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# x,y,z are consecutive integers (x<y<z). Which of the following

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Intern
Joined: 25 Oct 2016
Posts: 6
x,y,z are consecutive integers (x<y<z). Which of the following  [#permalink]

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02 Oct 2018, 05:40
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75% (hard)

Question Stats:

49% (01:53) correct 51% (01:26) wrong based on 55 sessions

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x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of $$z^2 - y^2 - x^2?$$

A. -12
B. -6
C. 0
D. 3
E. 4
Math Expert
Joined: 02 Aug 2009
Posts: 7100
Re: x,y,z are consecutive integers (x<y<z). Which of the following  [#permalink]

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02 Oct 2018, 05:53
1
trankimphuong wrote:
x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of $$z^2 - y^2 - x^2?$$

A. -12
B. -6
C. 0
D. 3
E. 4

$$z^2 - y^2 - x^2=(x+2)^2-(x+1)^2-x^2=x^2+4x+4-x^2-2x-1-x^2=2x-x^2+3=x(2-x)+3$$
let us see if the choices can fit in this equation..

A. -12.........$$x(2-x)+3=-12.....x(2-x)=-15 = -(3*5).. x = 5$$ YES
B. -6.........$$x(2-x)+3=-6.....x(2-x)=-9$$ x(2-x)=-{x(x-2)} so a multiple of two numbers with difference 2. can we write 9 as product of two such numbers..NO
C. 0.........$$x(2-x)+3=0.....x(2-x)=-3 = -(3*1).. x = 3$$ YES
D. 3.........$$x(2-x)+3=3.....x(2-x)=0 = -(2*0).. x = 2$$ YES
E. 4.........$$x(2-x)+3=4.....x(2-x)=1 = -(1*(-1)).. x = 1$$ YES

B
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x,y,z are consecutive integers (x<y<z). Which of the following  [#permalink]

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02 Oct 2018, 05:57
1
trankimphuong wrote:
x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of $$z^2 - y^2 - x^2?$$

A. -12
B. -6
C. 0
D. 3
E. 4

Another method to solve is going by process of elimination(substituting values for x,y,z s.t x<y<z)

A. -12 is possible for consecutive integers -3,-2, and -1
C. 0 is possible for consecutive integers -1,0, and 1
D. 3 is possible for consecutive integers 0,1, and 2
E. 4 is possible for consecutive integers 1,2, and 3

Therefore, Option B(-6) cannot be the value of $$z^2 - y^2 - x^2$$
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Director
Joined: 19 Oct 2013
Posts: 509
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
x,y,z are consecutive integers (x<y<z). Which of the following  [#permalink]

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02 Oct 2018, 12:25
Let x be first integer
y = x + 1
z = x + 2

So z^2 - y^2 - x^2 = x^2 +4x + 4 -(x^2 +2x + 1) - x^2

This would be = 2x + 3 - x^2
2x - x^2
X (2 - x ) + 3

Now if we put in values for x

X = 0 then the equation = 3 possible
X = 1 then the equation = 4 possible
X = 2 then the equation = 3 possible
X= 3 then the equation = 0 possible
X = 4 then the equation = - 5 possible
X = 5 then the equation = -12 possible

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VP
Joined: 09 Mar 2016
Posts: 1208
Re: x,y,z are consecutive integers (x<y<z). Which of the following  [#permalink]

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02 Oct 2018, 12:51
chetan2u wrote:
trankimphuong wrote:
x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of $$z^2 - y^2 - x^2?$$

A. -12
B. -6
C. 0
D. 3
E. 4

$$z^2 - y^2 - x^2=(x+2)^2-(x+1)^2-x^2=x^2+4x+4-x^2-2x-1-x^2=2x-x^2+3=x(2-x)+3$$
let us see if the choices can fit in this equation..

A. -12.........$$x(2-x)+3=-12.....x(2-x)=-15 = -(3*5).. x = 5$$ YES
B. -6.........$$x(2-x)+3=-6.....x(2-x)=-9$$ x(2-x)=-{x(x-2)} so a multiple of two numbers with difference 2. can we write 9 as product of two such numbers..NO
C. 0.........$$x(2-x)+3=0.....x(2-x)=-3 = -(3*1).. x = 3$$ YES
D. 3.........$$x(2-x)+3=3.....x(2-x)=0 = -(2*0).. x = 2$$ YES
E. 4.........$$x(2-x)+3=4.....x(2-x)=1 = -(1*(-1)).. x = 1$$ YES

B

hello chetan2u i didnt get how did you get from this $$z^2 - y^2 - x^2$$ you got this $$(x+2)^2-(x+1)^2-x^2$$

I didnt get why only xs i see whereas there is also $$z$$ and $$y$$ ? and where did you get +2 and +1 ? ;?

pls explain
thanks!
Director
Joined: 19 Oct 2013
Posts: 509
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
Re: x,y,z are consecutive integers (x<y<z). Which of the following  [#permalink]

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02 Oct 2018, 13:04
1
dave13 wrote:
chetan2u wrote:
trankimphuong wrote:
x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of $$z^2 - y^2 - x^2?$$

A. -12
B. -6
C. 0
D. 3
E. 4

$$z^2 - y^2 - x^2=(x+2)^2-(x+1)^2-x^2=x^2+4x+4-x^2-2x-1-x^2=2x-x^2+3=x(2-x)+3$$
let us see if the choices can fit in this equation..

A. -12.........$$x(2-x)+3=-12.....x(2-x)=-15 = -(3*5).. x = 5$$ YES
B. -6.........$$x(2-x)+3=-6.....x(2-x)=-9$$ x(2-x)=-{x(x-2)} so a multiple of two numbers with difference 2. can we write 9 as product of two such numbers..NO
C. 0.........$$x(2-x)+3=0.....x(2-x)=-3 = -(3*1).. x = 3$$ YES
D. 3.........$$x(2-x)+3=3.....x(2-x)=0 = -(2*0).. x = 2$$ YES
E. 4.........$$x(2-x)+3=4.....x(2-x)=1 = -(1*(-1)).. x = 1$$ YES

B

hello chetan2u i didnt get how did you get from this $$z^2 - y^2 - x^2$$ you got this $$(x+2)^2-(x+1)^2-x^2$$

I didnt get why only xs i see whereas there is also $$z$$ and $$y$$ ? and where did you get +2 and +1 ? ;?

pls explain
thanks!

Hi dave I’ll give my self the liberty to answer your question.

It says x,y,z are consecutive integers and (x<y<z)

If x is the least of these 3 consecutive integers

Then y = x + 1
And z = x + 2

He just changed the variables in a way to make it simpler to deal with.

I hope this clarifies your doubt.

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VP
Joined: 09 Mar 2016
Posts: 1208
x,y,z are consecutive integers (x<y<z). Which of the following  [#permalink]

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02 Oct 2018, 13:06
pushpitkc wrote:
trankimphuong wrote:
x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of $$z^2 - y^2 - x^2?$$

A. -12
B. -6
C. 0
D. 3
E. 4

Another method to solve is going by process of elimination(substituting values for x,y,z s.t x<y<z)

A. -12 is possible for consecutive integers -3,-2, and -1
C. 0 is possible for consecutive integers -1,0, and 1
D. 3 is possible for consecutive integers 0,1, and 2
E. 4 is possible for consecutive integers 1,2, and 3

Therefore, Option B(-6) cannot be the value of $$z^2 - y^2 - x^2$$

hey pushpitkc
Why do you pick negative numbers ? when squared negative number turns positive.
thanks

may be you know Salsanousi
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3325
Location: India
GPA: 3.12
Re: x,y,z are consecutive integers (x<y<z). Which of the following  [#permalink]

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02 Oct 2018, 23:10
dave13 wrote:
pushpitkc wrote:
trankimphuong wrote:
x,y,z are consecutive integers (x<y<z). Which of the following cannot be the value of $$z^2 - y^2 - x^2?$$

A. -12
B. -6
C. 0
D. 3
E. 4

Another method to solve is going by process of elimination(substituting values for x,y,z s.t x<y<z)

A. -12 is possible for consecutive integers -3,-2, and -1
C. 0 is possible for consecutive integers -1,0, and 1
D. 3 is possible for consecutive integers 0,1, and 2
E. 4 is possible for consecutive integers 1,2, and 3

Therefore, Option B(-6) cannot be the value of $$z^2 - y^2 - x^2$$

hey pushpitkc
Why do you pick negative numbers ? when squared negative number turns positive.
thanks

may be you know Salsanousi

Hey dave13

The reason I used negative numbers is that I needed a negative answer.
That was possible only if the magnitude of x was greater than y and y<z.

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Re: x,y,z are consecutive integers (x<y<z). Which of the following &nbs [#permalink] 02 Oct 2018, 23:10
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