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x and y consecutive perfect squares

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x and y consecutive perfect squares

by abhi332 » Thu Feb 25, 2010 1:34 pm
If x, y, and z are positive integers, where x > y and z=x^(1/2) , are x and y consecutive perfect squares? (A perfect
square is defined as the square of an integer. For example, 36 is a perfect square since it equals 6 squared, while
38 is not a perfect square since it is not equal to the square of any integer.)
(1) x + y = 8z +1
(2) x - y = 2z - 1

[spoiler]OA:B[/spoiler]
What you think, you become.
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Source: — Data Sufficiency |
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by vineetbatra » Thu Feb 25, 2010 5:55 pm
Can solve B using Plugging numbers, but have no clue on A.

Anyone?
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by hrishi19884 » Fri Feb 26, 2010 2:16 am
consider (2)

The rule is that - difference between any two consecutive perfect squares is equal to 2z - 1 when z =sqrt(x) and x>y

ex : x= 16 and Y = 9 ,so z will be naturally 4

x-y =16 -9 = 7

2z - 1 = 4*2 - 1 = 7

ex : x= 25 and Y = 16, so z will be 5

x-y =25 -16 = 9

2z - 1 = 5*2 - 1 = 9

Incase x,y are not consecutive - say x=16 and y=4, z=sqrt(x)=4

x-y =16 -4 = 12

2z - 1 = 4*2 - 1 = 7

Not equal, so x,y should be consecutive.

So B is sufficient

If we consider (1)

x + y = 16 + 9 =25

8z +1 = 4*8 + 1 = 33

not equal - hence (1)not sufficient

Ans = B
Hrishi

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by Fiver » Sat Feb 27, 2010 6:33 am
abhi332 wrote:If x, y, and z are positive integers, where x > y and z=x^(1/2) , are x and y consecutive perfect squares? (A perfect
square is defined as the square of an integer. For example, 36 is a perfect square since it equals 6 squared, while
38 is not a perfect square since it is not equal to the square of any integer.)
(1) x + y = 8z +1
(2) x - y = 2z - 1

[spoiler]OA:B[/spoiler]
When z = root (x)
and x > y

is root (y) + 1 = root (x)?
or is root (y) = root (x) - 1?
or is root (y) = z - 1 ?

A] x + y = 8z +1

Try replacing 'x' with z^2; it doesn't take us anywhere.
Insufficient.

B] x - y = 2z - 1
z^2 - y = 2z - 1
z^2 - 2z + 1 = y
(z - 1)^2 = y
Since given all are positive
(z - 1) = root (y)

And this is what we are looking for.
Hence B.
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by vertigo05 » Fri Mar 05, 2010 5:28 pm
But, if we put values in (A), we get that x and y are not consecutive perfect squares. So (A) should also be sufficient. Pls let me know where i'm wrong.
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by thephoenix » Fri Mar 05, 2010 8:18 pm
vertigo05 wrote:But, if we put values in (A), we get that x and y are not consecutive perfect squares. So (A) should also be sufficient. Pls let me know where i'm wrong.
a) X+Y=8Z+1

NOW IF x=16 and y=9 rhs=25 and lhs=33#rhs--------> x and y are not a perfect square

but if x=9 and y=16---->RHS=25 and LHS=25=RHS---------> x and y are perfect square

as bth the cases are possible so its inconclusive
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by kstv » Fri Mar 05, 2010 8:42 pm
If z = a , x = a², y = (a-1)² x and y are consecutive perrfect square
x+y = a²+(a²-2a+1)= 2a²-2a+1 = 8z+1
2a(a-1)= 8a or (a-1)= 4
so A is true only when a = 5
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by schumi_gmat » Sat Mar 06, 2010 1:16 am
IMO E

HEre is my solution -

Here is my solution -



x y z (i) (ii) ANS

25 16 5 Y Y

25 4 5 N N

36 25 6 N Y

Can anyone throw more light??
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by pharmxanthan » Fri Apr 30, 2010 5:22 pm
thephoenix wrote:
vertigo05 wrote:But, if we put values in (A), we get that x and y are not consecutive perfect squares. So (A) should also be sufficient. Pls let me know where i'm wrong.
a) X+Y=8Z+1

NOW IF x=16 and y=9 rhs=25 and lhs=33#rhs--------> x and y are not a perfect square

but if x=9 and y=16---->RHS=25 and LHS=25=RHS---------> x and y are perfect square

as bth the cases are possible so its inconclusive
If x=9 and y=16 then x is less than y. But in question stem, x should be more than y.
Statement A is conclusive and so is statement B. But statements A and B give different answers, i.e, per Statement A, x and y are not consecutive perfect squares, while per Statement B, x and y are! So, should answer be "E"?

Confusing!
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