GMATinsight wrote:What is the biggest positive integer which will always divide n(n+1)(n+2) evenly for any even integer value of n?

a) 2
b) 3
c) 6
d) 12
e) 24

Source: www.GMATinsight.com

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

regor60 wrote:

GMATinsight wrote:What is the biggest positive integer which will always divide n(n+1)(n+2) evenly for any even integer value of n?

a) 2
b) 3
c) 6
d) 12
e) 24

Source: www.GMATinsight.com

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

GMATinsight wrote:

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

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

GMATinsight wrote:What is the biggest positive integer which will always divide n(n+1)(n+2) evenly for any even integer value of n?

a) 2
b) 3
c) 6
d) 12
e) 24

Source: www.GMATinsight.com

\[n = 2M\,\,,\,\,\,\,M\,\,\operatorname{int} \]
\[\frac{{2M\left( {2M + 1} \right)\left( {2M + 2} \right)}}{{? = \max \,\,\operatorname{int} }}\,\,\, = \operatorname{int} \]

(1) Exactly one of the factors among 2M , (2M+1) and (2M+2) is divisible by 3. (They are three consecutive integers!)
We guarantee (at least) one factor 3. (Nine could be 2M+1, therefore more than one factor 3 is possible...)

(2) 2M , 2M+1 , 2(M+1) is even, odd, even :: in the product of these 3 factors there are:
two factors 2 plus another factor 2 (we have M and M+1 consecutive integers... one of them is even)!
We guarantee (at least) three factors 2.

We have already found 24 (one factor 3, three factors 2), and the maximum available alternative choice is exactly 24... we are done!

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.