeonhole here? – Rijul Saini Feb 16 '16 at 06:04
  • I see what you mean, but with pigeonhole as far as I understand I must be sure that all the possible values of n are associated with at least one of the 'holes' that at this case would be the members of the sucession you wrote, and that's what I actually want to prove, I mean, we don't know whether all the numbers will be able to get into a hole or not – Maximilian Feb 16 '16 at 07:55
  • This is an interesting question (with the exception that OP does not share any thoughts or efforts). Why voting to close it??? – barak manos Feb 16 '16 at 09:44

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Let me call $R(k)= (10^k-1)/9$ the number made of $k$ ones. So, for example, $R(3)=111$.

Now consider the set of $n+1$ numbers $R(1)$, $R(2)$, ..., $R(n+1)$ and take the remainder of these numbers modulo $n$.

Since the possible value of the remainder is between 0 and $n-1$ there will be at least two numbers, say $R(a)$ and $R(b)$, with $a>b$ whose remainder is equal modulo $n$, i.e., $R(a)\equiv R(b)\pmod n$. But this means that $$R(a)-R(b) = R(a-b)\cdot 10^b \equiv 0 \pmod n$$

So we have proved that $10^b\cdot R(a-b)$ is divisible by $n$. Since $n$ has no common factors with $10$, this implies that $n$ divides $R(a-b)$.

Giovanni Resta
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