Brent@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) 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
Answer:
E
Difficulty level: 700+
Source:
www.gmatprepnow.com
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
Brent Hanneson - Creator of GMATPrepNow.com
