Why is 6 a prime number?

Answer 1

It is the product, not the sum, of three consecutive integers that is divisible by 6. The three consecutive integers must be multiplied, not added.

For example, 2+3+4 = 9, which is not evenly divisible by 6.

If your first number is odd, then the statement works. Here's why:

The first number is odd, so it can be written as 2n+1 for some integer n.

The next two numbers are just 2n+2 and 2n+3.

Then the sum of the three numbers is:

(2n+1) + (2n+2) + (2n+3) = 6n + 6

which is obviously divisible by 6 regardless of the number you used for n.

For the 'Product of three consecutive integers..." see the Related Question below.

Answer 2

The same is true for any number. Given two numbers, x and y, to make them multiples of 6 they can be written as 6x and 6y. According to the distributive property, their sum (6x + 6y) can also be written as 6(x + y). That means that whatever the sum of x + y is, it will also be a multiple of 6.

Think of it this way: Multiplication is multiple addition.

2 x 6 = 6 + 6

3 x 6 + 6 + 6 + 6

It doesn't matter how many sixes each number has. Adding them together will just create a longer string of sixes.

Answer 3

2 x 3 = 6

That means that 2 and 3 are factors of 6. Any multiple of 6 will still have at least one 2 and one 3 in it.

Try it out.

1 x 2 x 3

2 x 2 x 3

3 x 2 x 3

4 x 2 x 3

Of course, any multiple of 6 will be divisible by six.

Any multiple of 6 will have 3 as a factor, which means it will be divisible by three.

Any multiple of 6 will have 2 as a factor, which means it will be divisible by two.

By definition, any number that is divisible by two is even.

Answer 4

All multiples of 6 have 2 as a factor.

All numbers that have 2 as a factor are even.

Answer 5

It is not prime - you can divide it by 2 or 3; a Prime number can only be divided by itself and 1

Answer 6

6 has 2 as a factor. All multiples of 6 have 2 as a factor. Numbers that have 2 as a factor are even. Even numbers have even numbers in the ones place.