jmc

For the greatest common divisor of 15 and $n$ to be equal to 3, $n$ must be divisible by 3 but not divisible by 5. In other words, $n$ is divisible by 3, but not by 15.

The greatest multiple of 3 that is less than or equal to 100 is 99, so there are $99/3=33$ multiples of 3 from 1 to 100. We must subtract from this the number of multiples of 15 from 1 to 100.

The greatest multiple of 15 that is less than or equal to 100 is 90, so there are $90/15=6$ multiples of 15 from 1 to 100. Therefore, there are $33-6=\boxed{27}$ numbers from 1 to 100 that are multiples of 3, but not 15.