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If m and n are integers, is mn an odd integer?

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If m and n are integers, is mn an odd integer?  [#permalink]

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New post 30 Sep 2018, 23:39
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A
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E

Difficulty:

  45% (medium)

Question Stats:

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[Math Revolution GMAT math practice question]

If \(m\) and \(n\) are integers, is \(mn\) an odd integer?

1) \(m(n+1)\) is even
2) \((m+1)n\) is even

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Re: If m and n are integers, is mn an odd integer?  [#permalink]

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New post 30 Sep 2018, 23:50
From statement 1:

m(n+1) is even.
Let's consider numbers for easy simplification.
m=1 n=3 then m(n+1) is even and mn = 3. Odd
If m=2 n=4 then m(n+1) is even but mn = 8. Even.
Insufficient.

From statement 2:
Same case as above.
Insufficient.

Combining both gives 2mn+m+n as even.
2mn will always be even. So m+n has to be even. If m=1 n=3. m+n is even Or If m=2 n=4. m+n is even. But mn will be even and odd.
Hence Combining also doesn't give a unique answer.

E is the answer.
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Re: If m and n are integers, is mn an odd integer?  [#permalink]

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New post 01 Oct 2018, 01:59
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(m\) and \(n\) are integers, is \(mn\) an odd integer?

1) \(m(n+1)\) is even
2) \((m+1)n\) is even


Question: Is mn odd?
Question: Are each if m and n odd?

Statement 1: \(m(n+1)\) is even
i.e. either m is even and n+1 is even OR m is odd and n+1 is even or m and n+1 both are even
i.e. either m is even and n is odd OR m is odd and n is odd
i.e. mn may be even or mn may be Odd
NOT SUFFICIENT

Statement 2: \((m+1)n\) is even
i.e. either n is even and m+1 is even OR n is odd and m+1 is even or m and n+1 both are even
i.e. either n is even and m is odd OR n is odd and m is odd
i.e. mn may be even or mn may be Odd
NOT SUFFICIENT

Combining the two statements
i.e. either n is even and m is odd OR n is odd and m is odd
i.e. mn may be even or mn may be Odd
NOT SUFFICIENT

Answer: Option E
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Re: If m and n are integers, is mn an odd integer?  [#permalink]

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New post 01 Oct 2018, 06:03
Top Contributor
MathRevolution wrote:
[Math Revolution GMAT math practice question]

If \(m\) and \(n\) are integers, is \(mn\) an odd integer?

1) \(m(n+1)\) is even
2) \((m+1)n\) is even


Some important rules:
#1. ODD +/- ODD = EVEN
#2. ODD +/- EVEN = ODD
#3. EVEN +/- EVEN = EVEN

#4. (ODD)(ODD) = ODD
#5. (ODD)(EVEN) = EVEN
#6. (EVEN)(EVEN) = EVEN


Target question: Is mn an odd integer?

Given: m and n are integers

Statement 1: m(n+1) is even
Let's test some values.
There are several values of m and n that satisfy statement 1. Here are two:
Case a: m = 1 and n = 1. Notice that m(n + 1) = 1(1+1) = 2, which is even. In this case, mn = (1)(1) = 1. So, the answer to the target question is YES, mn IS odd
Case b: m = 2 and n = 2. Notice that m(n + 1) = 2(2+1) = 6, which is even. In this case, mn = (2)(2) = 4. So, the answer to the target question is NO, mn is NOT odd
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: (m+1)n is even
Let's test some values (again).
IMPORTANT: When testing values a second time, check to see if you ran reuse either of the cases you used in statement 1.
If we do that here, we'll see that we can reuse both cases:
Case a: m = 1 and n = 1. Notice that (m+1)n = (1+1)1 = 2, which is even. In this case, mn = (1)(1) = 1. So, the answer to the target question is YES, mn IS odd
Case b: m = 2 and n = 2. Notice that (m+1)n = (2+1)2 = 6, which is even. In this case, mn = (2)(2) = 4. So, the answer to the target question is NO, mn is NOT odd
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statements 1 and 2 combined
Notice that we were able to use the same counter-examples to show that each statement ALONE is not sufficient.
So, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: m = 1 and n = 1. So, the answer to the target question is YES, mn IS odd
Case b: m = 2 and n = 2. So, the answer to the target question is NO, mn is NOT odd
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

Cheers,
Brent
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Re: If m and n are integers, is mn an odd integer?  [#permalink]

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New post 03 Oct 2018, 01:23
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

Modifying the question:
mn is odd only when both m and n are odd. So, the question asks if both m and n are odd.

Conditions 1) and 2), when applied together, tell us that either
both m and n are odd numbers or both m and n are even numbers.

Since we don’t have a unique solution, both conditions, taken together, are not sufficient.

Therefore, E is the answer.
Answer: E
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Re: If m and n are integers, is mn an odd integer? &nbs [#permalink] 03 Oct 2018, 01:23
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