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Given that m and n are positive integers such that mn − n − 3m

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Given that m and n are positive integers such that mn − n − 3m  [#permalink]

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New post Updated on: 06 Jan 2020, 04:59
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Question Stats:

69% (02:30) correct 31% (02:17) wrong based on 36 sessions

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Given that m and n are positive integers such that mn − n − 3m is even, which of the following MUST be odd?

A. mn − n − 1

B. 2m + 3n

C. (n − 3) + (m − 1)

D. m + n − 3m

E. 3nm




Correct answer is A. Can you please explain how to solve these type of questions where there are two variables (Odd and Even) questions.

Answer Explanation:

The entries should show whether the expression mn − n − 3m is even or odd for each of the four conditions shown above.
As you can see, the expression (mn − n − 3m ) is only even when both m and n are even.

So you have found out that both m and n are even. Now, plug in even values of m and n into each choice to see which one results in an odd number.

A : mn − n − 1= Odd. As both m and n are even, their product is even, and when an even number (n) is subtracted from another even number (mn), the result is even. When an odd number (1) is subtracted from an even number (mn − n), the result is ODD.

You have found the correct answer. For completeness’s sake, the following choices are discussed.

B : 2m + 3n = Even. As both m and n are even, any integer multiples of these numbers will be even and the sum of two even numbers is also even. Eliminate.

C : (n − 3) + (m − 1) = n + m − 3 + 1 = n + m − 2 = Even. As both m and n are even, m + n MUST be even. An even number (2) subtracted from another even number (n + m) equals an even number. Eliminate.

D : m + n −3m = Even. Since (1 − 3)m=−2m is even as m is even, and n is also even, the sum of two even numbers (n and −2m) is also even. Eliminate.

E : 3nm = Even. As both m and n are even, their product is even, and any even number multiplied by any integer is also even. Eliminate.[

Originally posted by arsalansyed21 on 05 Jan 2020, 23:06.
Last edited by Bunuel on 06 Jan 2020, 04:59, edited 1 time in total.
Edited the question.
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Re: Given that m and n are positive integers such that mn − n − 3m  [#permalink]

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New post 06 Jan 2020, 03:35
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Let's try and simplify the question
mn − n − 3m is even
=> mn - (n+3m) is Even
now, difference of two terms mn and (n+3m) is even => Either both the terms are even or both are odd
Case 1: Both are Even
mn = even and n+3m = even
if any one of m and n is odd then n+3m will become odd so not possible
=> both n and m are even

Case 2: Both are odd
if mn=odd => both m and n are odd, but then n+3m will become even, so not possible
From both cases => Both n and m are even

A. Even - Even - Odd = Odd
B. Even + Even = Even
C. (Even - Odd) + (Even - Odd) = Odd + Odd = Even
D. Even + Even - Even = Even
E. Odd*Even*Even = Even
So, Answer will be A
Hope it helps!

arsalansyed21 wrote:
Given that m and n are positive integers such that mn − n − 3m is even, which of the following MUST be odd?


A- mn − n − 1

B- 2m + 3n

C- (n − 3) + (m − 1)

D- m + n − 3m

E- 3nm




Correct answer is A. Can you please explain how to solve these type of questions where there are two variables (Odd and Even) questions.

Answer Explanation:

The entries should show whether the expression mn − n − 3m is even or odd for each of the four conditions shown above.
As you can see, the expression (mn − n − 3m ) is only even when both m and n are even.

So you have found out that both m and n are even. Now, plug in even values of m and n into each choice to see which one results in an odd number.

A : mn − n − 1= Odd. As both m and n are even, their product is even, and when an even number (n) is subtracted from another even number (mn), the result is even. When an odd number (1) is subtracted from an even number (mn − n), the result is ODD.

You have found the correct answer. For completeness’s sake, the following choices are discussed.

B : 2m + 3n = Even. As both m and n are even, any integer multiples of these numbers will be even and the sum of two even numbers is also even. Eliminate.

C : (n − 3) + (m − 1) = n + m − 3 + 1 = n + m − 2 = Even. As both m and n are even, m + n MUST be even. An even number (2) subtracted from another even number (n + m) equals an even number. Eliminate.

D : m + n −3m = Even. Since (1 − 3)m=−2m is even as m is even, and n is also even, the sum of two even numbers (n and −2m) is also even. Eliminate.

E : 3nm = Even. As both m and n are even, their product is even, and any even number multiplied by any integer is also even. Eliminate.[

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Re: Given that m and n are positive integers such that mn − n − 3m  [#permalink]

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New post 06 Jan 2020, 23:15
arsalansyed21 wrote:
Given that m and n are positive integers such that mn − n − 3m is even, which of the following MUST be odd?

A. mn − n − 1

B. 2m + 3n

C. (n − 3) + (m − 1)

D. m + n − 3m

E. 3nm




Correct answer is A. Can you please explain how to solve these type of questions where there are two variables (Odd and Even) questions.

Answer Explanation:

The entries should show whether the expression mn − n − 3m is even or odd for each of the four conditions shown above.
As you can see, the expression (mn − n − 3m ) is only even when both m and n are even.

So you have found out that both m and n are even. Now, plug in even values of m and n into each choice to see which one results in an odd number.

A : mn − n − 1= Odd. As both m and n are even, their product is even, and when an even number (n) is subtracted from another even number (mn), the result is even. When an odd number (1) is subtracted from an even number (mn − n), the result is ODD.

You have found the correct answer. For completeness’s sake, the following choices are discussed.

B : 2m + 3n = Even. As both m and n are even, any integer multiples of these numbers will be even and the sum of two even numbers is also even. Eliminate.

C : (n − 3) + (m − 1) = n + m − 3 + 1 = n + m − 2 = Even. As both m and n are even, m + n MUST be even. An even number (2) subtracted from another even number (n + m) equals an even number. Eliminate.

D : m + n −3m = Even. Since (1 − 3)m=−2m is even as m is even, and n is also even, the sum of two even numbers (n and −2m) is also even. Eliminate.

E : 3nm = Even. As both m and n are even, their product is even, and any even number multiplied by any integer is also even. Eliminate.[


Quote:
Can you please explain how to solve these type of questions where there are two variables (Odd and Even) questions.


Given: mn − n − 3m = Even
n(m - 1) - 3m = Even
We have two terms whose difference is even. So either both should be odd or both should be even.
If m is odd so that 3m is odd, (m - 1) will be even which makes the first term even. So this is not possible. Then m must be even and (m - 1) must be odd.
To make n(m - 1) even, then n must be even too.

So what do we know for sure? That both m and n are even. (and this was the toughest part of the question. Rest is easy)

A. mn − n − 1
- mn and n both are even. Even - 1 = Odd. So this MUST be odd.

B. 2m + 3n
- Even + even = even

C. (n − 3) + (m − 1) = n + m - 4
Both n and m are even; this is even.

D. m + n − 3m
- Both m and n are even; this will be even

E. 3nm
- Must be even since both m and n are even

Answer (A)
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Re: Given that m and n are positive integers such that mn − n − 3m   [#permalink] 06 Jan 2020, 23:15
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