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A "k-persistent" number is a number that when multiplied by any positive integer up to k still contains every digit from 0-10 (in any order with possible repeats). Show that no "infinitely persistent" numbers exist. (i.e, it will always have the 10 digits when multiplied by ANY integer.)

I got this question from a friend and I simply don't know where to start. Any help would be very appreciated, thanks c:

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Consider the numbers $a_i=\frac{10^i-1}9$, so $a_1=1, a_2=11, a_3=111, \ldots$. For any positive $n$, there must be 2 numbers among $\{a_1, a_2, \ldots, a_{n+1}\}$ that leave the same remainder when divided by $n$, so their difference is a multiple of $n$. But the the difference of any two $a_i,a_j$ consitsts of $1$'s and $0$'s only, no other digits.

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  • $\begingroup$ wait but how does this prove that there can't be an infinitely persistent number $\endgroup$ – threehundred Oct 23 '18 at 1:05
  • $\begingroup$ Because we have found a number $x$ that is multiple of $n$ (which can be any positive number) that only contains $1's$ and $0's$. Since it is a multiple of $n$, there is a factor $f$ such that $x=fn$. Unless I misunderstood your definition, this means $n$ cannot be infinitely persistent. $\endgroup$ – Ingix Oct 23 '18 at 6:56
  • $\begingroup$ Ah makes sense. Thanks! $\endgroup$ – threehundred Oct 23 '18 at 9:48

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