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Consequtive integers

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Source: — Data Sufficiency |
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by ssmiles08 » Wed Jun 17, 2009 11:56 am
IMO A.

Working backwards:

choice 2 is clearly insufficient. m can be 2 and n can be 3 or m can be 4 and n can be 3.

Choice 1 tells us that m - 1 and n + 1 are consecutive. The only way this could be possible if n becomes m and m becomes n.

lets take m = 10, n = 9

m-1 = 9
n+1 = 10

now we have 9 and 10 which are still consecutive integers. M has to be > n b/c it moves backwards one unit, otherwise it will move two units forward which will not make m and n consecutive.

(A)
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by abhinav85 » Wed Jun 17, 2009 11:58 am
Hey Ssmiles

lets take m = 0, n = -1

m-1 = -1
n+1 = 0

the question says "If m and n are consecutive positive integers, is m greater than n?
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by ssmiles08 » Wed Jun 17, 2009 12:02 pm
abhinav85 wrote:Hey Ssmiles

lets take m = 0, n = -1

m-1 = -1
n+1 = 0

the question says "If m and n are consecutive positive integers, is m greater than n?
yes abhinav85,

I changed my numbers before you could post your post. :) It still works out to be the same though.
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by abhinav85 » Wed Jun 17, 2009 12:11 pm
Ohh.......U changed :!:
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Is this correct

by rah_pandey » Thu Jun 18, 2009 1:10 am
m and n are consecutive

let m<n

m-1+1=n+1
m-0=n+1
=>n=m-1=>m-n=1 a contradiction
since m<n is assumption

let m>n
n+1=m-1-1=n+1
n=m-3
m-n=3
correct...

therefore A is sufficient

if m is even=> n is odd

cannot say which is greater

therefore A
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