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# If p, x, and y are positive integers, y is odd, and p = x^2

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If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

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20 Aug 2009, 12:29
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If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3

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20 Aug 2009, 21:17
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3

For (1), the minimum value for p should 13(since 8+5/8 = 5). In this case, x can only be 2 (since 2x2 + 3x3 = 13). so x is not divisible by 4. The next possible value for p would be 29(since 8x3+5/8 = 5). Here, x again can only be 3 (since 2x2 + 5x5 = 29). so x is not divisible by 4.

So (1) is sufficient.

For (2), if x-y = 3, it be x should atleast be 4 since y is odd integer(4-1=3). In this case, x is divisible by 4. However, x can be 6(since 6-3=3), in that case its not divisible.

So (2) is not sufficient.

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Joined: 29 Aug 2007
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20 Aug 2009, 21:25
1
gulatin2 wrote:
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3

(1) If the remainder is 5 when p is divided by 8, x (or x^2) has to be an even integer since y is already said to be an odd integer. However its not sure whether x could be 2 or 4 or 6 or 8 or so on...

If x = 2, y could be any odd integer. Lets say y = 1, p = x^2 + y^2 = 5, which has 5 reminder when it is divided by 8.
If x = 2 and y = 3, p = x^2 + y^2 = 13, which has 5 reminder when it is divided by 8.
If x = 2, and y = 5, p = x^2 + y^2 = 29, which has 5 reminder when it is divided by 8.

But if x = 4, none of the values of y generates 5 reminder when p is divided by 8. For ex:

If x = 4, and y = 1, p = x^2 + y^2 = 17, which has 1 reminder when it is divided by 8.
If x = 4, and y = 3, p = x^2 + y^2 = 25, which has 1 reminder when it is divided by 8.
If x = 4, and y = 5, p = x^2 + y^2 = 41, which has 1 reminder when it is divided by 8.
If x = 8, and y = 1, p has 1 reminder when it is divided by 8.
If x = 8, and y = 3, p has 1 reminder when it is divided by 8.
If x = 8, and y = 5, p has 1 reminder when it is divided by 8.

Any value for x i.e. divisible by 4 doesnot produce 5 reminder when p is divided by 8.

So it is sufficient to answer that x is not divisible by 4.

(2) If x = y+3, p = x^2 + y^2 = (y+3)^2 + y^2. Here y could be 1, or 3 or 5 or 7 or so on....
If y = 1, x = 4...........yes.
If y = 3, x = 6...........no.
If y = 5, x = 8........... yes..
If y = 7, x = 10.............no.................NSF...

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Joined: 12 Aug 2009
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22 Aug 2009, 08:49
OK.

(1) - Everyone agrees on this. Suff to prove not divisible by 4

(2) -> x is even and y is odd - square of even (x) is even and odd (y) is odd. sum of even + odd is always odd -> IT is not divisible by 4.

I will go for 'D'.
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23 Aug 2009, 09:46
dolly12 wrote:
(2) -> x is even and y is odd - square of even (x) is even and odd (y) is odd. sum of even + odd is always odd -> IT is not divisible by 4.

I will go for 'D'.

your statemet that "x is even" is not correct.

Update: your statemet that "x is even" is correct.
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23 Aug 2009, 11:51
GMAT TIGER wrote:
dolly12 wrote:
(2) -> x is even and y is odd - square of even (x) is even and odd (y) is odd. sum of even + odd is always odd -> IT is not divisible by 4.

I will go for 'D'.

your statemet that "x is even" is not correct?

I think the statement that x is even is correct. From the statement 2, x-y=3 => x=y+3. 3 is odd and y is odd, then the sum of two odds is even => x is even
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23 Aug 2009, 12:32
LenaA wrote:
GMAT TIGER wrote:
dolly12 wrote:
(2) -> x is even and y is odd - square of even (x) is even and odd (y) is odd. sum of even + odd is always odd -> IT is not divisible by 4.

I will go for 'D'.

your statemet that "x is even" is not correct?

I think the statement that x is even is correct. From the statement 2, x-y=3 => x=y+3. 3 is odd and y is odd, then the sum of two odds is even => x is even

agree X can be even ..but it is not sufficient.(2). Kudos to Gmat tiger for a brilliant explanation.
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Updated on: 26 Aug 2009, 11:04
1
yes, statement 2 alone is not sufficient because the only thing we can deduct from it is that x is even but an integer can still be an even but not a multiple of 8.
Statement 1 alone is sufficient:
Assumption: x is a multiple of 4.
If x is a multiple of 4, then x=4m=>x^2=16(m^2)
p=16(m^2)+y^2.
p/8=16(m^2)/8 + y^2/8
16(m^2) is divisible by 8. Therefore the remainder of p when divided by 8 will be equal to the remainer of y^2 when divided by 8.
y is odd => y=2n+1=> y^2=(2n+1)^2=4n^2+4n+1=4n(n+1)+1 We have two subsequent numbers n and n+1. So one of them is even and 4n(n+1) is divisible by 8. y^2=8k+1. When a square of an odd number is divided by 8, the remainder will be 1.
However, we are given that the remainder is 5 => the assumtion that x is a multiple of 4 is wrong. X cannot be a multiple of 4.

Originally posted by LenaA on 23 Aug 2009, 13:23.
Last edited by LenaA on 26 Aug 2009, 11:04, edited 1 time in total.
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Joined: 30 Jun 2009
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26 Aug 2009, 07:19
All,
I di agree with answer A
However I am having issues understanding the option that x may be 2

To me, p=x^2+y^2
as long as y is odd and positive, its min value is 1
to me x can note be 0 as it is not +ve or -ve
so p>=13
x^2 >= 12
so to me the min value of x is 4.

I am sure that I am missing something here.
Can anyone help?
Thx
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Joined: 24 Aug 2009
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Updated on: 26 Sep 2009, 12:31
Stmt I: Suff

Stmt II: InSuff

A

Originally posted by deepak115 on 26 Aug 2009, 10:06.
Last edited by deepak115 on 26 Sep 2009, 12:31, edited 1 time in total.
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26 Aug 2009, 11:00
defoue wrote:
All,
I di agree with answer A
However I am having issues understanding the option that x may be 2

To me, p=x^2+y^2
as long as y is odd and positive, its min value is 1
to me x can note be 0 as it is not +ve or -ve
so p>=13
x^2 >= 12
so to me the min value of x is 4.

I am sure that I am missing something here.
Can anyone help?
Thx

No you are not missing. I made a mistake. I will edit my previous post. x cannot be 2 given the fact that x-y=3. The answer is still A though.Thanks for a correction.
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Joined: 21 Aug 2009
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26 Aug 2009, 11:52
2
Given p,x,y>0; y is odd - of the form (2k-1); p = $$x^2+y^2$$
Is x divisible by 4?
Stat1: p = 8n+5 => 8n+5 = $$x^2 + (2k-1)^2$$ for different values of n and k
rearranging we get $$x^2 = 4(k-k^2+1+2n)$$
=> $$x^2$$ is a mulitple of 4 and x must be a multiple of 2
Suff. x is not divisible by 4 ( if it is $$x^2$$ should be a multiple of 16)

Stat2: (x-y) = 3 or x-(2k-1)= 3
x=2(k+1) => x will be divisible by 4 depending on the value of k

Insuff.

IMO A
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25 Sep 2009, 23:19
bml wrote:
Given p,x,y>0; y is odd - of the form (2k-1); p = $$x^2+y^2$$
Is x divisible by 4?
Stat1: p = 8n+5 => 8n+5 = $$x^2 + (2k-1)^2$$ for different values of n and k
rearranging we get $$x^2 = 4(k-k^2+1+2n)$$
=> $$x^2$$ is a mulitple of 4 and x must be a multiple of 2
Suff. x is not divisible by 4 ( if it is $$x^2$$ should be a multiple of 16)

Stat2: (x-y) = 3 or x-(2k-1)= 3
x=2(k+1) => x will be divisible by 4 depending on the value of k

Insuff.

IMO A

great explanation

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Re: If p, x, and y are positive integers, y is odd, and p = x^2  [#permalink]

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19 Apr 2018, 13:47
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Re: If p, x, and y are positive integers, y is odd, and p = x^2   [#permalink] 19 Apr 2018, 13:47
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