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

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m and n are positive integers. If mn + 2m + n + 1 is even,  [#permalink]

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New post 17 May 2018, 07:26
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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

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m and n are positive integers. If mn + 2m + n + 1 is even,  [#permalink]

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New post 17 May 2018, 10:31
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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


\(mn + 2m + n + 1=Even\)

\(=>mn+Even+n+Odd=Even\)

\(=> n(m+1)=Odd\). Hence \(n=Odd\) & \(m=Even\) because multiplication of two numbers is Odd only when both are Odd. So \(m+1=Odd =>m=Even\)

Now

i) \((2n + m)^2=>(Even+Even)^2=Even\) -----\(True\)

ii) \(n^2 + 2n – 11=Odd+Even-Odd=Even\) ------\(True\)

iii) \(m^2 – 2mn + n^2=Even-Even+Odd=Odd\) ------\(True\)

Option E
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Re: m and n are positive integers. If mn + 2m + n + 1 is even,  [#permalink]

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New post 19 May 2018, 06:31
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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 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 even,  [#permalink]

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New post 19 May 2018, 06:32
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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


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

Cheers,
Brent
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Re: m and n are positive integers. If mn + 2m + n + 1 is even,  [#permalink]

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New post 19 May 2018, 06:55
1
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


Given: m>0, n>0, mn + 2m + n + 1 = Even

Approach: We know that, e + e = e ; o + o = e
mn + 2m + n + 1 = Even
=> mn + n + 1 = Even - 2m = Even - Even = Even
=> mn + n + 1 = Even
or, mn + n = Even - 1= Even - odd = odd
=> mn + n = odd

For (mn + n) to be odd, n has to be odd.

1) (2n+m)^2 is even True because m is even.
So, (Even + Even )^2 = Even

2) n^2+2n–11 is even True
odd + Even - odd = odd - odd = Even

3) m^2–2mn+n^2 is odd True
( Even) ^2 - Even + (odd)^2 = Even - Even + odd = Even - odd = odd

Correct Answer= E
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Re: m and n are positive integers. If mn + 2m + n + 1 is even,  [#permalink]

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New post 28 Jun 2018, 00:44
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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

since mn + 2m + n + 1 is even
{n(m+1) } + {2m + 1 } = even
odd odd
n(m+1) = odd
n = odd , m+1 = odd
m = even
i) (2n+m)^2 = (even + even)^2 = even
ii) ( n^2+2n–11) = ( odd^2 + even - odd ) = (odd - odd) = even
iii) m^2–2mn+n^2 = (even^2 - even + odd ) =(even + odd) = odd
so all options are correct
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Re: m and n are positive integers. If mn + 2m + n + 1 is even,   [#permalink] 28 Jun 2018, 00:44
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