gmataspirant wrote:What is the greatest integer that will always evenly divide the sum of three consecutive even integers?
A)2
B)3
C)4
D)6
E)12
Your time and help is appreciated.
The numbers are small; in all likelihood, the quickest way to do this is trial and error.
2+4+6 = 12
4+6+8 = 18
6+8+10 = 24
So far, 6 looks like a good answer.
However, we should be careful and try some negatives.
-4 + (-2) + 0 = -6
-2 + 0 + 2 = 0
Ok, looks like we still get (d) 6 as the biggest number that goes into all possible sums.
We could prove it mathematically if we really wanted to, but it would almost certainly be a lot slower:
Let x be any integer.
Therefore, 2x will be even.
So, we can write our series as:
2x + (2x + 2) + (2x +4) = 6x + 6 = 6(x+1)
As we can see, 6 will always be a factor of 6(x+1).
