Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 25 May 2017, 05:16

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Senior Manager
Joined: 05 May 2003
Posts: 424
Location: Aus
Followers: 2

Kudos [?]: 10 [0], given: 0

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

### Show Tags

22 Dec 2003, 09:12
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I am not able to find out whether y is perfect square. Is the answer E ?
Director
Joined: 13 Nov 2003
Posts: 961
Location: Florida
Followers: 1

Kudos [?]: 144 [0], given: 0

### Show Tags

22 Dec 2003, 09:37
to be continuous perfect sq, ain't there sq roots need to be continuous intergers too?

D ??
Senior Manager
Joined: 05 May 2003
Posts: 424
Location: Aus
Followers: 2

Kudos [?]: 10 [0], given: 0

### Show Tags

22 Dec 2003, 09:52
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 ?
Manager
Joined: 26 Aug 2003
Posts: 233
Location: United States
Followers: 1

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

### Show Tags

22 Dec 2003, 09:53
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, 10:59, edited 1 time in total.
Director
Joined: 13 Nov 2003
Posts: 961
Location: Florida
Followers: 1

Kudos [?]: 144 [0], given: 0

### Show Tags

22 Dec 2003, 10:05
oops.. m sorry.
this should be C only.
Director
Joined: 14 Oct 2003
Posts: 583
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 53 [0], given: 0

### Show Tags

22 Dec 2003, 10:47
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
Director
Joined: 13 Nov 2003
Posts: 961
Location: Florida
Followers: 1

Kudos [?]: 144 [0], given: 0

### Show Tags

22 Dec 2003, 11:37
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, 12:03, edited 1 time in total.
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA
Followers: 1

Kudos [?]: 52 [0], given: 0

### Show Tags

22 Dec 2003, 11:48
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?
Director
Joined: 13 Nov 2003
Posts: 961
Location: Florida
Followers: 1

Kudos [?]: 144 [0], given: 0

### Show Tags

22 Dec 2003, 11:58
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.
CEO
Joined: 15 Aug 2003
Posts: 3454
Followers: 67

Kudos [?]: 874 [0], given: 781

### Show Tags

22 Dec 2003, 12:27
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
CEO
Joined: 15 Aug 2003
Posts: 3454
Followers: 67

Kudos [?]: 874 [0], given: 781

### Show Tags

22 Dec 2003, 12:31
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.
Director
Joined: 14 Oct 2003
Posts: 583
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 53 [0], given: 0

### Show Tags

22 Dec 2003, 12:38
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, 13:10, edited 2 times in total.
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA
Followers: 1

Kudos [?]: 52 [0], given: 0

### Show Tags

22 Dec 2003, 12:42
But X=25, Y=16 is solution to the equations of st1) and st2) , shouldn't the answer be C?
Director
Joined: 13 Nov 2003
Posts: 961
Location: Florida
Followers: 1

Kudos [?]: 144 [0], given: 0

### Show Tags

22 Dec 2003, 12:43
what if X=36
Z=6

Z^2 - 8Z - 1 = -Y
36-48-1=-Y
Y=13
Director
Joined: 13 Nov 2003
Posts: 961
Location: Florida
Followers: 1

Kudos [?]: 144 [0], given: 0

### Show Tags

22 Dec 2003, 12:44
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.
Director
Joined: 14 Oct 2003
Posts: 583
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 53 [0], given: 0

### Show Tags

22 Dec 2003, 13:24
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).
22 Dec 2003, 13:24
Display posts from previous: Sort by