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If x, y, and z are positive integers, where x > y and z = √x

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If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 31 Jan 2012, 16:24
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If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?

(1) x + y = 8z +1
(2) x – y = 2z – 1

OA is B. But I got stuck after a while. Please see below my solution.

From question stem

z = \sqrt{x}
i.e. x = \(z^2\) ---------------------------------------------(1)

Considering Statement 1

y = 8z-x+1 -----------------------------------------------------(2)

Putting some values of x and z

When x =2 , z =4

Substituting these values in Equation 2 we will get y = 13. But this cannot be applied as x >y. So I tried another values which are

x = 25 then z =5

Substituting these values in Equation 2 we will get y = 16. Therefore , YES x and y are perfect squares.

I tried one more value which was

when x = 36, z =6

Substituting these values in Equation 2 we will get y = 13. Therefore , NO x and y are NOT perfect squares.
Two conclusions, therefore this statement alone is NOT sufficient.

Considering Statement 2

After simplifying we will get y = \((z-1)^2\)
Therefore z-1 = \sqrt{y}

But from statement 1 we can say z = \sqrt{x}

Therefore, \sqrt{x} - 1 = \sqrt{y}

\sqrt{x} = \sqrt{y} + 1

I am stuck after this. Can someone please help?

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Re: Consecutive Perfect Sqaures  [#permalink]

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New post 31 Jan 2012, 16:41
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If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?

If \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{y}\) and \(\sqrt{x}\) must be consecutive integers. So, the question becomes: is \(\sqrt{y}=\sqrt{x}-1\)? --> square both sides: is \(y=x-2\sqrt{x}+1\)?

(1) x + y = 8z +1 --> \(x+y=8\sqrt{x}+1\) --> \(y=8\sqrt{x}+1-x\). Now, question becomes is \(8\sqrt{x}+1-x=x-2\sqrt{x}+1\)? --> is \(10\sqrt{x}=2x\)? --> is \(5\sqrt{x}=x\)? --> is \(\sqrt{x}=5\)? --> is \(x=25\)? But we don't know that, hence insufficient.

(2) x – y = 2z – 1 --> \(y=x-2\sqrt{x}+1\), which is exactly what we needed to know. Sufficient.

Answer: B.

Hope it's clear.
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Re: If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 10 Mar 2012, 02:13
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Isn't this question a good candidate for plugging in simple numbers? I plugged in x=4 and y=1 and got the answer pretty quickly.
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Re: If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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Re: If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 12 Jun 2013, 22:33
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enigma123 wrote:
If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?

(1) x + y = 8z +1
(2) x – y = 2z – 1



Excellent explanation by Bunuel, but i will try to give some alternative solution. Since the question askes whether x and y are consecutive perfect squares, if yes they must be perfect numbers in all cases, so we can plug some numbers and check. While plugging in our task is to find numbers when x and y are consecutive and when they are not.

Lets take 4, 9, 16 and 25 - consecutive perfect squares.

(1) if x=9, y=4, z=3, then according to the statement we have 9+4=24+1 which is not true, in case of x=16, y=9 and z=4 we have 16-9=8-1 which is again true. Lets try x=25, y=16, z=5 so we have 25-16=10-1, true again. So in all cases have the same conclusions based on the statement. Sufficient - B.

Hope that helps.

(2) if x=9, y=4, z=3, then according to the statement we have 9-4=6-1 which is true, in case of x=16, y=9 and z=4 we have 16-9=+1 which is again not true. When plug the numbers it is advisable to check at least 3 numbers. So lets try x=25, y=16, z=5 so we have 25+16=40+1 - Bingo! In this case the statement supports that X and Y are consecutive. So we have two different conclusions based on the statement. Not sufficient.
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Re: If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 03 Sep 2013, 09:36
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ziko wrote:
enigma123 wrote:
If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?

(1) x + y = 8z +1
(2) x – y = 2z – 1



Excellent explanation by Bunuel, but i will try to give some alternative solution. Since the question askes whether x and y are consecutive perfect squares, if yes they must be perfect numbers in all cases, so we can plug some numbers and check. While plugging in our task is to find numbers when x and y are consecutive and when they are not.

Lets take 4, 9, 16 and 25 - consecutive perfect squares.

(1) if x=9, y=4, z=3, then according to the statement we have 9+4=24+1 which is not true, in case of x=16, y=9 and z=4 we have 16-9=8-1 which is again true. Lets try x=25, y=16, z=5 so we have 25-16=10-1, true again. So in all cases have the same conclusions based on the statement. Sufficient - B.

Hope that helps.

(2) if x=9, y=4, z=3, then according to the statement we have 9-4=6-1 which is true, in case of x=16, y=9 and z=4 we have 16-9=+1 which is again not true. When plug the numbers it is advisable to check at least 3 numbers. So lets try x=25, y=16, z=5 so we have 25+16=40+1 - Bingo! In this case the statement supports that X and Y are consecutive. So we have two different conclusions based on the statement. Not sufficient.


Solution is truly great but the way it has been explained is confusing !Also there are typos !

in the highlighted part above we are testing statement 2 so 16-9=8-1 = 7 which is true !



More simply :


1) x + y = 8z +1

Work backwards take values of Z find x, and then find y. ( \(x= z^{2}\hs{3}\) put x and z in statement and find y )

If z=4 ,x= 16 we get y = 17 cannot take this case as condition is that x>y ( for all positive z<5, y>x , hence we cannot take those cases )
If z=5, x=25,y=16, does satisfy, here x and y are consecutive perfect squares
If z=6,x=36, y=13, also does satisfy, here x and y are not consecutive perfect squares.In fact y is not even a perfect square.
If z=7, x=49,y=9 also does satisfy, here x and y are not consecutive perfect squares.

Two cases hence insufficient.



(2) x – y = 2z – 1

Again work backwards take values of Z find x, and then find y. ( \(x= z^{2}\hs{3}\) put x and z in statement and find y )


if Z=1 then x=1 on solving we get y =0 cannot take this case as y is supposed to be positive.

if Z= 2 then x=4 on solving and finding y we get y =1 , here 4 and 1 are consecutive perfect squares
if Z=3 then x=9 then y = 4 , here also x and y are consecutive perfect squares
if z=4 then x= 16 then y = 9 also satisfies the equation and x and y are perfect squares
. ........
Take any positive integer for z and find x and y we will see both x and y are consecutive perfect squares
Lets take z= 10 then x= 100 and we get y =81 here also x and y are consecutive perfect squares,

hence B is sufficient.

Hope this helps.
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If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 27 Jun 2015, 07:40
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Can we just consider the consecutive perfect squares as \(n^2\) and \((n-1)^2\) and the difference could then be expressed as \((n^2)\)- \((n^2-2n+1)\)which is the same as 2n-1 which indicates that Statement 2 is sufficient?
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If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 17 Oct 2017, 08:47
1
Bunuel wrote:
If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?

If \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{y}\) and \(\sqrt{x}\) must be consecutive integers. So, the question becomes: is \(\sqrt{y}=\sqrt{x}-1\)? --> square both sides: is \(y=x-2\sqrt{x}+1\)?

(1) x + y = 8z +1 --> \(x+y=8\sqrt{x}+1\) --> \(y=8\sqrt{x}+1-x\). Now, question becomes is \(8\sqrt{x}+1-x=x-2\sqrt{x}+1\)? --> is \(10\sqrt{x}=2x\)? --> is \(5\sqrt{x}=x\)? --> is \(\sqrt{x}=5\)? --> is \(x=25\)? But we don't know that, hence insufficient.

(2) x – y = 2z – 1 --> \(y=x-2\sqrt{x}+1\), which is exactly what we needed to know. Sufficient.

Answer: B.

Hope it's clear.


Hi Bunuel

How did u get? 5[square_root]x=x----[square_root]x=5?

Pls explain thanks
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If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 17 Oct 2017, 08:52
zanaik89 wrote:
Bunuel wrote:
If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?

If \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{y}\) and \(\sqrt{x}\) must be consecutive integers. So, the question becomes: is \(\sqrt{y}=\sqrt{x}-1\)? --> square both sides: is \(y=x-2\sqrt{x}+1\)?

(1) x + y = 8z +1 --> \(x+y=8\sqrt{x}+1\) --> \(y=8\sqrt{x}+1-x\). Now, question becomes is \(8\sqrt{x}+1-x=x-2\sqrt{x}+1\)? --> is \(10\sqrt{x}=2x\)? --> is \(5\sqrt{x}=x\)? --> is \(\sqrt{x}=5\)? --> is \(x=25\)? But we don't know that, hence insufficient.

(2) x – y = 2z – 1 --> \(y=x-2\sqrt{x}+1\), which is exactly what we needed to know. Sufficient.

Answer: B.

Hope it's clear.


Hi Bunuel

How did u get? 5x=x----x=5?

Pls explain thanks


Reduce by \(\sqrt{x}\):

\(5\sqrt{x}=x\);

\(5\sqrt{x}=(\sqrt{x})^2\);

\(5=\sqrt{x}\).
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Re: If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 17 Oct 2017, 10:00
Bunuel wrote:
zanaik89 wrote:
Bunuel wrote:
If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?

If \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{y}\) and \(\sqrt{x}\) must be consecutive integers. So, the question becomes: is \(\sqrt{y}=\sqrt{x}-1\)? --> square both sides: is \(y=x-2\sqrt{x}+1\)?

(1) x + y = 8z +1 --> \(x+y=8\sqrt{x}+1\) --> \(y=8\sqrt{x}+1-x\). Now, question becomes is \(8\sqrt{x}+1-x=x-2\sqrt{x}+1\)? --> is \(10\sqrt{x}=2x\)? --> is \(5\sqrt{x}=x\)? --> is \(\sqrt{x}=5\)? --> is \(x=25\)? But we don't know that, hence insufficient.

(2) x – y = 2z – 1 --> \(y=x-2\sqrt{x}+1\), which is exactly what we needed to know. Sufficient.

Answer: B.

Hope it's clear.


Hi Bunuel

How did u get? 5x=x----x=5?

Pls explain thanks


Reduce by \(\sqrt{x}\):

\(5\sqrt{x}=x\);

\(5\sqrt{x}=(\sqrt{x})^2\);

\(5=\sqrt{x}\).


Hi Bunuel,

How is \sqrt{x}=5 not sufficient?
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Re: If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 17 Oct 2017, 10:04
1
j.shivank wrote:
Bunuel wrote:
If x, y, and z are positive integers, where x > y and z = √x , are x and y consecutive perfect squares?

If \(x\) and \(y\) are consecutive perfect squares, then \(\sqrt{y}\) and \(\sqrt{x}\) must be consecutive integers. So, the question becomes: is \(\sqrt{y}=\sqrt{x}-1\)? --> square both sides: is \(y=x-2\sqrt{x}+1\)?

(1) x + y = 8z +1 --> \(x+y=8\sqrt{x}+1\) --> \(y=8\sqrt{x}+1-x\). Now, question becomes is \(8\sqrt{x}+1-x=x-2\sqrt{x}+1\)? --> is \(10\sqrt{x}=2x\)? --> is \(5\sqrt{x}=x\)? -->is \(\sqrt{x}=5\)? --> is \(x=25\)? But we don't know that, hence insufficient.
(2) x – y = 2z – 1 --> \(y=x-2\sqrt{x}+1\), which is exactly what we needed to know. Sufficient.

Answer: B.

Hope it's clear.


Hi Bunuel,

How is \sqrt{x}=5 not sufficient?


We did not get that \(\sqrt{x}=5\),the question became: is \(\sqrt{x}=5\)?
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Re: If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 17 Oct 2017, 11:40
It is given that x and y are 2 consecutive squired. This means they are squires of 2 integers whose mod will give 2 consecutive integers.

Let

x = \(n^2\)
y = \((n-1)^2\) As x > y

Option 1

x+y =8z+1 => \(n^2\) + \((n-1)^2\) = 8n+1 => 2\(n^2\) - 2n +1 = 8n +1 => \(2n^2\) - 10n=0 => n=0,5
The equation satisfied for 2 values of n, while for other values of n it do not satisfy.
Hence, INSUFFICIENT

Option 2

x-y = 2n-1 => \(n^2\) - \((n-1)^2\) = 2n-1 => \(n^2\) - (\(n^2\) +1 -2n) = 2n -1 => 2n -1 = 2n -1
Both LHS and RHS are equal. The equation will satisfy for all values of n.
Hence, SUFFICIENT

Answer "B"
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If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 24 Oct 2017, 12:06
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In order to determine that x and y are consecutive squares, we need to plug in y=(z-1)^2 with x=z^2 as given. If, in an equation, LHS=RHS, then its Correct.

St.1: x+y=8z+1
z^2+(z-1)^2=8z+1
z^2+z^2+1-2z=8z+1
2z^2-10z=0
Therefore, for LHS=RHS, Z must be 5. (Z cannot be 0, being a positive interger).We are not given value of Z.
Hence, Insufficient

St.2: x-y=2z-1
z^2-(z-1)^2=2z-1
z^2-(z^2+1-2z)=2z-1
2z-1=2z-1
LHS=RHS
Sufficient.

Hence, Ans B

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Re: If x, y, and z are positive integers, where x > y and z = √x  [#permalink]

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New post 06 Mar 2018, 10:13
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Bumping for an alternative approach -

Statement I: This statement can be written as

\(z = \sqrt{17-y} + 4\)
When y = 1, x = 64, z = 8 ---> X,Y Not Consecutive Squares
When y = 16, x = 25, z = 5 --> x, y Consecutive squares.


Statement II: This Statement can be written as

\(y = (z-1)^2\) ----> X, Y are Consecutive Squares.

Hence, B.
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Re: If x, y, and z are positive integers, where x > y and z = √x &nbs [#permalink] 06 Mar 2018, 10:13
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