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

Expert replies
Source: — Data Sufficiency |

by Jay@ManhattanReview » Mon Dec 30, 2019 10:44 pm
BTGmoderatorDC wrote:If m and n are positive integers, is n even?

(1) m(m + 2) + 1 = mn
(2) m(m + n) is odd.

OA D

Source: Official Guide
Let's take each statement one by one.

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

Say m = even, thus, mn, the right-hand side = even, thus, m(m + 2) + 1, the left-hand side is even. Or m(m + 2) = Even - 1 = Odd. We know that the product of two odd integers is odd, thus, for m(m + 2) to be odd, m as well as (m + 2) must be odd; however, it invalidates our assumption that m is even. Thus, must be odd.

With m = odd, we have m(m + 2) = Odd and m(m + 2) + 1 = m(m + 2) + Odd = Even. Thus, mn, the right-hand side = Even. Since m is odd, for mn to be even, n must be even. Sufficient.

(2) m(m + n) is odd.

We know that the product of two odd integers is odd, thus, for m(m + n) to be odd, m as well as (m + n) must be odd. This is the same condition that we discussed in Statement 1. Sufficient.

The correct answer: D

Hope this helps!

-Jay
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by GMATGuruNY » Wed Jan 01, 2020 10:33 pm
BTGmoderatorDC wrote:If m and n are positive integers, is n even?

(1) m(m + 2) + 1 = mn
(2) m(m + n) is odd.
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 = 9/2 = ODD/EVEN
Not viable, since 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.

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.

The correct answer is D.
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