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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?

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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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Max@Math Revolution 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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Max@Math Revolution 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
\[m,n\,\,\,{\text{ints}}\,\,\,{\text{ }}\left( * \right)\]
\[mn\,\,\mathop = \limits^? \,\,{\text{odd}}\,\,\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( * \right)} \,\,\,\,\,\boxed{\,\,\,?\,\,\,\,:\,\,\,\,\,m,n\,\,\,{\text{both}}\,\,{\text{odd}}\,\,\,}\]
\[\left( {1 + 2} \right)\,\,\,\,\left\{ \begin{gathered}
{\text{Take}}\,\,\left( {m,n} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\
{\text{Take}}\,\,\left( {m,n} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\,\,\, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\text{INSUFF}}{\text{.}}\]


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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=>

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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Only $99 for 3 month Online Course
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Unlimited Access to over 120 free video lessons-try it yourself
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