There will be (200-100)/2+1 = 51 numbers divisible by 2.
There will be Floor[(200 - 100)/3] = 33 numbers divisible by 3.
Multiples of 6 from the 17th through the 33rd will be counted twice by the sum of these numbers (51+33). There are 17 of them.
That means the count of numbers you are interested in is 51+33-17 = 67.
There will be Floor[(200 - 100)/3] = 33 numbers divisible by 3.
Multiples of 6 from the 17th through the 33rd will be counted twice by the sum of these numbers (51+33). There are 17 of them.
That means the count of numbers you are interested in is 51+33-17 = 67.
