Multiples of 60.
There are no prime numbers as all multiples of 6 are divisible by 2, 3, and 6
All multiples of 6 are evenly divisible by 6, such as: 6, -6, 12, -12, 18, -18, and so on. Numbers divisible by six must also be divisible by both 2 and 3.
No. All of the multiples of 6 are even numbers.
Any numbers that are multiples of 6 or any numbers that are divisible by bith the numbers 2 and 3
If the numbers are even, they are divisible by 2. So, the set of numbers must be divisible by both 2 and 3, which is the same as being divisible by 6. The set is {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, ...}
6 and its multiples are divisible by 2 and 3.
Itself and any of its multiples
All multiples of 6 Numbers are evenly divisible by 6 if they are evenly divisible by both 2 AND 3. Even numbers are always evenly divisible by 2. Numbers are evenly divisible by 3 if the sum of all the individual digits is evenly divisible by 3. For example, the sum of the digits for the number 3627 is 18, which is evenly divisible by 3 but 3627 is an odd number so the number 3627 is not evenly divisible by 6
All numbers divisible by 9 are divisible by 3; since 9 = 3 x 3 all multiples of 9 are also multiples of 3. However, all numbers divisible by 3 are not divisible by 9, eg 6 = 2 x 3 but 6 is not divisible by 9 (since 6 is not a multiple of 9) - it only takes one counter example to disprove a theory.
The number 2 is divisible by the multiples of 2 (which are infinite), including these: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.
Oh, dude, numbers divisible by both 6 and 2 are basically multiples of the least common multiple of 6 and 2, which is 6. So, any number that's divisible by 6 is also divisible by 2 because 6 is just 2 times 3. It's like a two-for-one deal in the world of divisibility.