If m and n are consecutive positive integers, is m greater than n ?

Imagine the numbers on the number line extending towards infinity.
m and n are consecutive means they are right next to each other on the number line.
When both take one step back, they will still be consecutive (m-1 and n -1 will be consecutive too)
When one number takes a step forward and the other takes a step back, they are still consecutive. This will happen only if the one in front takes a step back and the one in back takes a step in front. So m-1 and n+1 will be consecutive when m was in front and n back initially. So m > n.
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SOLUTION

If m and n are consecutive positive integers, is m greater than n?

(1) m-1 and n+1 are consecutive positive integers

If \(m\) were less than \(n\), then \(m-1\) (an integer less than \(m\)) and \(n+1\) (an integer greater than \(n\)) wouldn't be consecutive. Hence, \(m\) must be greater than \(n\). Sufficient.

Or look at this in another way: the stem says that the distance between \(m\) and \(n\) is 1. Now, if \(m < n\), then the distance between \(m-1\) and \(n+1\) would be 3, and they couldn't be consecutive as (1) states. Thus, it must be true that \(m > n\).

(2) m is an even integer. Clearly insufficient.

Answer: A.
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St1: m-1 and n+1 are consecutive positive integers.

Case 1:
m is even, n is odd and m < n
Let m = 2, n = 3
m-1 = 1, n+1 = 4 -> not consecutive positive integers.

Case 2:
m is even, n is odd and m > n
Let m = 2, n = 1
m-1 = 1, n+1 = 2 -> consecutive positive integers.

Case 3:
m is odd, n is even and m < n
Let m = 3, n = 4
m-1 = 2, n+1 = 5 -> not consecutive positive integers.

Case 4:
m is odd, n is even and m > n
Let m = 3, n = 2
m-1 = 2, n+1 = 3 -> consecutive positive integers.

Not sufficient. Dow to B, C or E.

St2: m is an even integer. Not sufficient. If m = 2 and n = 1 -> m > n but if m = 2 and n = 3 -> m < n.

St 1 + St2: Not sufficient as illustrated by case 1 and case 2.

Answer (E).

I think you are making a mistake somewhere !! Well from the different cases you have illustrated it is clear that "A" is sufficient !! Because both cases where m-1,n+1 were consecutive you had m>n therefore statement "1" is clearly sufficient. Hope that helped

The question states that \(m\) &\(n\) are consecutive positive integers. The question can then be boiled down to whether the order is \(m,n\) or \(n,m\).

(1) m-1 and n+1 are consecutive positive integers.

We know that \(m-1\) and \(n+1\)are consecutive positive integers. If the order is \(m,n\) then this statement clearly fails. If the order is \(n,m\) then this statement is true, hence \(A\) is sufficient.

(2) m is an even integer.

If \(m\) is an even integer, \(n\) must be odd. \(n\) can still fall to the left or right of \(m\), and therefore statement (2) is insufficient.

Answer: A

Hi,
we take ABSOLUTE MOD because we are not aware whether x is big or y is big..
if x is bigger, x-y=1..
if y is bigger, x-y=-1..
so |x-y|=1..

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I have one small confusion. The question says that "if m and n are consecutive positive integers..." Doesn't this imply automatically that the order is m and then n? How are we unsure that its not m and n rather it could be n and m? Please help me.

Thanks!

If it were the way you are saying, we won't have this question. So, x and y are consecutive integers does NOT necessarily mean that y > x.

P.S. Also, note that this is an official question.
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VeritasKarishma chetan2u MentorTutoring nick1816



I find the highlighted part and St 1 to be conflicting (usually I take the approach if I could SOLVE q stem by using the statements)
If m and n are two consecutive no, are not m-1 and n-1 ALSO consecutive by default?

To avoid, confusion, can I safely use number picking as below:
Case 1: I take m-1 less than n+1
m-1 = 2
n+1 = 3
hence m=3, n=2 m>n

In next case, I purposely take m-1 greater than n+1
m-1= 4
n+1= 3
Hence m=6, n=2 Still m>n

Clearly St 1 is suff.

St 2 is clearly insuff. Ans: A
Can you share your two cents?
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Yes, if m and n are consecutive, m-1 and n-1 would surely be consecutive, but this is a peculiar case where m-1 and n+1 are also consecutive.
This now should tell you that n+1>m-1 or n+1=(m-1)+1...n+1=m...m>n
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Hello, adkikani. My in-line responses are below.


I do not understand the conflict. Statement (1) presents the information in a distorted way, making it such that the larger value has changed places. That is, it is clear that m is greater than n within the constraints of the question stem and this statement, but n + 1 yields the greater value than m - 1. This is the sort of confusion the GMAT™ likes to create, just enough to make test-takers doubt themselves.

This second case does not conform to the given constraints. If m and n are consecutive positive integers, then m - 1 cannot be greater than n + 1. Assume, for the sake of argument, that you had originally picked n to be the greater of the two integers. For instance, let n = 2 and m = 1.

m - 1 = 1 - 1 = 0
n + 1 = 2 + 1 = 3

These values do not conform to the information given in Statement (1), so you would know you had leaned in the wrong direction.

Yes, clearly Statement (1) is indeed SUFFICIENT, but be careful about testing, willy-nilly, whatever numbers you want, since that sort of method could lead to incorrect conclusions.

I hope that helps. Thank you for bringing my attention to the question.

- Andrew
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Question:- If m and n are consecutive positive integers, is m greater than n?
What I inferred> Question is basically asking, is M=N+1? (as M and N are consecutive, Difference between them is 1)

Statement 1: m-1 and n+1 are consecutive positive integers.

Assume; M=6 & N=5
> M-1 = N = 5
> N+1 = M = 6 ...... (Yes, M>N) .. Thus, Statement sufficient

Statement 2: m is an even integer.
>If m is an even integer, n must be odd. m can still fall to the left or right of n, and therefore statement (2) is insufficient.

Answer: A

Is my approach, Correct?
Bunuel VeritasKarishma chetan2u

Yes, though if it holds for one set of values it doesn't mean it will hold for all. Understand the logic I discussed above to know why it will hold for all.
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let's choose one random positive integer as m: exp: m=5 => n could be: 4 or 6

(1) (m-1) and (n+1) are positive consecutive integers
(m-1)= 5-1=4
#1: n=4 => n+1= 5
#2: n=6 => n+1= 6
Keep #1 (4 and 5 are consecutive positive integers) and eliminate #2 based on the (1) statement, we could conclude m>n : SUFFICIENT

(2) m is even => n must be odd. But cannot determine m is consecutively after n or not : INSUFICIENT

===> A

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