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M19-26

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M19-26 [#permalink]

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Official Solution:


Statement (1) by itself is insufficient. Both \(\{-1, -3, -5, -7\}\) and \(\{1, 3, 5, 7\}\) are possible sets.

Statement (2) by itself is sufficient. The median of four consecutive odd integers always equals their mean, therefore, the sum of the integers is \(4*4 = 16\).


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New post 10 Nov 2016, 14:20
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hi,

Are negative integers considered for odd and even? I have this doubt because Prime numbers don not include negative numbers, and i believe that same rule holds for odd/even as well. Please help me with this. Thanks.

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New post 10 Nov 2016, 19:20
@Saurav Arora
(1). It is not written in the question that we have to take prime numbers.
(2). 3, 5 and 7 are prime numbers, but 1 is not a prime number.
(3). These terms exist, such as negative odds/evens and positive odds/evens. So, it is right.

I hope you'll find this useful.

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M19-26 [#permalink]

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New post 28 Aug 2017, 17:15
Bunuel wrote:
Official Solution:

Statement (2) by itself is sufficient. The median of four consecutive odd integers always equals their mean, therefore, the sum of the integers is \(4*4 = 16\).



Can someone explain why the bolded part is true? I tried using numbers and it is indeed the case, but I am curious as to why it's true and how I can predict similar patterns.

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M19-26 [#permalink]

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New post 12 Sep 2017, 12:09
mba.2020 wrote:
Bunuel wrote:
Official Solution:

Statement (2) by itself is sufficient. The median of four consecutive odd integers always equals their mean, therefore, the sum of the integers is \(4*4 = 16\).



Can someone explain why the bolded part is true? I tried using numbers and it is indeed the case, but I am curious as to why it's true and how I can predict similar patterns.


The median of 4 terms will be average of 2nd and 3rd term.
Since the numbers are evenly spaced i.e. consecutive odd numbers, average of middle 2 numbers or last 2 terms will give the mean hence, it is same as median.

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M19-26   [#permalink] 12 Sep 2017, 12:09
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