An arithmetic sequence whose members do not contain the digit ‘9’ There is a non-constant arithmetic progression made of natural numbers only; none of them contains the digit $9$. Prove that such an arithmetic progression has no more than $72$ terms.

We have a sequence $a_n = b+ns$, where we call $b$ the "base element" and $s$ the step. We notice for first that if $s\equiv 1,3,7,9\pmod{10}$ all the ten residue classes $\pmod{10}$ are covered by $a_0,\ldots,a_9$, hence the sequence cannot avoid the digit $9$ for more than $9$ terms. For a similar reason, we cannot have $s\equiv 2,4,6,8\pmod{10}$, since otherwise we fall into one of the residue classes $90+[0,9]\pmod{100}$ before the $45$th term. If $s\equiv 0\pmod{10}$, hence $s=10s'$, the digits represented by the sequence, apart from the last one that is constant, are the same that the digits represented by the sequence $a_n' = \lfloor b/10\rfloor + ns'$. So we may assume without loss of generality that $s\equiv 5\pmod{10}$. If now we focus on the sequence $b_n=a_{2n}$, we can easily deduce that $s\equiv 25\pmod{100}$ by the same way as above. By focusing now on $c_n = a_{4n}$, it is easy to check that, in order to avoid the residue classes $900+[0,99]\pmod{1000}$ in the first $40$ terms, the only chances are $s\equiv 125,625\pmod{1000}$. Now it is just tedious, by focusing on $d_n=a_{8n}$, to check that even such sequences cannot avoid the residue classes $9000+[0,999]\pmod{10^4}$ for more than $72$ terms, hence we're done.
