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.

This is only a partial solution. I thought I could extend it to a full solution, but that failed. The argument may be instructive, and could get somebody else on the track, so I will leave it here.
We cover the case where the arithmetic progression starts with 0,
so our sequence is $0,x,2x,\ldots$.
Assume that $x$ has $k$ decimal digits.
If we are counting in multiples of $x$ there is no way thay we can skip the
entire range from $[9\cdot 10^{k+1},10^{k+2})$, but if we want an arithmetic
progression of more then 10 terms avoiding 9, then we must clearly skip the
entire range $[9\cdot 10^k,10^{k+1})$.
This gives us the requirement that we need to find a one-digit number $b$
such that $bx\geq10^{k+1}$ and $(b-1)x<9\cdot 10^k$.
We will show that $b=9$ cannot work and that $b=8$ gives us exactly what we want.
Smaller values are not interesting: e.g. if $b=7$ this would imply that
$60x<90\cdot 10^k$ and $70x\geq 10^{k+2}$, so we get smaller limits.
When $b=9$ we get $9x\geq10^{k+1}$ and $8x<9\cdot 10^k$.
$9x\geq10^{k+1}$ implies $8x\geq\frac{80}{9}\cdot 10^k$,
so $\frac{80}{9}\cdot 10^k\leq 8x<9\cdot 10^k$.
This implies that already $8x$ contains a 9 (write down the decimal expansion of $\frac{80}{9}$ and realize that it is not an integer!), which is not what we are looking for.
When $b=8$ we get $8x\geq10^{k+1}$ and $7x<9\cdot 10^k$.
$8x\geq10^{k+1}$ implies that $x\geq\frac54\cdot 10^k$.
But then $72x\geq 90\cdot10^k$, so our sequence can have at most 72 terms
(note that we started with 0). The case $x=125$ shows that this result is sharp.
