Math problem?

Question

Prove that:
(n+2)x(n2+n+6) is divisible by 6.

n is every number from N, so, n is an integer n>0 or n=0.

n2 is not n x 2, is n^2.

Answer 1

(n+2)(n2+n+6) mod 6
= (n+2)(n2+n) mod 6
= (n+2)(n+1) n mod 6
= 0 because the product of any three consecutive numbers is divisible by 6.

Answer 2

Use induction

I liked the method of the former answerer. You may do this too:

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

(n+2)[n(n+1) + 6] = (n+2)n(n+1)+ 6 (n+2)

6 (n+2) is a multiple of 6

So, if you can prove that n(n+1)(n+2) is a multiple of 6, youre done.

And, in fact, it is. If n is a multiple of 3, then it is, because either n or n+1 is a multiple of 2

If n+1 is a multiple of 3, it is, because of the same reason.

And, if n+2 is a multiple of 3, it is, because of the same reason

And remember, a number only can be a multiple of 3, a multiple of 3 + 1 or a multiple of 3 + 2.



Ana

Answer 3

rewrite it as

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

n(n+1)(n+2) + 6(n+2)

Now we know that 6(n+2) is divisible by 6 so we need only to prove that

n(n+1)(n+2) is divisible by 6

Obviously n(n+1) is divisible by 2 since if n is odd then n+1 is even
Similarly if n is not divisible by 3, neither is n+3 but between n+3 and n, there must be a number divisible by 3 and it must be either n+1 or n+2.

Answer 4

take n=any real no. say 1 or 2 or 3...

let n=1:-

=(1+2)(1^2+1+6)
=3(8)
=24 which is completely divisible by 6.
use same for another numbers also.

Answer 5

You did not specify the constraints of n.
Is n an integer? Can n be zero, less than zero?

There is at least one case to show that this proof does not exist.

If n=5

(5+2)x(5x2+5+6) =
7x21 = 147 which is not divisible by 6

Answer 6

(n+2)x(n2+n+6) = (we multiply both expressions)
n³+3n²+6n+12

If we divide this expresion by 6 we get residue equal to zero.

So it is divisible by 6

Answer 7

use an example.
n=2

(2+2)x(2(2) + 2+6)

4 X 12 =48

48/6= 8

Answer 8

another example

n=3

(3+2)x(3x2+3+6)

5 x 15= 75

75 / 6 = 12 & 1/2

so it depends on which number to use for 'n'.....some numbers will work (like 2)....some will not.