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This question already has an answer here:

Prove that $$\underbrace{11111\ldots1}_{91\text{ times}}$$ is a composite number and not a prime. Please give full steps of proving.

I tried and found that it is divisible by $1111111$ and $1111111111111$ but I can't prove it.

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marked as duplicate by Jyrki Lahtonen May 12 '17 at 11:02

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  • $\begingroup$ Showing it is divisible by 1111111 proves it, how did you figure that out?. $\endgroup$ – marshal craft May 12 '17 at 10:06
  • $\begingroup$ Multiply 53 × 79 × 239 × 547 × 4649 × 14197 × 17837 × 4 262077 × 265 371653 × 43442 141653 × 316877 365766 624209 × 110 742186 470530 054291 318013 $\endgroup$ – Henry May 12 '17 at 10:09
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Since $7|91$ we see that we can write $1111111$-($7\;$ $1$'s). Next to itself $13$ times and get the result of $111...$ (-$91$ times.)
This is equivalent to saying that $$1111111+(10^7\times 1111111) + (10^{14}\times 1111111) + ... + (10^{84}\times 1111111) = 1111... -91 \text{times}$$.

Which we can also see using a geometric series. Taking a factor of $1111111$ out of the LHS gives your result that $111111$ divides $1111...$(-$91$ times)

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  • $\begingroup$ I dont know this proves the result, perhaps it could be more clear. $\endgroup$ – marshal craft May 12 '17 at 11:44
  • $\begingroup$ @marshalcraft yeah the answer below goes in to more detail regarding the geometric series I mention; and is definitely better! $\endgroup$ – SEWillB May 12 '17 at 11:46
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$91=7\times13$. So the number is $\sum_{k=0}^{12}(1111111\times10^{7k})=1111111\sum_{k=0}^{12}10^{7k}$.

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    $\begingroup$ So $1111111 \times 100000010000001\ldots10000001$ $\endgroup$ – Henry May 12 '17 at 10:16

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