Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Re: m and n are positive integers. If mn + 2m + n + 1 is [#permalink]
17 May 2018, 11:33

1

This post received KUDOS

Expert's post

GreenlightTestPrep 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) n² + 2n – 11 is even iii) m² – 2mn + n² 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

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

First 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 EVEN If (n + 2) is ODD, then n must be ODD

Now 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
_________________

Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Dayemails

Re: m and n are positive integers. If mn + 2m + n + 1 is [#permalink]
17 May 2018, 11:50

1

This post received KUDOS

Expert's post

GreenlightTestPrep 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) n² + 2n – 11 is even iii) m² – 2mn + n² 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

Here's a different solution:

Since m and n can each be either even or odd, there are 4 possible cases to consider:

case a) m is EVEN and n is EVEN case b) m is ODD and n is EVEN case c) m is EVEN and n is ODD case d) m is ODD and n is ODD

Now let's test each case as we examine \(mn + 2m + n + 1\) To make things super easy, let's plug in 0 as a nice EVEN number, and we'll plug in 1 as a nice ODD number.

case a) m is EVEN and n is EVEN mn + 2m + n + 1 = (0)(0) + 2(0) + (0) + 1 = 1 (an ODD number) We're told that mn + 2m + n + 1 is EVEN, so case a is NOT POSSIBLE

case b) m is ODD and n is EVEN mn + 2m + n + 1 = (1)(0) + 2(1) + (0) + 1 = 3 (an ODD number) We're told that mn + 2m + n + 1 is EVEN, so case b is NOT POSSIBLE

case c) m is EVEN and n is ODD mn + 2m + n + 1 = (0)(1) + 2(0) + (1) + 1 = 2 (an EVEN number) We're told that mn + 2m + n + 1 is EVEN, so case c IS POSSIBLE

case d) m is ODD and n is ODD mn + 2m + n + 1 = (1)(1) + 2(1) + (1) + 1 = 5 (an ODD number) We're told that mn + 2m + n + 1 is EVEN, so case d is NOT POSSIBLE

Since case c is the ONLY possible case, we know that m is EVEN and n is ODD

Now check the 3 statements (using the same strategy that we applied above):

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

RELATED VIDEO

_________________

Brent Hanneson – Creator of greenlighttestprep.com Sign up for my free GRE Question of the Dayemails

Re: m and n are positive integers. If mn + 2m + n + 1 is [#permalink]
14 Dec 2018, 16:57

1

This post received KUDOS

GreenlightTestPrep 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) n² + 2n – 11 is even iii) m² – 2mn + n² 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

Here is a faster way. Create a table, there are only 2 variables so it shouldn't be too hard.

Let E be even and O be odd.

m|n|m^2 + 2m + n + 1 --------------------------- E|E| E^2 + 2 E + E + 1 = E + E + E + 1 = O E|O| EO + E + O + 1 = E O|E|OE + 2O + E + 1 = O O|O|O^2 + 2O + O + 1 = O

Where we used the rules odd +- odd = even odd+- even = odd even+-even = even odd*odd = odd odd*even = even even*even = even

You don't need to memorize these, just keep 1 and 2 as the specific cases if you forget, i.e. 1*2 = 2 so odd*even = even.

Now, notice that the only one that satisfies the criteria that mn + 2m + n + 1 is the second option. Therefore m is even and n is odd.

We check all three cases. i) (2n + m)² is even ii) n² + 2n – 11 is even iii) m² – 2mn + n² is odd