# Show that no ∞-persistent numbers exist.

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:

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.
• 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. – Ingix Oct 23 '18 at 6:56