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# I am not able to find out whether y is perfect square. Is

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Senior Manager
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I am not able to find out whether y is perfect square. Is [#permalink]  22 Dec 2003, 08:12
I am not able to find out whether y is perfect square. Is the answer E ?
Director
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[#permalink]  22 Dec 2003, 08:37
to be continuous perfect sq, ain't there sq roots need to be continuous intergers too?

D ??
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[#permalink]  22 Dec 2003, 08:52
dj wrote:
to be continuous perfect sq, ain't there sq roots need to be continuous intergers too?

D ??

Can you please give example and elaborate ?
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[#permalink]  22 Dec 2003, 08:53
B

Actually I put answer to wrong question before. The answer shouldn't be C.

(1) More than one value of x and y that satisfy this condition.
(2) Way way many values of x and y that satisfy this condition.
(combined) x = 25, y = 16 (just another subset of what Stmnt2 gives us, so Stmnt2 is sufficient)

Last edited by wonder_gmat on 22 Dec 2003, 09:59, edited 1 time in total.
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[#permalink]  22 Dec 2003, 09:05
oops.. m sorry.
this should be C only.
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[#permalink]  22 Dec 2003, 09:47
I get B

1) rearrange as z^2+y=8z+1
=z^2-8z+y-1=0 = z(z-8)+y-1=0

choose a few numbers starting with x
x=100; therefore z=10 thus: 10(2)+21-1=0 y = 21 and x=100 (no)
x=25 z=5 5(-3)+16-1=0 y=16 x=25 (yes)

Insufficient

2) x-y=2z-1

rewrite: z^2-2z-y+1=0

z(z-2)-y+1=0
x=121 z=11; thus y=100 (yes consecutive)
x= 100 z =10 y=81 (yes consecutive)
x=16 z=4 y= 9 (yes consecutive)
x=81 z=9 y=64 (yes x and y are consecutive square numbers)
x=4 z=2 y=-1 (doesn't apply b/c x,y,z are all positive integers)

Sufficient

B
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[#permalink]  22 Dec 2003, 10:37
i need to do this from the beginning...

A) in terms of Z:
Z^2 - 8Z - 1 = -Y
for X=25, Z=5 --> Y = 16. Yes
for X=6561, Z=81 --> Y = -5812. nope
for X=0, Z=0 --> Y = 1. Yes

A, not sufficient.

B) Z^2 -2Z +1 = Y
(Z-1)^2 = Y
for X=25, Z=5 --> Y = 16. Yes
for X=6561, Z=81 --> Y = 6400. Yes
for X=0, Z=0 --> Y = 1. Yes

B suffice.
furthermore, (sqrt(X)-1)^2 = Y says it all, ain't it?? X and Y will always be consecutive.

ans, B

....it's difficult to work on, this kinda stuff, from office

Last edited by dj on 22 Dec 2003, 11:03, edited 1 time in total.
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[#permalink]  22 Dec 2003, 10:48
In MO when x=4,z=2,y=1 not -1 since after transformation of b) (z-1)^2=y and when z=2 then y=1. 1 raised to any power is 1 so 1 is always perfect square. Can we assume that in this case B) is correct?
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[#permalink]  22 Dec 2003, 10:58
BG wrote:
In MO when x=4,z=2,y=1 not -1 since after transformation of b) (z-1)^2=y and when z=2 then y=1. 1 raised to any power is 1 so 1 is always perfect square. Can we assume that in this case B) is correct?

yes, (sqrt(X)-1)^2 or (Z-1)^2 = Y, will hold good, always.
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[#permalink]  22 Dec 2003, 11:27
dj wrote:
i need to do this from the beginning...

A) in terms of Z:
Z^2 - 8Z - 1 = -Y
for X=25, Z=5 --> Y = 16. Yes
for X=6561, Z=81 --> Y = -5812. nope
for X=0, Z=0 --> Y = 1. Yes

A, not sufficient.

B) Z^2 -2Z+1 = Y
(Z-1)^2 = Y
for X=25, Z=5 --> Y = 16. Yes
for X=6561, Z=81 --> Y = 6400. Yes
for X=0, Z=0 --> Y = 1. Yes

B suffice.
furthermore, (sqrt(X)-1)^2 = Y says it all, ain't it?? X and Y will always be consecutive.

ans, B

....it's difficult to work on, this kinda stuff, from office

x y z are positive integers.

and x>y

I think its D.

the only solution is x = 25 , y =16
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[#permalink]  22 Dec 2003, 11:31
Titleist wrote:
I get B

1) rearrange as z^2+y=8z+1
=z^2-8z+y-1=0 = z(z-8)+y-1=0

choose a few numbers starting with x
x=100; therefore z=10 thus: 10(2)+21-1=0 y = 21 and x=100 (no)
x=25 z=5 5(-3)+16-1=0 y=16 x=25 (yes)

Insufficient

2) x-y=2z-1

rewrite: z^2-2z-y+1=0

z(z-2)-y+1=0
x=121 z=11; thus y=100 (yes consecutive)
x= 100 z =10 y=81 (yes consecutive)
x=16 z=4 y= 9 (yes consecutive)
x=81 z=9 y=64 (yes x and y are consecutive square numbers)
x=4 z=2 y=-1 (doesn't apply b/c x,y,z are all positive integers)

Sufficient

B

titliest,

could you check the bold part. the solution does not equal 0.

x= 25 , y =16 is the only solution.
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[#permalink]  22 Dec 2003, 11:38
praetorian123 wrote:
Titleist wrote:
I get B

1) rearrange as z^2+y=8z+1
=z^2-8z+y-1=0 = z(z-8)+y-1=0

choose a few numbers starting with x
x=100; therefore z=10 thus: 10(2)+21-1=0 y = 21 and x=100 (no)
x=25 z=5 5(-3)+16-1=0 y=16 x=25 (yes)

Insufficient

2) x-y=2z-1

rewrite: z^2-2z-y+1=0

z(z-2)-y+1=0
x=121 z=11; thus y=100 (yes consecutive)
x= 100 z =10 y=81 (yes consecutive)
x=16 z=4 y= 9 (yes consecutive)
x=81 z=9 y=64 (yes x and y are consecutive square numbers)
x=4 z=2 y=-1 (doesn't apply b/c x,y,z are all positive integers)

Sufficient

B

titliest,

could you check the bold part. the solution does not equal 0.

x= 25 , y =16 is the only solution.

OOPS! Yes - you're right Praetorian = X=25, Y=16 is the only solution - I agree with DJ - it is tough doing this in the office.

I stand corrected again! The answer is still B - I was right by being wrong! As DJ pointed out, X=36, Z=6, Y=13 (This is a solution but X and Y are not consecutive squares).

Last edited by Titleist on 22 Dec 2003, 12:10, edited 2 times in total.
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[#permalink]  22 Dec 2003, 11:42
But X=25, Y=16 is solution to the equations of st1) and st2) , shouldn't the answer be C?
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[#permalink]  22 Dec 2003, 11:43
what if X=36
Z=6

Z^2 - 8Z - 1 = -Y
36-48-1=-Y
Y=13
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[#permalink]  22 Dec 2003, 11:44
BG wrote:
But X=25, Y=16 is solution to the equations of st1) and st2) , shouldn't the answer be C?

this, actually, is the first hit. but, without using two eqs also you can solve the problem.
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[#permalink]  22 Dec 2003, 12:24
BG wrote:
But X=25, Y=16 is solution to the equations of st1) and st2) , shouldn't the answer be C?

No. This is a "Yes/No" problem. You have both yes and no for (1) and only yes for (2).
[#permalink] 22 Dec 2003, 12:24
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