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Re M2107 [#permalink]
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16 Sep 2014, 01:10



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Re: M2107 [#permalink]
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12 Apr 2015, 00:43
Bunuel wrote: Official Solution:
If \(x\), \(y\), and \(z\) are even positive integers, each of the following must be an even integer EXCEPT:
A. \(\frac{xyx}{4}\) B. \(x^{(y + z)}\) C. \((x+y)^{z1}\) D. \(xyz\) E. \(y^{zx}\)
If \(z=x\), then option E gives: \(y^{zx}=y^0=1=odd\). All the other options are necessarily even.
Answer: E Sir Question says "Must be even integer". So either it can be odd or it can be 0 ( as it doesnt say the resultant will be positive) If we take option D i.e. xyz (let x=4 y=2 and z=2), the resultant would be 0 this also satisfies the equation in my opinion. Please clarify this doubt. Thanks



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Re: M2107 [#permalink]
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12 Apr 2015, 04:34
saharshbagaria wrote: Bunuel wrote: Official Solution:
If \(x\), \(y\), and \(z\) are even positive integers, each of the following must be an even integer EXCEPT:
A. \(\frac{xyx}{4}\) B. \(x^{(y + z)}\) C. \((x+y)^{z1}\) D. \(xyz\) E. \(y^{zx}\)
If \(z=x\), then option E gives: \(y^{zx}=y^0=1=odd\). All the other options are necessarily even.
Answer: E Sir Question says "Must be even integer". So either it can be odd or it can be 0 ( as it doesnt say the resultant will be positive) If we take option D i.e. xyz (let x=4 y=2 and z=2), the resultant would be 0 this also satisfies the equation in my opinion. Please clarify this doubt. Thanks ZERO:1. 0 is an integer. 2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.3. 0 is neither positive nor negative integer (the only one of this kind). 4. 0 is divisible by EVERY integer except 0 itself. Check more here: tipsandhintsforspecificquanttopicswithexamples172096.html#p1371030
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Re: M2107 [#permalink]
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08 Nov 2015, 20:30
Quote: If \(x\), \(y\), and \(z\) are even positive integers, each of the following must be an even integer EXCEPT:
A. \(\frac{xyx}{4}\) B. \(x^{(y + z)}\) C. \((x+y)^{z1}\) D. \(xyz\) E. \(y^{zx}\)
Hi Bunuel, I have a doubt about choice C. In case z=0, it turns out to be the reciprocal of (x+y), which may not be an even integer, isn't it? Pls clarify. I had this Q as my 1st Q in quant section in a GmatClub test. Spent 2 full minutes and got demotivated Pls help Thanks, Binit



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Re: M2107 [#permalink]
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08 Nov 2015, 21:38
binit wrote: Quote: If \(x\), \(y\), and \(z\) are even positive integers, each of the following must be an even integer EXCEPT:
A. \(\frac{xyx}{4}\) B. \(x^{(y + z)}\) C. \((x+y)^{z1}\) D. \(xyz\) E. \(y^{zx}\)
Hi Bunuel, I have a doubt about choice C. In case z=0, it turns out to be the reciprocal of (x+y), which may not be an even integer, isn't it? Pls clarify. I had this Q as my 1st Q in quant section in a GmatClub test. Spent 2 full minutes and got demotivated Pls help Thanks, Binit You should read questions more carefully: If \(x\), \(y\), and \(z\) are even POSITIVE integers... 0 is not positive.
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Re: M2107 [#permalink]
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08 Nov 2015, 22:33
Quote: You should read questions more carefully: If x, y, and z are even POSITIVE integers... 0 is not positive. Yes. I should be more careful in reading. I unnecessarily got nervous on the first Q and wasted time. Thanks so much for ur suggestion. Binit.



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HI BUNUEL ! this is my 1st post in gmatclub !.. i also answered E ; my thought process was : if z>x then option E would not be an integer is it ok ?



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Re: M2107 [#permalink]
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13 Dec 2015, 06:16



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Re: M2107 [#permalink]
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14 Dec 2016, 06:03
Hi bunuel,
My question might be ridiculous but pls. reply. If x and z are same integers then why would the author name it differently? In questions like these, i should also consider the fact that all integers might be same?



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Re: M2107 [#permalink]
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14 Dec 2016, 06:06



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M2107 [#permalink]
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Updated on: 13 Nov 2017, 09:59
...
Originally posted by vrkr on 13 Nov 2017, 09:55.
Last edited by vrkr on 13 Nov 2017, 09:59, edited 1 time in total.



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Re: M2107 [#permalink]
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13 Nov 2017, 09:58










