What is the biggest positive integer which will always divid

To find the biggest number, we must minimize the expression, so let's use 1, 2 and 3.

1x2x3 = 6 is the largest number that will ALWAYS divide the expression.

We know that 6 will always divide the expression regardless of choice of n as follows;

To be divisible by 6 it needs to be divisible by both 2 and 3.

Let's assume n is even. That means it can be written as n=2K. Rewriting the expression:

2K(2K+1)(2K+2) > this is clearly divisible by 2. But is it divisible by 3 ? . Let's assume 2K is not divisible by 3, because if it were, then we would be done. So 2K equals 3 times some number X plus a remainder of 1 or 2, or 3X + 1 or 2

Try 3X+1: (3X+1)(3X+2)(3X+3) = 3(3X+1)(3X+2)(X+1) > clearly divisible by 3

Now 3X+2: (3X+2)(3X+3)(3X+4). Notice that same 3X+3 as above, so also divisible by 3.

So now we know that with n being even, the expression is divisible by 3 and 2, or 6.

Testing n being odd means n = 2K+1 . Rewriting the original expression:

(2K+1)(2K+2)(2K+3) = 2(2K+1)(K+1)(2k+3) > Clearly divisible by 2. But is it divisible by 3 ? Following logic above, (2K+1)= 3X plus a remainder of 1 or 2.

and the expression is the same as above, establishing that with an odd n, it is also divisible by 3

So n(n+1)(n+2) is always divisible by 2 and 3, or 6

Read the question carefully... n is even integer hence answer is Option E

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Yes, thanks, read the two evens in an odd manner