# Thread: When n Is A Positive Integer

1. ## When n Is A Positive Integer

If n is a positive integer, then n(n + 1)(n + 2) is

A. Even only when n is even
B. Even only when n is odd
C. Odd whenever n is odd
D. Divisible by 3 only when n is odd
E. Divisible by 4 whenever n is even

I decided to let n = 2. I got 24, which is even.
I then let n = 3. This produced 60, which is also even.

When n is 3, 60 is the answer, which is divisible by 4.
However, when n is 2, the answer is 24, which is also divisible by 4.

Question:

Why is the answer E and not B?

2. ## Re: When n Is A Positive Integer

Originally Posted by harpazo
If n is a positive integer, then n(n + 1)(n + 2) is
A. Even only when n is even
B. Even only when n is odd
C. Odd whenever n is odd
D. Divisible by 3 only when n is odd
E. Divisible by 4 whenever n is even
Question: Why is the answer E and not B?
$(\forall n\in \mathbb{Z}^+)~[n(n+1)(n+2)\text{ is an even integer.}]$ That makes B. FALSE.

3. ## Re: When n Is A Positive Integer

Originally Posted by Plato
$(\forall n\in \mathbb{Z}^+)~[n(n+1)(n+2)\text{ is an even integer.}]$ That makes B. FALSE.

4. ## Re: When n Is A Positive Integer

Originally Posted by harpazo
$(\forall n\in \mathbb{Z}^+)~[n(n+1)(n+2)\text{ is an even integer.}]$ the upside down A is read "for all" n that is a positive integer ($n\in\mathbb{Z}^+$) the product $[n(n+1)(n+2)\text{ is an even integer.}]$

5. ## Re: When n Is A Positive Integer

Originally Posted by Plato
...the upside down A is read "for all"
So if there is no charge for an upside down A, then it's a "free for all"

6. ## Re: When n Is A Positive Integer

Of three consecutive integers, there must be at least one even integer: "odd, even, odd" or "even, odd, even". And if there is at least one even integer in a product then there is a factor of 2 so the product is even.

7. ## Re: When n Is A Positive Integer

Originally Posted by HallsofIvy
Of three consecutive integers, there must be at least one even integer: "odd, even, odd" or "even, odd, even". And if there is at least one even integer in a product then there is a factor of 2 so the product is even.
Sorry but I don't get it.

8. ## Re: When n Is A Positive Integer

Originally Posted by Plato
$(\forall n\in \mathbb{Z}^+)~[n(n+1)(n+2)\text{ is an even integer.}]$ the upside down A is read "for all" n that is a positive integer ($n\in\mathbb{Z}^+$) the product $[n(n+1)(n+2)\text{ is an even integer.}]$