jmc

If $m$ is any integer, then $\gcd (m,100)$ is one of the positive divisors of $100$: $$1,2,4,5,10,20,25,50,100.$$We note that the numbers on this list with more than one digit are all multiples of $10$, with the exception of $25$. Thus, $m$ has a single-digit $\gcd$ with $100$ if and only if $m$ is not a multiple of either $10$ or $25$. Thus, we just need to count integers $m$ between $0$ and $100$ which are not multiples of $10$ or of $25$.

There are $99$ integers $m$ such that $0<m<100$. These include nine multiples of $10$ ($10,20,30,\dots ,80,90$) and two more multiples of $25$ ($25$ and $75$; we don't count $50$ because we already counted it). So, that leaves $99-9-2=\boxed{88}$ integers whose greatest common divisor with $100$ has a single digit.