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29 Jun 2023, 06:00
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
59% (01:56) correct
41%
(02:05)
wrong
based on 138
sessions
History
Date
Time
Result
Not Attempted Yet
If \(m\) and \(n\) are integers, is \(mn\) an odd integer?
(1) \(m(n+1)\) is even
(2) \((m+1)n\) is even
(1) \(m(n+1)\) is even
(2) \((m+1)n\) is even
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
29 Jun 2023, 13:29
Kudos
Bookmarks
Bunuel
Statement 1
(1) \(m(n+1)\) is even
Case 1: m = even; n = odd / even
Is \(mn\) an odd integer → No
Case 2: m = odd; n = odd
Is \(mn\) an odd integer → Yes
As we have two contradicting answers, the statement alone is not sufficient. We can eliminate A and D.
Statement 2
(2) \((m+1)n\) is even
Case 1: n = even; m = odd / even
Is \(mn\) an odd integer → No
Case 2: m = odd; n = odd
Is \(mn\) an odd integer → Yes
As we have two contradicting answers, the statement alone is not sufficient. We can eliminate B.
Combined
Case 1: n = even; m = even
- \(m(n+1)\) is even
- \((m+1)n\) is even
Is \(mn\) an odd integer → No
Case 2: n = odd; m = odd
- \(m(n+1)\) is even
- \((m+1)n\) is even
Is \(mn\) an odd integer → Yes
The statements combined are not sufficient.
Option E.
08 Aug 2023, 04:17
Kudos
Bookmarks
Bunuel
Solution:
Pre Analysis:
- \(m\) and \(n\) are integers
- We are asked if \(mn\) is an odd integer or not
Statement 1: \(m(n+1)\) is even
- According to this statement, \(mn+m=even\)
- Or, \(mn=even-m\)
- If \(m=even\), then \(mn=even-even=even\)
- Else if \(m=odd\), then \(mn=even-odd=odd\)
- So, it depends on the even/odd nature of m which we don't know
- Thus, statement 1 alone is not sufficient and we can eliminate options A and D
Statement 2: \((m+1)n\) is even
- According to this statement, \(mn+n=even\)
- Or, \(mn=even-n\)
- If \(n=even\), then \(mn=even-even=even\)
- Else if \(n=odd\), then \(mn=even-odd=odd\)
- So, it depends on the even/odd nature of n which we don't know
- Thus, statement 2 alone is also not sufficient and we can eliminate option B
Combining:
- Even after combining, we do not get the even/odd nature of m or n
Hence the right answer is Option E
