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Re: m and n are positive integers. If mn + 2m + n + 1 is [#permalink]
17 May 2018, 11:33
1
This post received KUDOS
Expert's post
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)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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Re: m and n are positive integers. If mn + 2m + n + 1 is [#permalink]
17 May 2018, 11:50
1
This post received KUDOS
Expert's post
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)² = EVEN - EVEN + ODD = EVEN + ODD = ODD So, statement iii is TRUE
Answer: E
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Brent Hanneson – Creator of greenlighttestprep.com Sign up for GRE Question of the Dayemails
Re: m and n are positive integers. If mn + 2m + n + 1 is [#permalink]
14 Dec 2018, 16:57
1
This post received KUDOS
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 is a faster way. Create a table, there are only 2 variables so it shouldn't be too hard.
Let E be even and O be odd.
m|n|m^2 + 2m + n + 1 --------------------------- E|E| E^2 + 2 E + E + 1 = E + E + E + 1 = O E|O| EO + E + O + 1 = E O|E|OE + 2O + E + 1 = O O|O|O^2 + 2O + O + 1 = O
Where we used the rules odd +- odd = even odd+- even = odd even+-even = even odd*odd = odd odd*even = even even*even = even
You don't need to memorize these, just keep 1 and 2 as the specific cases if you forget, i.e. 1*2 = 2 so odd*even = even.
Now, notice that the only one that satisfies the criteria that mn + 2m + n + 1 is the second option. Therefore m is even and n is odd.
We check all three cases. i) (2n + m)² is even ii) n² + 2n – 11 is even iii) m² – 2mn + n² is odd
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50% (01:35) correct
