karthikpandian19 wrote:If n is an integer greater than 6, which of the following must be divisible by 3?
a. n(n+1)(n-4)
b. n(n+2)(n-1)
c. n(n+3)(n-5)
d. n(n+4)(n-2)
e. n(n+5)(n-6)
The expression will be divisible by three if at least one of its factors is divisible by 3. One approach is to plug in numbers. Note that every integer is either a multiple of 3, one more than a multiple of 3, or two more than a multiple of 3. All of the expressions contain 'n', so plugging in a multiple of 3 for n won't be helpful. So try plugging in a number that is one more than a multiple of 3. Let's say 10:
a. 10(11)(6)---divisible by 3
b. 10(12)(9)---divisible by 3
c. 10(13)(5)---not divisible by 3
d. 10(14)(8)---not divisible by 3
e. 10(15)(4)---divisible by 3.
So we can eliminate c and d. Now for the tiebreaker, plug in a value of n that is TWO more than a multiple of 3: 11.
a. 11(12)(7)---divisible by 3
b. 11(13)(10---not divisible by 3
e. 11(16)(5)---not divisible by 3
So the answer must be
A.
You could also approach this from a more theoretical perspective. If you examine choice A, note that n and n+1 are consecutive integers. If one of them is a multiple of 3, then the product is a multiple of 3. If neither of them is a multiple of 3, then n-1 must be a multiple of 3. But if n-1 is a multiple of 3, then n-4 is also a multiple of 3. Thus, if n and n+1 are not multiples of 3, then n-4 is a multiple of 3, so one of the factors of A is a multiple of 3 regardless of the value of n.
