If m and n are positive integers, is n even?

If m and n are positive integers, is n even?

STATEMENT (1) m(m + 2) + 1 = mn
let m be even
then m(m+2) = even
m(m+2)+1 = even+odd = odd
mn = even (since m is even )
this case is not possible since both sides have to be odd
but m(m+2)+1--odd and mn = even

now let m = odd
--m(m+2) = odd*(odd+even) = odd
m(m+2)+1 = odd+odd = even
now mn to be equal to = m(m+2)+1-- has to be even
mn = odd *even (m odd--n has to be even for mn to be even)

so, from this statement, we can say that n is even
SUFFICIENT

STATEMENT (2) m(m + n) is odd.

m(m+n) = odd
this implies that m = odd and (m+n) = odd
since (m+n) = odd
odd + n = odd
n = even (n has to be even for (m+n) to be odd---because if n = odd than m+n = even )
so, from here we can say that n is even
SUFFICIENT

D is the answer

Statement 1-

\(n=m+2 +(\frac{1}{m})\)
As n is a positive integer, 1/m must be an integer. It's only possible if m is 1 .

Hence m is odd.

n is O+E+O= even

Sufficient

Statement 2-

m(m+n) is odd

Hence m and m+n are both odd; Hence, n must O-O= even

\(Sufficient\)

We know that m and n are positive integers. The original question: Is n=E ?

1) After rearranging the terms, we can make some inferences about the parities of the variables.

m(m + 2) - mn = -1
m(m + 2 - n) = O, so m=O and
m + 2 - n = O
O + E - n = O
n = (O - O) + E = E + E = E

Thus, the answer to the original question is a definite Yes. \(\implies\) Sufficient

2) We know that

m(m + n) = O, so m=O and
m + n = O
O + n = O
n = O - O = E

Thus, the answer to the original question is a definite Yes. \(\implies\) Sufficient

Answer: D
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Statement 1:
Left side we have \(m^2 + 2m + 1 = (m + 1)^2 = m*n\). So if we plug in m = even, we get odd = even*n which is not possible. If we plug in m = odd, we have even = odd*n, and n must be even. We can do this odd/even test without transforming the left side but doing so makes the odd/even evaluation slightly faster. Sufficient.

Statement 2:
If a product is odd, all the components of the product must be odd. So m is odd, m + n is odd. Because m + n = odd + n = odd, n must be even. Sufficient.

Ans: D
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Forget the conventional way to solve DS questions.

We will solve this DS question using the variable approach.

Remember the relation between the Variable Approach, and Common Mistake Types 3 and 4 (A and B)[Watch lessons on our website to master these approaches and tips]

Step 1: Apply Variable Approach(VA)

Step II: After applying VA, if C is the answer, check whether the question is key questions.

StepIII: If the question is not a key question, choose C as the probable answer, but if the question is a key question, apply CMT 3 and 4 (A or B).

Step IV: If CMT3 or 4 (A or B) is applied, choose either A, B, or D.

Let's apply CMT (2), which says there should be only one answer for the condition to be sufficient. Also, this is an integer question and, therefore, we will have to apply CMT 3 and 4 (A or B).

To master the Variable Approach, visit https://www.mathrevolution.com and check our lessons and proven techniques to score high in DS questions.

Let’s apply the 3 steps suggested previously. [Watch lessons on our website to master these 3 steps]

Step 1 of the Variable Approach: Modifying and rechecking the original condition and the question.

We have to find Is 'n' even -where 'm' and 'n' is positive integers.

Second and the third step of Variable Approach: From the original condition, we have 2 variables (m and n). To match the number of variables with the number of equations, we need 2 equations. Since conditions (1) and (2) will provide 2 equations, C would most likely be the answer.

But we know that this is a key question [Integer question] and if we get an easy C as an answer, we will choose A or B.

Let’s take a look at each condition.

Condition(1) tells us that m(m + 2) + 1 = mn.

=> If m = even, then m(m + 2) = even and even + 1 = odd. On the other hand, 'mn' will be even as 'm' is even.Lhs (ODD) and RHS(EVEN) - not possible.

=> If m = odd, then m(m + 2) = odd and odd + 1 = even. On the other hand, 'mn' will be even when 'n' is even.

=> Is 'n' even - YES

Since the answer is a unique YES, the condition(1) is alone sufficient by CMT 1.


Condition(2) tells us that m( m + n) = odd .

=> Either m = odd or m + n = odd

=> Odd + n = Odd - This is possible when 'n' is even because odd + even = odd

=> Is 'n' even - YES

Since the answer is a unique YES, the condition(2) alone is sufficient by CMT 1.


Each condition alone is sufficient.

So, D is the correct answer.

Answer: D
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Very high quality question! I missed this question initially. This all hinges on testing cases

(1) M(M+2) + 1 = mn

First, assume M = odd

then, odd*(odd+2=odd) +1 = mn
so odd*odd +1 = mn
so even = mn. we have already assumed m odd, so n must be even

Now, assume M = even

even*(even+2=even) +1 = mn

even +1 = even*n
odd = even *n
odd = even impossible, so M cannot be even. M must be odd. thus N must be even sufficient

(2) m (m+n) is odd

assume m odd

odd (odd + n) = odd

only odd * odd = odd, so odd +n must be odd so n must be even


assume m odd
odd (odd +n) = odd

only odd*odd = odd, so odd +n must be odd, so n must be even

either way m must be even. sufficient
OA is D

Very high quality question! I missed this question initially. This all hinges on testing cases

(1) M(M+2) + 1 = mn

First, assume M = odd

then, odd*(odd+2=odd) +1 = mn
so odd*odd +1 = mn
so even = mn. we have already assumed m odd, so n must be even

Now, assume M = even

even*(even+2=even) +1 = mn

even +1 = even*n
odd = even *n
odd = even impossible, so M cannot be even. M must be odd. thus N must be even sufficient

(2) m (m+n) is odd

assume m odd

odd (odd + n) = odd

only odd * odd = odd, so odd +n must be odd so n must be even


assume m odd
odd (odd +n) = odd

only odd*odd = odd, so odd +n must be odd, so n must be even

either way m must be even. sufficient
OA is D

If m and n are positive integers, is n even?

(1) \(m(m + 2) + 1 = mn\)
\(m^2 + 2m + 1 = mn\)
\((m+1)^2 = mn\)
if m = odd
even * even = odd * even; m = even

if m = even
odd * odd = even * odd; not possible.

SUFFICIENT.

(2) \(m(m + n) = odd\)

This mean m and (m+n) are both odd. As (m+n) is odd, we can conclude m is odd and n is even. SUFFICIENT.
_________________

This question can be solved without testing cases.

Fact 1: \(m(m + 2) + 1 = mn\)

Solving:
\(m^2+2m-mn = -1\)
\(m(m+2-n)= -1\)

This means that m and (m+2-n) must be odd. We know this because -1 is odd and only the product of two odd numbers will give an odd result.

Since we now know that (m+2-n) and m are odd, then n must be even because only a combination of Odd+2(Even)-Even will give an odd number.

Fact 1 is sufficient, knock out B,C,E

Fact 2: m(m + n) is odd

Similar idea, m and (m+n) have to be odd for their product to be odd

We know that m is odd and for (m+n) to be odd, n has to be even

Fact 2 is also sufficient. Knock out A

Best answer is D

Statement 1:
Case 1: m is ODD
Plugging m=1 into m(m + 2) + 1 = mn, we get:
1(1+2) + 1 = 1(n)
4 = n
In this case, the answer to the question stem is YES.

Case 2: m is EVEN
Plugging m=2 into m(m + 2) + 1 = mn, we get:
2(2+2) + 1 = 2n
9 = 2n
\(n = \frac{9}{2} = \frac{ODD}{EVEN}\)
Not viable, since \(\frac{ODD}{EVEN} =\) noninteger, violating the condition that n must be an integer.

Since only Case 1 is viable, the answer to the question stem is YES.
SUFFICIENT.

Statement 2:
Case 1: m is ODD
Plugging m=1 into m(m + n) is odd, we get:
1(1+n) = ODD
1+ n = ODD
n = ODD - 1 = EVEN
In this case, the answer to the question stem is YES.

Case 2: m is EVEN
Plugging m=2 into m(m + n) is odd, we get:
2(2+n) = ODD
EVEN = ODD
Not viable.

Since only Case 1 is viable, the answer to the question stem is YES.
SUFFICIENT.


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Tried solving with random numbers. I just did not get it right!

I thought the answer was B.

Because I could not find pair of numbers that fit into the equation! (Under a min)

The O/E method did not strike me

Statement - A
m(m+2) + 1 = mn;
\(m^2 + 2m - mn = -1\);
m(m+2+n) = -1

now, as mentioned m and n are positive int.
let m = 1 (because product of any two int not equal to 1 can never be equal to +/- 1)
then, 1+n+2 = -1; then n = -4 (even)

let m = -1
then, -1+n+2 = -1; then n = 2(even)
hence, A is sufficient.

Statement - B
m(m+n) is odd; as O * O = O
then m is odd
and m+n is odd. for m+n to be odd, n has to be even, else the summation will become even if n is odd too.
hence, B is sufficienty

Rearranging (1) we get m^2 -mn +2m =-1, so m(m-n+2) = odd, so m-n +2 must be odd since odd*odd = odd

So m-n = odd since odd + even =odd, so n even, otherwise m would be even and m(m-n+2) would be even which is not allowed. sufficient

(2) m(m+n) odd, so m odd and m+n odd since odd*odd = odd, and n even since odd + even = odd and if m even m(m+n) even. sufficient

OA is D

Hi all!

Ooops..That "seems" like a GMAT question from the tough segment with an agreed difficulty level of 700 but for someone who has a good grip on the topic of numbers from the Arithmetic this question would be at an intermediate level.
Lets zoom it a little more and we can see that the question tests on your understanding of the properties of evens and odds.

Lets summarize the properties before we deep dive into the solution!

1.An even number added or subtracted with an even and an odd number added or subtracted with an odd number will ALWAYS generate an even number whereas an even added or subtracted with an odd number will always generate an odd number.

2.Similarly, an even number multiplied with an odd or even will always generate an even number whereas an odd number multiplied with an odd number will always generate an odd number.

3. An even number divided by an even number doesnt give us any definite conclusion. For example,6 / 2 = 3 which is an odd number but 4/2 =2 which is an even number. An odd number divided by another odd number will give us an odd value as the result if the result is an integer.

Well hold on now. Knowing this properties is half the story but applying them gives you the GMAT grip on application!

Here is where we would try our hand at the properties on this question!

Since its a DS question, we shall first look at the Q. stem and see if we can break it and simplify further before we proceed to the statements.

We know m and n are positive integers from the Q. stem and so the quick inference is m and n cannot be 0 and belong to the set {1,2,3,4,......}

We are now asked if n is even? So can n be expressed as 2k where k is an integer and/or does n satisfy the properties of the ODD/EVEN nos is the alternate thought process.

Lets now strike at the statements;

St(1) says m(m + 2) + 1 = mn

There are two ways to look at the statement.

First way- Use the properties you just learnt above!

Case 1: If m is odd, then m+2 is also odd(say m=1 then m+2=1+2=3 which is odd). Hence the product m * (m+2) is odd * odd and this will give us
an odd result. m * (m+2) + 1 thus becomes odd + 1 which is even.
Since its an equation, the value mn at the right side of the equation also needs to be EVEN. This gives us an inference that n has to be EVEN for mn to be EVEN with m being odd. The definite answer to the question stem is YES. ​

Summarizing it ; [O* O ] + 1 = O + 1=mn=E and since m is Odd n has to be even

Case 2:If m is even then m+2 is also even (say m=2 then m+2=4 which is even). Hence the product m * (m+2) is even * even and this will give us an even result (following the properties explained above)
This implies m*(m+2) + 1 = even + 1 =odd
iAs its an equation,t he value mn at the right side of the equation also needs to be ODD.
This is NOT POSSIBLE with m being even and the only possibility is m is ODD as discussed in Case 1 and n is even giving us a definite YES as answer to the question stem.

So we have a definite answer to the Q.stem of a YES from statement 1.
You can right away eliminate options B,C,E as statement 1 now is sufficient.

Lets now attack statement 2.

St(2) says m(m + n) is odd.
Lets again take the cases here.

Case 1: m is even and so we have even * (even+ n) is odd which is NOT POSSIBLE as even * any value will always be even.

Case 2 : m is odd and this gives us odd * (odd + n) as odd.
This implies (odd + n ) has to be odd
This again gives us n as even as odd + even is odd.
We again have a definite answer from statement 2. So eliminate answer B and D becomes the appropriate answer choice for this question.

[Refer to attached image]

As a second approach,you can alternately look at statement one and isolate n as n is being asked in the Q.stem.
In that case,
St(1): m(m + 2) + 1 = mn

This gives us n = (m+2) + 1/m

The only possible value for 1/m to be a positive integer is when m =1.

This gives us n = (1+2)+1 = 4 and 4 is even.

So on GMAT we can also isolate variables tested and analyze their properties to save time.

Number system is a lengthy and detailed topic. So practice a lot of official questions from this topic and "think" through as you question yourself "how to i eliminate the answer choices", "which smart and simple value could i use to test" , "how do i break down the stub and make it simpler"!

As you question yourself,you shall gain confidence since you would gain clarity on what not to do!

Hope you have gained some insights here and we could be of help

Devmitra Sen
GMAT Quant SME

Stmt 1:
\(m(m + 2) + 1 = mn\)
\(m^{2}+2m+1=mn\)
\((m+1)^{2}/m=n\)

Lets assume some values for m
m=1, n=4 (even)
m=2, n=9/2 (not valid since n is an integer)
m=3, n=16/3 (not valid since n is an integer)
m=4, n=25/4 (not valid since n is an integer)
Only 1 solution, n = even
Sufficient

Stmt 2:
m(m + n) is odd.
m = odd and (m+n)=odd
Considering m+n=odd
m=1, n=2 m+n=3 odd OK
m=1, n=4, 3+4=7=odd OK
m=3 n=3, 3+3= even Not valid
So n is even
Sufficient

Ans . D

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