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show that if $N$ is relatively prime to 10, then there exists a multiple that consists only of 1s.

The multiple can be expressed as :

$$\frac{(10^a-1)}{9}$$

thus if $\gcd(N,3)$ is not 3, using the totient function it would be easy to find $a$ such that this is $0\bmod N$

If $\gcd(N,3)$ is 3, then we would need to find $10^a-1$ divisible by $3N$ which is also possible.

from https://artofproblemsolving.com/articles/files/SatoNT.pdf. It is said that:

"If N is relatively prime to 10, then we may divide out all powers of 10, to obtain an integer of the form 111...1 that remains divisible by N."

which I'm not sure I follow. I would want to know what this means since the answer seems concise

Bernard
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SuperMage1
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    what the quote means: using the pigeonhole principle, they find an integer multiple of $N$ of the form $111...000$, and then divide by a power of $10$ to get an integer multiple of $N$ consisting of only $1$s – J. W. Tanner Jan 27 '20 at 15:14
  • The arguments in the dupe work generally. If you have further questions please ask in comments there (or here). – Bill Dubuque Jan 27 '20 at 17:16

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