VeritasPrepKarishma wrote:

I am providing the theoretical explanation below. Once you get it, you can solve such questions in a few seconds in future!

Notice a few things about integers:

-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16......

Every number is a multiple of 1

Every second number is a multiple of 2

Every third number is a multiple of 3

Every fourth number is a multiple of 4 and so on...

So if I pick any 3 consecutive integers, one and only one of them will be a multiple of 3: e.g. I pick 4, 5, 6 (6 is a multiple of 3) or I pick 11, 12, 13 (12 is a multiple of 3) etc..

x(x - 1)(x - k) will be evenly divisible by 3 if at least one of x, x - 1 and x - k is a multiple of 3. We know from above, (x - 2)(x - 1)x will have a multiple of 3 in it. Also, (x-1)x(x + 1) will have a multiple of 3 in it because they both are products of 3 consecutive integers. So k can be 2 or -1. Eliminate these options.

Now let me write down consecutive integers around x:

(x-5), (x - 4), (x - 3), (x - 2), (x - 1), x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5) etc

(x - 2)(x - 1)x will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x - 2) could be the multiple of 3, in which case (x - 5) will also be a multiple of 3.

So in any case, (x - 5)(x - 1)x will have a multiple of 3 in it. So k can be 5.

Similarly, (x-1)x(x + 1) will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x + 1) could be the multiple of 3, in which case (x + 4) will also be a multiple of 3.

So in any case, (x - 1)x(x + 4) will have a multiple of 3 in it. So k can be -4.

We cannot say whether (x-1)x(x + 2) will have a multiple of 3 in it and hence if k = -2, we cannot say whether the product is evenly divisible by 3.

Answer (B).

**Quote:**

Plz Could you please explain how x-5 will also be a multiple of 3. I couldnot understand that part.

If (x - 2) is a multiple of 3, (x - 5), a number 3 places away from (x - 5) will also be divisible by 3.

Say (x - 2) = 9 (a multiple of 3)

then (x - 5) = 6 (previous multiple of 3)

Hi Karishma, Initially I was trying the same method only, though I could not figure out the following scenario (x-1) x (x+1) (x+2), now say (x-1) is divisible by 3, in that case (x+2) i.e. k = -2 will also be divisible by 3 and hence in that case (x-1) x (x+2) would be evenly divisible by 3. Could you please tell me what am I missing here?