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MBA.Aspirant: Master | Next Rank: 500 Posts; Posts: 401; Joined: Tue May 24, 2011 1:14 amFollowed by: 5 members; Thanked: 37 times

by MBA.Aspirant » Tue Aug 09, 2011 5:54 pm

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

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Source: Beat The GMAT — Problem Solving | Tue Aug 09, 2011 5:54 pm

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Aug 09, 2011 8:25 pm

MBA.Aspirant wrote: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.

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gmatboost: Master | Next Rank: 500 Posts; Posts: 312; Joined: Tue Aug 02, 2011 3:16 pmFollowed by: 33 members; Location: New York City; Thanked: 130 times; GMAT Score:780

by gmatboost » Tue Aug 09, 2011 10:10 pm

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.

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