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Prove that for every integer $n$, divisible by neither $2$ nor $5$, there exists an integer $m$, whose decimal representation consists only of $1$s, which is divisible by $n$ (e.g. for $n=13$, we have $m=111111$ divisible by $13$).

Sorry if the format is not up to standard.

Chris Whitehead
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RyanY
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  • I presume that by "not made up of 1s" you mean "not 1 or 11 or 111 or 1111, etc." Is that correct? – user247327 Sep 03 '19 at 00:29
  • Hint: Do you remember how when you were in elementary school you were taught how to turn an infinitely repeating decimal into a fraction? Did you every notice those fractions only had $9$s in the denominator? – fleablood Sep 03 '19 at 00:29
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    Consider this. $\frac 17 = 0.14285714285714285714285714285714...$ so $1000000\frac 17 = 142857.14285714285714285714285714285714...$ so $999999\frac 17 = 142857$. And so $17142857 = 999999= 1111119$. And so $7\frac {142857}9 =715873=111111$. Can you generalize that? – fleablood Sep 03 '19 at 00:40

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