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This question already has an answer here:

What is the largest number such that the number formed by the first $n$ digits is divisible by $n$?

For example, if we have a number $$abcdefghijklm,$$ and all of these leters stand for digits, then $a$ is divisible by $1$, $ab$ is divisible by $2$, $abc$ is divisible by $3$, and so on. Also, what is allowed is (besides $a$) the digits can be $0$ and digits can repeat.

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marked as duplicate by Zander, Start wearing purple, Lord_Farin, Mark Bennet, Davide Giraudo Jun 8 '13 at 12:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Cool question.. $\endgroup$ – Lord Soth May 22 '13 at 22:04
  • $\begingroup$ If you're gready, it feels like you get to $98765\ldots$? $\endgroup$ – Jeppe Stig Nielsen May 22 '13 at 22:11
  • $\begingroup$ As a side note, numberphile just did a video where they mentioned a (the only) pan-digital polydivisible number, $3,\!816,\!547,\!290$. $\endgroup$ – Arthur May 22 '13 at 22:26
  • $\begingroup$ Then, the earth will be destroyed the day 3/8/1654729 !! $\endgroup$ – Lord Soth May 22 '13 at 22:33
  • $\begingroup$ I agree it's a duplicate, but the title is so much better... $\endgroup$ – Erick Wong Jun 8 '13 at 12:21
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By copying and pasting the question, it turns out that these numbers are called polydivisible numbers, and there are finitely many of them, with the largest being:

$3608528850368400786036725$

Source: http://en.wikipedia.org/wiki/Polydivisible_number

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