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$N$ is a positive integer such that no two of its digits are equal and each digit is also its factor. What is the largest value of $N$?

So far, I've determined that $0$ cannot be the last digit, and $5$ cannot appear in the number. By divisibility by 9, then $1+2+3+4+6+7+8+9=36$ would be divisible by $9$ if $N$ is a $7$-digit number. By divisibility by $8$, then the last three digits might be $312$ or $712$. The last digit can be, $2,4,6,8$. It is so hard to use the divisibility by $7$. So I guess $\overline{abcd312}$ or $\overline{abcd712}$.

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  • $\begingroup$ What have you tried? Can you determine if any digits cannot appear in certain positions? $\endgroup$ – Milo Brandt Dec 13 '14 at 16:57
  • $\begingroup$ Zero can not be in the last position, and digit 5 can not appear in the number. $\endgroup$ – Thumbolt Dec 14 '14 at 1:46
  • $\begingroup$ From then on, the digits left for us include $\{1,2,3,4,6,7,8,9\}$. By divisibility by 9, then $1+2+3+4+6+7+8+9=36$ would be divisible by 9 if $N$ is a 7-digit number. By divisibility by 8, then the last three digits might be $312$ or $712$. The last digit can be, ${2,4,6,8}$. It is so hard to use the divisibility by 7. So I guess $\overline{abcd312}$ or $\overline{abcd712}$. $\endgroup$ – Thumbolt Dec 14 '14 at 13:34
  • $\begingroup$ Why can't $5$ appear? It would have to be the ones digit, but $15$ is an acceptable number. It is not the greatest. $0$ cannot be a digit at all. $\endgroup$ – Ross Millikan Dec 18 '14 at 17:34
  • $\begingroup$ $0$ cannot appear anywhere in the number since $0$ is only a factor of $0$. $\endgroup$ – steven gregory Nov 15 '15 at 16:31
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These are the Lynch-Bell numbers, listed in OEIS A115569. The largest is $9867312$. I found the link by finding some small ones by hand and entering $9,12,15,24,36,48$ into the search box. This was the first that came up.

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