A number $n$ is called a Lynch-Bell number if its digits are all distinct and $n$ is divisible by each digit (the sequence is presented HERE). It can be found in many places that the biggest Lynch-Bell number is 9867312 and it can be proved with computer easily.

Bet let's do the same by doing some elementary math. Obviously, digit 0 must be excluded. And digits 2 and 5 cannot be both included in $n$ because such $n$ would end in zero. And then... what?

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    $\begingroup$ That sounds like a task for a computer, why do you want to do it without? $\endgroup$ – Henrik May 4 '18 at 7:50
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    $\begingroup$ @Henrik - maybe, as Mallory said, "because it's there" ? $\endgroup$ – gandalf61 May 4 '18 at 9:15
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    $\begingroup$ @Henrik - Because it would be like watching a forklift winning in a weightlifting competition. Thinking is healthy for our brains. $\endgroup$ – Oldboy May 4 '18 at 11:51


A fitting reply to people using the computer on this one.

No nine or ten digit Lynch Bell Numbers

To start with, we claim that there is no eight,nine or ten digit Lynch Bell number. Anything with $11$ or more digits must have two digits as the same by the pigeonhole principle.

Any ten digit Lynch Bell number contains zero as a digit, contradiction.

Consider a nine digit Lynch Bell number. It certaily must avoid zero, and therefore has the digits $123456789$ in some order. This contradicts the fact that $2 \times 5 = 10$ is a multiple of the number.

No eight digit Lynch Bell Numbers

Consider an eight digit Lynch bell number. If it has a zero, we have a contradiction. On the other hand, if it has a five, then it has at least one of $2$ or $4$ , since we can exclude only two digits, and $0$ is already out. But then it is a multiple of $10$ either way. In short, such a number must not contain zero or five. Therefore, its digits are precisely $12346789$.

Which is an issue, since the sum of the digits of this number is $40$, so by our usual divisibility test , it is not a multiple of $3$ (or $9$).

Hence, no eight digit Lynch bell number exists!

Seven digit Lynch Bell - What to Exclude?

Coming to seven digit Lynch Bell numbers, we know they have seven different digits. By our previous logic, $0$ must be excluded. Now, note that if $5$ is in the number, then at least one of $2,4$ or $6$ is in the number, since all these cannot be excluded. However, each of these gives that our number is a multiple of $10$ so we have a contradiction.

Hence, $0$ and $5$ are out. The question is of the third digit to be excluded.

At this stage, all we need to do is note that the divisibility test comes to our rescue again. Indeed, we have to exclude some number from $12346789$, whose sum of digits is $40$. Excluding $9$ gives rise to a number whose sum of digits is $31$ and which is not divisible by $3$, a contradiction. Hence, $9$ cannot be excluded.

But then, the number must be a multiple of $9$, since $9$ is one of the digits of the number! Hence, the sum of digits must be a multiple of $9$. Indeed, one sees that $40-\color{blue}{4} = 36$ and hence $4$ must be the excluded one, since if anything else(bar $9$) is excluded we won't get a multiple of $9$.

Where are we, and what now?

In conclusion, any seven digit Lynch Bell number must consist precisely of the digits $1236789$. Fantastic. Now, life is actually quite simple.

A logical point would be to start from the largest possible candidate for a Lynch Bell number, namely $9876321$.

Why the very large ones don't work

We now eliminate all numbers of the form $987....$ from being a Lynch Bell number.

Recall the criteria of divisibility by eight. Indeed, a number is divisible by $8$ if and only if its last three digits are. Therefore, what combinations of $1236$ give numbers divisible by $8$? Indeed, only $136,216,312,632$ give a positive result.

At this stage, we note that $987....$ is divisible by $7$ if and only if $....$ in that order is divisible by $7$. Therefore, all we need to do is check whether any of $2136,3216,6312,1632$ are divisible by $7$. Indeed, none of them are, and you don't even need to do the actual division in some cases. For example , $2136$ is not divisible by $7$ because $36$ is not, and $21$ is.

Therefore, it follows that $987....$ is not a Lynch Bell number for any combination of $....$


The numbers which come next in the descending order is $9867...$ Indeed, note that $...$ must end with $2$, since we have a multiple of $2$. We now see that $9867312$ is a Lynch Bell number, by seeing that it has distinct digits and is divisible by each of them.

We have already shown that all numbers larger than $9867312$ are not Lynch Bell numbers. This allows us to conclude that $9867312$ is the largest Lynch Bell number, with a handful of divisions by hand and no computer.

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    $\begingroup$ We are asked not to post comments like "thanks" or "+1", but I cannot resist. So thanks, I enjoyed every chapter of your proof. And +1, of course :) $\endgroup$ – Oldboy May 4 '18 at 12:07
  • $\begingroup$ You are welcome! Great question, kept me at bay for an hour or so. $\endgroup$ – астон вілла олоф мэллбэрг May 4 '18 at 15:21

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