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

Question

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

niks18 · Answer

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

BrentGMATPrepNow · Answer

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 =  ODD ) +  EVEN )]²
=  ²
=  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)    
No problem. In the next solution, you'll see another way to handle this question.

Cheers,
Brent

BrentGMATPrepNow · Answer

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 =  ODD ) +  EVEN )]²
=  ²
=  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

Cheers,
Brent

PP777 · Answer

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

shridhar786 · Answer

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

fireagablast · Answer

m=1, n=1
1+2+1+1 = 5; no good

m=2 n=1
2+4+1+1 = 8; good

m=1 n=2
2+2+2+1 = 7; no good

m=even, n = odd
I, II, III true

bumpbot · Answer

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

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