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The question is as stated in the title. I feel like the question is geared towards some kind of case that yields to the divisibility test for $9$, but I think an argument can be made for other numbers as well. For example in the sequence $100$ ,...,$117$ we have of course $3 \mid 102$ and $9 \mid 117$, but we also have $2 \mid 110$ and $4 \mid 112$. And secondly in sequences where the digit sum is above $9$, we still have divisors that are multiples of $9$. For example $18 \mid 990$.

So if anyone could help me with a proof, I'd be really grateful. Thank you for your help.

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  • $\begingroup$ If all else fails, compile a list those of the 3-digit numbers that satisfy the property, and check that no gap is larger than 17 :-) $\endgroup$ – Henning Makholm Aug 1 '16 at 8:40
  • $\begingroup$ @HenningMakholm Honestly, that is probably also the easiest way to do it :) $\endgroup$ – 5xum Aug 1 '16 at 8:41
  • $\begingroup$ @HenningMakholm Only if all else fails though :P $\endgroup$ – Airdish Aug 1 '16 at 8:45
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    $\begingroup$ Tis easily remedied, however, by appending, "then generalize to base $b$" to the problem. $\endgroup$ – Henning Makholm Aug 1 '16 at 8:45
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    $\begingroup$ Interestingly the value of 18 is actually the smallest possible. There are no numbers with that required property in the 17 long ranges: $[559,575]$, $[667,683]$, $[739,755]$,$[847,863]$, $[937,953]$, $[973,989]$. $\endgroup$ – Ian Miller Aug 1 '16 at 8:53
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We can use your idea: there is at least one multiple of $9$ in this sequence, so the sum of its digits will be $9$, $18$ or $27$. Let's separate some cases:

  1. The sum is $27$: in this case, the number is $999$, so $990$ is also in the sequence and satisfies the properties.

  2. The sum is $9$, and there is nothing to do.

  3. The sum is $18$ and the number is even, and there is nothing to do again.

  4. The sum is $18$ and the number is odd: in this case, let $x$ be this multiple of $9$. Then either $x+9$ or $x-9$ will be in the sequence, and it will fall in case 2. or 3. above. Anyway we are done.

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  • $\begingroup$ Perhaps it's clearer to say "every multiple of 18 has this property", so you need to deal with only case 2 and 3 (and 1 just to show that it's impossible). $\endgroup$ – Henning Makholm Aug 1 '16 at 9:02
  • $\begingroup$ $999=27\times37$. $\endgroup$ – Gerry Myerson Aug 1 '16 at 9:23
  • $\begingroup$ @GerryMyerson I hadn't noticed that. Thanks $\endgroup$ – Luiz Cordeiro Aug 1 '16 at 22:53

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