Multiple

[MBA.Aspirant]

If n is a positive integer and (n+1)*(n+3) is odd, then (n+2)*(n+4) must be a multiple of?

A) 3
B) 5
C) 6
D) 8
E) 16

in problems like these, to determine the multiple do you put it in prime factor form or do you see its divisibility? because in other problems like if z is a multiple of x and y, then it must be a multiple of, it's solved by the LCM

Thanks ahead

[Anurag@Gurome]

If n is a positive integer and (n+1)*(n+3) is odd, then (n+2)*(n+4) must be a multiple of?

As (n + 1)*(n + 3) is odd, both (n + 1) and (n + 3) must be odd. Which implies both (n + 2) and (n + 4) are even, i.e. (n + 2) and (n + 4) are consecutive integers.

Hence, either (n + 2) or (n + 4) must be divisible by 4 and the other one by 2.
Therefore, (n + 2)*(n + 4) must be divisible by 2*4 = 8.

The correct answer is D.

[gmatboost]

Also a good question for plugging in if you get stuck.

The first two numbers you might plug in that do make (n+3)(n+1) odd are n=2, n=4.
In each case, the result, 24 or 48, will only eliminate 2 answers.

But if you try again with n=6, you get 80, which should help you narrow down the answer choices to just 1.

This is a good example of the fact that sometimes you have to work a little bit when you plug in, by which I mean try a few different things to identify the 1 answer that always works. In this case, 8.