13
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Take a number $(x)$, then create the complete list of the numbers formed by deleting single digits from its base ten representation $(d_1,d_2,...,d_n)$. If the sum of those new numbers equals $x$ we call the number a special number.

Example :

1729404 = 729404 (delete 1) + 129404 (delete 7) + 179404 + 172404 + 172904 + 172944 + 172940

13758846 = 3758846 + 1758846 + 1358846 + 1378846 + 1375846 + 1375846 + 1375886 + 1375884

(in the 2nd example we see 1375846 twice.)

Question :

  • Is the list of these kind of numbers infinite?
  • If not, find the largest number (without the aid of a computer) with this property, then prove it!
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  • $\begingroup$ Did you mean infinite on your first question? Or perhaps start the second question by "if so,"? $\endgroup$ – stack reader Apr 20 '17 at 9:57
  • $\begingroup$ @stackreader : thanks for correction $\endgroup$ – Jamal Senjaya Apr 20 '17 at 10:01
  • $\begingroup$ Are we allowed to use any base? $\endgroup$ – Jonathan Allan Apr 20 '17 at 10:05
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    $\begingroup$ @JonathanAllan : Use base 10 $\endgroup$ – Jamal Senjaya Apr 20 '17 at 10:13
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    $\begingroup$ Related OEIS sequence (spoiler alert) $\endgroup$ – Ankoganit Apr 20 '17 at 10:23
11
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If one considers numbers that are less than / greater than rather than equal to it becomes clear that the list must be finite. The sums of the list of numbers of length $12$ or more must be greater than the requirement even if all trailing digits are zeros ($11$ times a tenth of a number is greater than the original number, and the multiple only increases as we add more digits).

If the largest number does have $11$ digits the last ten digits must all be zero (to make a single $000000000=0$ in the sum).
Working backwards with $x$ as this digit:
$x\times 10,000,000,000 = 10\times x \times 1,000,000,000 + 0$
implies $x$ can be any digit, hence the largest possible is when $x=9$, which is:
$90,000,000,000$

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  • $\begingroup$ Cripes, I wrote a solution only to see it was pretty much the same as yours. $\endgroup$ – Nautilus Apr 20 '17 at 12:46
5
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simple proof that such numbers are finite.

11 = 1 + 1 = 2
111 = 11 + 11 + 11 = 33
1111 = 111 + 111 + 111 + 111 = 444
11111 = 5555
111111 = 66666
1111111 = 777777
11111111 = 8888888
111111111 = 99999999
1111111111 = 1111111110 (possible around this range of numbers)
11111111111 = 12222222221 (from here an higher will always have too big results).
The result is highly dependent on the number of digits. The number of digits that allow a correct answer is very limited and vary slightly depending on the first digit.

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  • $\begingroup$ I like the inductive layout :) $\endgroup$ – Jonathan Allan Apr 20 '17 at 10:35
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    $\begingroup$ Just a note - you missed the all trailing zeros trick allowing 8 solutions higher than your upper bound. $\endgroup$ – Jonathan Allan Apr 20 '17 at 10:53

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