It's all about 9

First note that any multiple of an even number will be even and thus cannot end in $9$ and that any multiple of a multiple of $5$ will end in $5$ or $0$ and cannot end in $9$ either.

Now take any number $N$ such that $(N,10)=1$ and consider the numbers $\{10^0,10^1,\dots 10^{N-1}\}$ . Since $(N,10)=1$ then clearly none of these numbers is a multiple of $N$ (For $N>1$ but this case is trivial).

Since there are $N-1$ possible non-zero remainders mod $N$ (namely $1,2,\dots, N-1)$ and we have $N$ numbers, then two of the numbers must leave the same remainder mod $N$ . Say $10^i \equiv 10^j$ mod $N$ , for $j > i$ .

Then $N \mid 10^j-10^i \implies N \mid 10^i\cdot(10^{j-i}-1)$

But since $(N,10)=1$ then $N$ has no common factor with $10^i$ . Hence $N \mid 10^{j-i}-1$ , a number of the form $999\dots9$

So any $N$ such that $(N,10)=1$ has a multiple whose digits are all $9$ (we have also found that the smallest such multiple has at most $N-1$ digits).

Finally there are $100-\frac{100}{2}-\frac{100}{5}+\frac{100}{10}=40$ such integers between $1$ and $100$ .

We can write 999......999 as $10^n-1$ .Now,for numbers coprime to 10 we can simply take n to be $\phi (of that number)$ .

Hence,we only need to exclude multiples of 2 or 5.