If x is an integer

[muktarashmi]

If x is an integer, then x(x - 1)(x - k) must be evenly divisible by three when k is any of the following values EXCEPT
-4
-2
-1
2
5
(x) and (x - 1) are consecutive, so the three terms would be consecutive if (x - k) is either the lowest of the three, or the greatest of the three:

(x - k), (x - 1), and (x) are consecutive when (x - k) = (x - 2), or k = 2
(x - 1), (x), and (x - k) are consecutive when (x - k) = (x + 1), or k = -1 ....I am convinced about this

but could you please explain this:

Note that the difference between k = -1 and k = 2 is 3. Every third consecutive integer would serve the same purpose in the product x(x - 1)(x - k): periodically serving as the multiple of three in the list of consecutive integers.

[mithunsam]

Another way to write this question is... for what value of k, x(x - 1)(x - k) is not a multiple of 3.

If x(x - 1)(x - k) is not a multiple of 3, then x and (x-1) cannot be a multiple of 3.

Now consider 1,2,X,4,5,X,7,8,X,10,11,X... here X is a multiple of 3

Now consider (x-1) & x are 4 & 5 respectively (keep in mind, x=5). Now go back to the answer choices

A) k = -4 => (x-k) = (5--4) = 5 + 4 = 9 - - - - > multiple of 3
B) k = -2 => (x-k) = (5--2) = 5 + 2 = 7 - - - - > not a multiple of 3, correct answer
C) k = - 1 => (x-k) = (5--1) = 5 + 1 = 6 - - - - > multiple of 3
D) k = 2 => (x-k) = (5 - 2) = 3 - - - - > multiple of 3
E) k = 5 => (x-k) = (5 - 5) = 0 - - - - > yes, 0 is a multiple of 3

I hope this would explain your question...

[muktarashmi]

Thank you that helps!!

[oz_gurses]

Hi

my 2 cents:

when added 3 together, it should not be divided to 3.

So, x+x-1+x-k = 3x-1-k
Then plug in numbers in the options.

Only B does satisfy the condition.

[jnelson0612]

That works too!