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Find the largest 7-digit number without its digits repeating, such that the number is divisible by all of its digits

I found some constraints: • 0,4 and 5 cannot be included • The number found should be even • It is enough to check whether the number formed with the digits I excluded be divisible by 7 and 8.

I am not able to proceed any further.

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  • $\begingroup$ Why must it be even? $\endgroup$ – greenturtle3141 Sep 29 '16 at 14:44
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    $\begingroup$ @greenturtle3141 Because if you're using 7 distinct digits, you need to include at least one even digit. This digit must be a divisor of the number as a whole, implying the number itself is even. (Odd numbers don't have even divisors.) $\endgroup$ – Ian MacDonald Sep 29 '16 at 15:08
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    $\begingroup$ @Marius Those aren't the same question at all... $\endgroup$ – Sconibulus Sep 29 '16 at 15:14
  • $\begingroup$ @Sconibulus The linked question is a mathematical formulation of this one. $\endgroup$ – Aza Sep 29 '16 at 20:43
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    $\begingroup$ @Sconibulus Oh, whoops, you're totally right. $\endgroup$ – Aza Sep 29 '16 at 21:07
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The answer is

9867312

You can check the divisibility rules

Even- divisible by 2, sum of digits=36, divisible by 3 and 9. Divisibility with 2 and 3 shows that it is divisible by 6. Last 3 digits show divisible by 8. You can simply check the divisibility by 7 by simply dividing. P.S. Here's the link for the answer https://math.stackexchange.com/questions/1362741/numbers-divisible-by-all-of-their-digits-why-dont-4s-show-up-in-6-or-7-digi

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    $\begingroup$ ^vote with an note: Perfect cross reference but no mention of why this number is the largest. $\endgroup$ – humn Sep 29 '16 at 17:36
  • $\begingroup$ @humn Well, that I wasn't too sure about, myself. I supposed that someone would probably write a code for it. Although, I could somewhat defend it by pointing that it is probably close to the largest 7 digit number.. $\endgroup$ – Sid Sep 29 '16 at 17:40
  • $\begingroup$ @humm, LCM of 6, 7, 8, 9 is 504, so there can't be a 7 digit number that fits the criteria better. $\endgroup$ – TheChetan Sep 29 '16 at 17:58
  • $\begingroup$ @TheChetan So, how does that prove anything? A bit further explanations for me at least, so that I can edit in your explanation.. $\endgroup$ – Sid Sep 29 '16 at 18:00
  • $\begingroup$ @humn, the link to math.stackexhange proves that the number should contain 6, 7, 8 and 9. $\endgroup$ – TheChetan Sep 29 '16 at 18:03
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Sid already told us what the answer is, but let's see if we can prove it without a computer search. Knowing what answer we're looking for will help to guide our choice of things to prove, but I won't exploit this too much.

So, let's see whether our number can begin 987. If so, it looks like 987xxxx and must be a multiple of 9x8x7=504.

It can't contain a 0 because the only thing divisible by 0 is 0. It can't contain a 5 because it's even, and being divisible by 0 and 5 means being divisible by 10 which means ending in 0, which we just saw was impossible.

So our remaining four digits are four out of {1,2,3,4,6}. The sum of all the digits is going to have to be a multiple of 9, so the sum of these four has to be 3 mod 9, whereas the sum of all five is 7 mod 9, so the missing one needs to be 4. So now we need 987xxxx where the xxxx are 1236 in some order; and the number (hence the xxxx alone) needs to be a multiple of 7 and of 8. From the fact that it's a multiple of 4 we deduce that it ends [odd][even], and then the multiples of 8 are 3216 2136 6312 1632. None of these is a multiple of 7.

So, our number cannot begin 987. Let's next try 986. The best possible outcome would be for the number to begin 9867; can it?

Well, the same reasoning from a couple of paragraphs ago indicates that the remaining digits must then be 123 in some order. Again, they must form a multiple of 8. Even says xx2, and of these only 312 is a multiple of 8. So if the number begins 986 then the biggest it could possibly be is 9867312. And we can readily check that this is a multiple of all its digits, and then we're done.

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