That's "divisible". 3 x 66 = 198 so 66 is your answer.
Numbers that are divisible by 80 are itself and any of its other multiples
Smallest composite number dividable by prime numbers is product of them all. In this case it is: 2x3x5x7x11 = 2310.
Any number dividable by six must also be dividable by 3 as well as by 2. Nice Question!
5 of them.
They are the numbers that end with a 0
That's "divisible". 3 x 66 = 198 so 66 is your answer.
4,8,12,16,20,24,28,32,36
No, it is not.
I'm not quite sure about that theory, but if all the numbers in the number add up to a number that's dividable by 3, then the number itself would then be divisible by 3. Ex. 141 1+4+1= 6 6 is divisible by 3, so 141 is too. 141/3= 47.
Numbers that are divisible by 80 are itself and any of its other multiples
Smallest composite number dividable by prime numbers is product of them all. In this case it is: 2x3x5x7x11 = 2310.
An interesting question. The answer is 80%. A number must end in 0 or 5 to be dividable by 5. From 100 to 109 (10 numbers in total), there will be 2 numbers that are dividable by 5. If I then consider the entire set from 100 to 999 is 900 numbers, you have 90 sets of 10 numbers, each set with 2 dividable number, or 180/900 = 2/10 =20%. If 20% are dividable, 80% aren't. Suppose I extended the question, and say my set is all whole positive numbers and zero with n digits or less, where I pick n randomly from 1 to a million. If from this set, a number is picked randomly, what is the chance that it is dividable by 5. Answer: 20%.
Any number dividable by six must also be dividable by 3 as well as by 2. Nice Question!
2 is the only prime number dividable by 2. to be a prime number, it must only be dividable by 1 and itself. This is the reason why 2 is the only prime number dividable by 2. If any other number was dividable by two then it would not be prime any more
3/42 = 1/14
No as it is dividable by 3 (123/3=41)