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M04-03

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M04-03  [#permalink]

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New post 16 Sep 2014, 00:22
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87% (00:47) correct 13% (00:47) wrong based on 153 sessions

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


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

(2) \(m\) is an even integer

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Re M04-03  [#permalink]

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New post 16 Sep 2014, 00:22
1
Official Solution:


(1) \(m-1\) and \(n+1\) are consecutive positive integers. If \(m\) were less than \(n\) than \(m-1\) (integer less than \(m\)) and \(n+1\) (integer more than \(n\)) wouldn't be consecutive, so \(m\) is greater than \(n\). Sufficient.

Or look at this in another way: stem says that the distance between m and n is 1. Now, if \(m \lt n\) 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 \(m \gt n\).

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


Answer: A
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Re M04-03  [#permalink]

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New post 31 Jan 2015, 12:35
I think this question is good and helpful.
Hi,
M an N are consecutive positive integers (like 5 and 6).
But could be 6 and 5 consecutive positive intergers?
Thanks,
Mike
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Re: M04-03  [#permalink]

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New post 01 Feb 2015, 04:44
Mikeruz wrote:
I think this question is good and helpful.
Hi,
M an N are consecutive positive integers (like 5 and 6).
But could be 6 and 5 consecutive positive intergers?
Thanks,
Mike


The order there does not matter: 5 and 6, or 6 and 5, are consecutive integers.
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M04-03  [#permalink]

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New post 24 Jun 2015, 15:57
I looked at this in the following way. I first grouped n and n+1 in a consecutive manner. Since n is positive, n+1 cannot come before n. Thus, n listed consecutively is:
n, n+1

M is similar. Since m is positive, the grouping cannot be m, m-1. So it must be:
m-1, m

Thus, since n+1 and m-1 are consecutive, the consecutive grouping must be n, n+1, m-1, m, meaning that m is always greater than m, which means (1) is sufficient.

(2) This cannot be sufficient because n is not mentioned.

I liked doing it this way, but I don't know if this would be Bunuel approved.
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Re: M04-03  [#permalink]

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New post 05 Jul 2018, 07:12
Easiest way is to give m+1 and n-1 consecutive values. Two scenarios:

1. m>n (e.g. m=3,n=2)
2. n>m (e.g. n=3, m=2)

You'll notice that when you solve both both (1) and (2) the answer is the same.
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Re: M04-03 &nbs [#permalink] 05 Jul 2018, 07:12
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