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,\ldots,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.