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.
Divisible by both? 2 numbers... 45 and 90
Exactly 50
Twenty of them.
To be divisible by both 4 and 5, the number has to be divisible by 20. Therefore, there are 12 numbers between 1 and 240, inclusive, which are divisible by 4 and 5: 20, 40, 60, 80, 100, 120, 140, 160, 180, 200, 220, 240.
There are 8 of them and they are all multiples of 12
There are 1,000 positive integers between 1,000 and 9,999, inclusive, that are divisible by nine.
There are 39 natural numbers between 1,000 and 1,500, inclusive, that are divisible by 13.
193 of them are divisible by one (or more) of the given numbers.
between 1 and 600 inclusive there are:300 numbers divisible by 2200 numbers divisible by 3100 numbers divisible by both 2 and 3400 numbers divisible by 2 or 3.
There are 10 palindromes divisible by 9 between 1000 and 9999.
Divisible by both? 2 numbers... 45 and 90
Seven of them.
To find the numbers between 1 and 100 inclusive that are divisible by either 9 or 4, we first determine how many numbers are divisible by 9 and how many are divisible by 4. There are 11 numbers divisible by 9 (9, 18, 27, ..., 99) and 25 numbers divisible by 4 (4, 8, 12, ..., 100). However, we must be careful not to double-count numbers divisible by both 9 and 4 (36, 72). Therefore, the total number of numbers divisible by 9 or 4 between 1 and 100 inclusive is 11 + 25 - 2 = 34.
Exactly 50
Twenty of them.
128 of them.
343!