The product of two consecutive even intergers is divisible by?

Answer 1

those 2 integers and any of either of their factors or any product of their factors

Answer 2

2

Answer 3

The average of those two numbers and 2. For example 2+4 = 6. Divisible by 3 (the average of 2 & 4) and 2.

Answer 4

Meaninglessly, each of the two integers.
Trivially, 1.
Very obviously, 2 since they are even integers.
Obviously, 4 since they are both even integers.
Less obviously, by a previous answer, 8, since one of them must be a multiple of 4.
And, this is what I offer:
2n(2(n+1))
= 4n(n+1)
= 4n^2 + 4n
= 4(n^2 + n)
Thus, n^2+n is a factor.
E.g. 8x10=80, n=4, n^2+n = 16+4=20, and 80/20 = 4.

So, we have 1, 2, 4, 8, factors of n, factors of n+1, some products of these factors provided the results are still a factor, 4(n^2+n), 2(n^2+n), (n^2+n), (n^2+n)/2, and should probably have some more..

Redundantly, product of -1 and any of the above.

Answer 5

8. If you start with two consecutive even integers, one of them is divisible by 4...hence their product is at least divisible by 8.

Answer 6

1, 2, and both of the integers used.

Answer 7

8, since
(2n)(2n+2)=(2)(2)(n)(n+1), and since n or n+1 has to be even, then the product can be divided by 2^3=8

Answer 8

1,2,4 & 8 and the numbers being multiplied :)

Answer 9

those integers