Interesting Recreational Sequence

[Hans Havermann]

I wrote:

> I've just added more terms and made a comment-correction to Amarnath 
> Murthy's A078249 (Smallest multiple of n using all the digits other 
> than used by n, or 0 if no such number exists.). The first 125 terms 
> are:

<snip>

> The reason I'm bringing this sequence to the list is because i feel 
> strongly that there ought to be more qualifiers, above and beyond 
> those already stated, that will enable an n-th term not to exist. I'm 
> approaching this by brute force, trying to find the smallest index-n 
> for which I cannot find a(n) ...

The current qualifiers are: "a(n) = 0 if n==0 (mod 10), n ends in 5 and 
uses 0, n uses all even digits, or n uses all 9 non-zero digits."

It hasn't been as speedy as I'd imagined it would be, but here's what 
I've got: As a preliminary, I tried the first million multiples of all 
numbers less than a million and ended up with 1143 unresolved cases. I 
whittled these down to 1054 unresolved cases wherein the index contains 
the digits 2, 4, 6, and 8, all but one of them even. I resolved 186 of 
these before I ran into 282624. After a couple of days on this number 
alone, I decided to look a little closer...

Now, the multiple has to be a multiple of 5 in order to force a final 0.

     5 ->     ...20
    10 ->     ...40
    15 ->     ...60
    20 ->     ...80
    25 ->     ...00, and the cycle repeats. So the multiple has to be a 
multiple of 25 to force a final 00.

    25 ->    ...600
    50 ->    ...200
    75 ->    ...800
   100 ->    ...400
   125 ->    ...000, and the cycle repeats. So the multiple has to be a 
multiple of 125 to force a final 000.

   125 ->   ...8000
   250 ->   ...6000
   375 ->   ...4000
   500 ->   ...2000
   625 ->   ...0000, and the cycle repeats. So the multiple has to be a 
multiple of 625 to force a final 0000.

   625 ->  ...40000
  1250 ->  ...80000
  1875 ->  ...20000
  2500 ->  ...60000
  3125 ->  ...00000, and the cycle repeats. So the multiple has to be a 
multiple of 3125 to force a final 00000.

  3125 -> ...200000
  6250 -> ...400000
  9375 -> ...600000
12500 -> ...800000
15625 -> ...000000, and the cycle repeats.

And at this point, because the digits to the left of the zeros repeat 
as they did for multiples of 5, I feel I can conjecture that there will 
never be an odd digit to the left of the zeros. So, would somebody 
verify that the 282624th term of this sequence is 0 (and thus 
represents the first member of a family that requires a new qualifier)?