How many integers m are there such that 0<m<100 and gcd(m,100) is a single-digit number?

Answer 1

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?

Answer 2

I don't get what you are trying to say. Can you rephrase it?

Answer 3

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.

Answer 4

oops. I meant, there are 19 multiples of 5 (not multiples of 3, which is not a factor of 100)

Answer 5

Aren't all multiplies of 4 also a multiple of 2?

Answer 6

Ugh! This is going to get very messy. There must be a better way...

Answer 7

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...