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M21-07

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Math Expert
Joined: 02 Sep 2009
Posts: 42597

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16 Sep 2014, 00:10
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Question Stats:

50% (00:56) correct 50% (00:59) wrong based on 129 sessions

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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. $$(x+y)^{z-1}$$
D. $$x-y-z$$
E. $$y^{z-x}$$
[Reveal] Spoiler: OA

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Kudos [?]: 135562 [1], given: 12699

Math Expert
Joined: 02 Sep 2009
Posts: 42597

Kudos [?]: 135562 [1], given: 12699

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16 Sep 2014, 00:10
1
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Expert's post
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. $$(x+y)^{z-1}$$
D. $$x-y-z$$
E. $$y^{z-x}$$

If $$z=x$$, then option E gives: $$y^{z-x}=y^0=1=odd$$. All the other options are necessarily even.

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Joined: 28 Sep 2014
Posts: 24

Kudos [?]: 4 [0], given: 43

Schools: HBS '19

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11 Apr 2015, 23:43
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. $$(x+y)^{z-1}$$
D. $$x-y-z$$
E. $$y^{z-x}$$

If $$z=x$$, then option E gives: $$y^{z-x}=y^0=1=odd$$. All the other options are necessarily even.

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. x-y-z (let x=4 y=2 and z=2), the resultant would be 0
this also satisfies the equation in my opinion.

Kudos [?]: 4 [0], given: 43

Math Expert
Joined: 02 Sep 2009
Posts: 42597

Kudos [?]: 135562 [1], given: 12699

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12 Apr 2015, 03:34
1
KUDOS
Expert's post
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. $$(x+y)^{z-1}$$
D. $$x-y-z$$
E. $$y^{z-x}$$

If $$z=x$$, then option E gives: $$y^{z-x}=y^0=1=odd$$. All the other options are necessarily even.

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. x-y-z (let x=4 y=2 and z=2), the resultant would be 0
this also satisfies the equation in my opinion.

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: tips-and-hints-for-specific-quant-topics-with-examples-172096.html#p1371030
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Kudos [?]: 135562 [1], given: 12699

Manager
Joined: 02 Nov 2014
Posts: 215

Kudos [?]: 136 [0], given: 75

GMAT Date: 08-04-2015

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08 Nov 2015, 19:30
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. $$(x+y)^{z-1}$$
D. $$x-y-z$$
E. $$y^{z-x}$$

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

Kudos [?]: 136 [0], given: 75

Math Expert
Joined: 02 Sep 2009
Posts: 42597

Kudos [?]: 135562 [0], given: 12699

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08 Nov 2015, 20:38
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. $$(x+y)^{z-1}$$
D. $$x-y-z$$
E. $$y^{z-x}$$

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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Kudos [?]: 135562 [0], given: 12699

Manager
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Posts: 215

Kudos [?]: 136 [0], given: 75

GMAT Date: 08-04-2015

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08 Nov 2015, 21:33
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.

Kudos [?]: 136 [0], given: 75

Intern
Joined: 13 Oct 2014
Posts: 2

Kudos [?]: 3 [0], given: 13

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13 Dec 2015, 05:05
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 ?

Kudos [?]: 3 [0], given: 13

Math Expert
Joined: 02 Sep 2009
Posts: 42597

Kudos [?]: 135562 [0], given: 12699

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13 Dec 2015, 05:16
priya1987 wrote:
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 ?

______________
Yes, that;s correct.
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Kudos [?]: 135562 [0], given: 12699

Intern
Joined: 18 Nov 2013
Posts: 28

Kudos [?]: 16 [0], given: 11

Location: India
WE: Engineering (Transportation)

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14 Dec 2016, 05:03
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?

Kudos [?]: 16 [0], given: 11

Math Expert
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Posts: 42597

Kudos [?]: 135562 [0], given: 12699

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14 Dec 2016, 05:06
Klenex wrote:
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?

Unless it is explicitly stated otherwise, different variables CAN represent the same number.
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13 Nov 2017, 08:55
...

Last edited by vrkr on 13 Nov 2017, 08:59, edited 1 time in total.

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Math Expert
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Kudos [?]: 135562 [0], given: 12699

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13 Nov 2017, 08:58
vrkr wrote:
I guess the answer is A. if x=2, y=6, z=10 (non-multiples of 4), then (x+y+z)/4 = 18/4 = 4.5 which is not an integer.

The correct answer is E, not A.

Option A is xyz/4, NOT (x+y+z)/4.
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