# Find the greatest integer $N$ such that no two of its digits are equal and each digit is also its factor

$N$ 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 $\overline{abcd712}$.

• What have you tried? Can you determine if any digits cannot appear in certain positions? Dec 13, 2014 at 16:57
• Zero can not be in the last position, and digit 5 can not appear in the number.
– Epa
Dec 14, 2014 at 1:46
• 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 $\overline{abcd712}$.
– Epa
Dec 14, 2014 at 13:34
• 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. Dec 18, 2014 at 17:34
• $0$ cannot appear anywhere in the number since $0$ is only a factor of $0$. Nov 15, 2015 at 16:31

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.