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m and n are positive integers. If mn + 2m + n + 1 is

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m and n are positive integers. If mn + 2m + n + 1 is [#permalink] New post 17 May 2018, 06:30
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Question Stats:

40% (01:10) correct 60% (02:52) wrong based on 10 sessions
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
[Reveal] Spoiler: OA

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1 KUDOS received
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Kudos [?]: 1055 [1] , given: 6

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Re: m and n are positive integers. If mn + 2m + n + 1 is [#permalink] New post 17 May 2018, 11:33
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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
= 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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Re: m and n are positive integers. If mn + 2m + n + 1 is [#permalink] New post 17 May 2018, 11:50
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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
= EVEN - EVEN + ODD
= EVEN + ODD
= ODD
So, statement iii is TRUE

Answer: E

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