Numbers up to 200 divisible by both 2 and 3 = numbers to 200 divisible by 2*3 = 6 which is int(200/6) = int(33.33) = 33
Just count up all the even numbers which will all be divisible by 2
There are 25 prime numbers up to 100.
it is 2,3,6because 4152 the 2 is divisible by 2and if you add up all the numbers it = 12 and 12 is divisible by 2&3 and if a number is divisible by 2 and 3 then its divisible by 6
The even numbers up to 100 are: 2 468101214161820222426283032343638404244464850525456586062646668707274767880828486889092949698100.
There are 14 numbers between 1 & 100 that are divisible by 7.
The numbers divisible by nine up to a 100 are 9,18,27,36,45,54,63,72,81,90,and,99
Any of its multiples up to 100
100 is divisible by 5. The next is 105, the next is 110, and so on forever.
Any number ending in a 5 or 0 !
Numbers up to 200 divisible by both 2 and 3 = numbers to 200 divisible by 2*3 = 6 which is int(200/6) = int(33.33) = 33
Just count up all the even numbers which will all be divisible by 2
There are 80 such numbers.
Depending on how you look at it, there are either 99 or 100 whole numbers up to 100. If you mean up to and including 100, then the answer is 100 whole numbers, if not, then the answer is 99 whole numbers.
There are 25 prime numbers up to 100.
Counting up from 1, there is a number divisible by 3 every 3 numbers. Thus to find out how many numbers are divisible by 3, we just have to divide the number we're counting to by 3. In this case, the top number is 50, so do: 50/3 = 16 with two remainder. Because the numbers divisible by 3 come last in every set of 3 numbers, we can discard the remainder. Therefore there are 16 numbers between 1 and 50 that are divisible by 3.
A number is divisible by 9 if all the digits add up to 9 or are a factor of. IF the digits add up to 9 (or a factor of 9), they will automatically be a factor of 3.Yes. Test a couple numbers divisible by 9--27, 81--and they are all divisible by 3.