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.