Since we are looking for always, look for the least condition...
At least one will surely be div by 3 as there are 3 consecutive integers. Also at least one will be divisible by 2...
So if it is ALWAYS, Ans will be 6...

But if n is even then ...
One of n and n+2 will be div by 2 and other by 4 and one by 3 as there are 3 consecutive numbers..
So product is always divisible by 2*3*4=24..

But here as per the wordings answer will be 6

GMATinsight, please check your OA,
Let n be 13, so 13*14*15 will be just divisible by 6 and NOT by 12 or 24.
Hence Ans will not be 24.
Please change OA or the wordings

The product of 3 consecutive integers is ALWAYS divisible by 2 and 3 , so the possible values are - 6, 12 , 18 , 24

Now, among the given options , (E) is the biggest positive integer, hence answer must be (E)
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Thank you for highlighting mistake. I meant n to be an even integer. Have modified the question accordingly
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\(n = 2M\,\,,\,\,\,\,M\,\,\operatorname{int}\)

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

(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.
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n is any even integer.
If n=2, n(n+1)(n+2) = 2.3.4=24
If n=4, 4.5.6=120
Biggest integer which will always divide n(n+1)(n+2) is 24.
Answer E.

Since the product of m consecutive integers is divisible by m!, and since n(n + 1)(n + 2) represents a product of 3 consecutive integers, n(n + 1)(n + 2) is divisible by 3! = 6. However, we are also given that n is an even integer. In that case, we can express n as 2k where k is a integer. Therefore, we have

n(n + 1)(n + 2)

2k(2k + 1)(2k + 2)

2k(2k + 1)[2(k + 1)]

4k(k + 1)(2k + 1)

Notice that k(k + 1) is a product of 2 consecutive integers, therefore, k(k + 1) is divisible by 2! = 2 and hence n(n + 1)(n + 2) = 4k(k + 1)(2k + 1) is divisible by 4 x 2 = 8.

Recall that if a number is divisible by a and b, then it will also divisible by the LCM of a and b. Here we have n(n + 1)(n + 2) is divisible by 6 and 8; therefore, it’s also divisible by the LCM of 6 and 8, which is 24.

Alternate Solution:

Since n is even, n + 2 is also even and furthermore, since n and n + 2 are consecutive even numbers, one of them must be divisible by 4. Since either n or n + 2 is divisible by 4 and the other number is even, we have at least three factors of 2, which means the product of n and n + 2 is divisible by 8.

Since we have three consecutive integers, precisely one of these integers must be divisible by 3.

Since n(n + 1)(n + 2) is divisible by both 3 and 8, it is divisible by their LCM, which is 24.

Answer: E
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