150
150
17 of them.
It is 120, 150, and 180.
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.
Yes, it becomes 100.
100 ÷ 6 = 16 r 4 so the first whole number between 100 and 200 divisible by 6 is 6 x 17 (= 102) 200 ÷ 6 = 33 r 2 so the last whole number between 100 and 200 divisible by 6 is 6 x 33 (= 198) So the whole numbers between 100 and 200 divisible by 6 are the multiples of 6 from 17 to 33 which are: 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192 and 198.
The numbers between 100 and 200 that are divisible by three but not by two are: 105, 111, 117, 123, 129, 135, 141, 147, 153, 159, 165, 171, 177, 183, 189 and 195.
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.
They are all the numbers of the form 1n5 where n is any integer.
2 times 3 times 5 is 30 so you just have to have a number divisible by 30 so 120.
No. It is only divisible by 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200.
All of them but I assume you mean evenly divisible. 150. 150/2 is 75 150/3 is 50 and 150/10 is 15