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 EVENIf (n + 2) is ODD, then
n must be ODDNow 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
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