Prove that for any fixed string of digits $S$ (each digit from $0$ to $9$), there are at most $o(N)$ integers in $\{1,2,\ldots,N\}$ such that their decimal representation does not contain $S$.
This is easy to show if $S$ consists of a single digit (roughly, from $10^n$ available integers with at most $n$ digits there are at most $9^n$ which do not contain the particular digit). Is there a similar argument if $S$ consists of several digits? Considering each block of $|S|$ digits gives a bad bound if we ignore intersections, but how to deal with intersections then?
Any help appreciated!
Suppose $S$ has length $k$ and call an integer bad if its representation does not contain $S$. An integer $N$ has at most $\left \lceil \log_b N\right \rceil + 1$ digits in base $b$, so for each of $1$, $2$, $\ldots$, $N$ we may consider only the last $k\left \lceil \frac{\log_b N + 1}{k}\right \rceil$ digits in the representation (include some leading zeroes if necessary). Divide these digits into groups of $k$. Then for every bad integer none of these groups coincides with $S$. Thus for each group there are at most $b^k - 1$ possibilities (for each digit there are $b$ possibilities), hence the amount of bad integers is at most
$$(b^k-1)^{\left \lceil \frac{\log_b N + 1}{k}\right \rceil} \leq C(b^k-1)^{\frac{\log_b N}{k}} $$ where $C$ is a constant, depending only on the fixed $b$ and $k$. To argue this is $o(N)$ just note that $k$ is fixed and $$ \frac{(b^k-1)^{\log_b N}}{N^k} = \frac{(b^k-1)^{\log_b N}}{{(b^k)^{\log_b N}}} = \left(1-\frac{1}{b^k}\right)^{\log_b N} \to 0 $$ as $N\to \infty$.