These two numbers: 135 and 180.
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.
The list of whole numbers that are divisible by XXX is infinite. The first four are: 100, 200, 300, 400 . . .
There are 5 multiples of 18 between 200 and 300. (An even number divisible by 9 is any number divisible by 18).
105 110 115 120 125 130 135 Etc.
12. 13, if you count 200
14 of them
These two numbers: 135 and 180.
17 of them.
14:105,112,119,126,133,140,147,154,161,168,175,182,189,196.
A number that is divisible by 4 and 6 must be divisible by the LCM of 4 and 6 which is 12. 100/12 = 8.33 200/12 = 16.67 So, there are 16 numbers below 200 that are divisible by 4 and 6 and, of these, 8 are smaller than 100. That gives 16 - 8 = 8 between 100 and 200.
There are 22 such numbers.
There are 43 natural numbers between 200 and 500 that are divisible by seven.
There are 33 numbers between 200 and 300 that are divisible by three.
There are (500-100)/2 = 200 numbers divisible by 2 between 100 and 500 counting 100 but not 500. Of these (500-100)/8 = 50 are divisible by 8. So there are 150 numbers between 100 and 500 divisible by two but not by 8. By relative primeness exactly 50 out of these 150 are divisible by 3 and therefore these 50 are exactly the ones divisible by 6 but not by 8.
I think the answer is: 66 numbers between 1 and 200 are divisible by 3. :) That's what I think the answer is.
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.