GMATPrepNow wrote:

m and n are positive integers. If \(mn + 2m + n + 1\) is even, which of the following MUST be true?

i) \((2n + m)^2\) is even

ii) \(n^2 + 2n – 11\) is even

iii) \(m^2 – 2mn + n^2\) is odd

A) ii only

B) ii and iii only

C) i and ii only

D) i and iii only

E) i, ii and iii

*kudos for all correct solutions

There are several ways to approach this question. Here's one approach:

First recognize that mn + 2m + n + 1 is ALMOST factorable. If the expression were mn + 2m + n + 2, then we COULD factor it.

Next, recognize that, if mn + 2m + n +

1 is even, then mn + 2m + n +

2 must be ODD

Now recognize that we can factor mn + 2m + n + 2 to get: mn + 2m + n + 2 = (m + 1)(n + 2)

So, if mn + 2m + n + 2 is ODD, then it must be true that (m + 1)(n + 2) is ODD

If (m + 1)(n + 2) is ODD, then we know that (m + 1) is ODD,

AND (n + 2) is ODD

If (m + 1) is ODD, then

m must be EVENIf (n + 2) is ODD, then

n must be ODDNow check the 3 statements:

i) (2n + m)² is even.

(2n + m)² = [2(

ODD) +

EVEN)]²

= [EVEN + EVEN]²

= [EVEN]²

= EVEN

So, statement i is TRUE

ii) n² + 2n – 11 is even

n² + 2n – 11 = (

ODD)² + 2(

ODD) – ODD

= ODD + EVEN - ODD

= ODD - ODD

= EVEN

So, statement ii is TRUE

iii) m² – 2mn + n² is odd

m² – 2mn + n² = (

EVEN)² – 2(

EVEN)(

ODD) + (

ODD)²

= EVEN - EVEN + ODD

= EVEN + ODD

= ODD

So, statement iii is TRUE

Answer: E

IMPORTANT: Let's say you didn't see that mn + 2m + n + 2 factors nicely into (m + 1)(n + 2)

[most students will NOT see that]No problem. In the next solution, you'll see another way to handle this question.

Cheers,

Brent

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