How many integers m are there such that 0<m<100 and gcd(m,100) is a single-digit number?
Since the only 1-digit divisors of 100 are 2,4 and 5,
how many numbers contain a single power of 2,4 or 5, multiplied by combinations of other prime powers?
I don't get what you are trying to say. Can you rephrase it?
hmmm. that gets messy. How about this?
There are 24 multiples of 4 < 100
There are 49 multiples of 2, but we have already counted the 24 multiples of 4, so that leaves 25 multiples of 2 only
There are 33 multiples of 3, but we have already counted some of them as multiples of 2 and 4.
See what you can do with that.
oops. I meant, there are 19 multiples of 5 (not multiples of 3, which is not a factor of 100)
Aren't all multiplies of 4 also a multiple of 2?
Ugh! This is going to get very messy. There must be a better way...
GCD=2
2,6,14,18,22,26,34,...
all the multiples of 2 which are not also multiples of 4 or 5. Basically, all the odd multiples of 2 except 10,20,30,...
GCD=4
4,8,12,16,24,... (multiples of 4 which are not also multiples of 5)
GCD=5
5,10,15,20,... multiples of 5
There aren't that many...