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Let's call a positive integer N self-indulgent of degree K>2 if for every positive integer k<K the following is true:

More than half of the first k multiples N,2N,...,kN of N contain with multiplicity all the digits of N. So, if, for example, the digit 4 occurs three times in the decimal representation of N, then it has to occur three or more times in that of the multiple.

Question:

Do self-indulgent numbers exist? If yes, can you give one of degree at least 10?

Attribution: Mine I think but wouldn't be too surprised if to coin a phrase I independently rediscovered it.

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The answer is

that self-indulgent numbers of arbitrary degree exist

because

if we pick a prime p such that the decimal expansion of 1/p has a repeating period of "full" length p-1, then the integer N formed by that repeating period has the property that N, 2N, ..., (p-1)N all use the exact same digits. (They are cyclic permutations of one another.) And there are infinitely many primes with this property.

For instance,

if we take p=19 we get the number N=52631578947368421 (it "really" has a leading zero but this does us no harm) and the numbers 2N,...,18N all have the same set of digits as N, plus one zero.

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  • $\begingroup$ Wow, I hate to admit it but I wasn't aware that there is such a smart answer to my humble question! $\endgroup$ – loopy walt Apr 6 at 1:21
  • $\begingroup$ Does that mean you have a simpler lower-brow solution? If so, and if it goes beyond "here is one example", I'd be interested to see it. $\endgroup$ – Gareth McCaughan Apr 6 at 14:17
  • $\begingroup$ I'm not sure it goes much beyond "here are two examples" but I'll show you anyway: 1234567890 (works for 1,2,4,5,7,8,..,22,23,25,26) and to a lesser extent 9876543210 (works for 1,2,4,5,7,8,10). I felt these are guessable enough to make an acceptable puzzle (for example, multiplying by 2 creates (viewed digit-by-digit) two copies of each even digit and 5 carries, just the right number to recreate the 5 odd digits, so trying permutations of 0123456789 is sort of educated guessing). I'd love to hear if you can see any deeper reason, also for the rather intriguing hit-hit-miss pattern. $\endgroup$ – loopy walt Apr 6 at 17:12
  • $\begingroup$ Oh, those are rather cute. Beyond the observation that 1/81 = 0.012345679 recurring I don't see anything that looks like a fancy mathematical reason why they work, but I could well be missing something. $\endgroup$ – Gareth McCaughan Apr 6 at 22:34
  • $\begingroup$ Actually, this 1/81 thing is good enough for me! Goes quite a bit of way to make it all plausible. For example, for k = 8 we get an almost clean argument: 8/81 = 1/9-1/81 = 0.1111111 - 0.012345, thus each digit d is replaced with 9-d. But how did you find it? WIth the benefit of hindsight, yes, of course, (1/9)^2 = 0.1111111^2 = 0.012345 etc. but it would have taken me forever to think of that myself. $\endgroup$ – loopy walt Apr 7 at 0:10

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