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If p, x, and y are positive integers, y is odd, and p = x^2
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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 == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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Re: Divisibility
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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 xy = 3, it be x should atleast be 4 since y is odd integer(41=3). In this case, x is divisible by 4. However, x can be 6(since 63=3), in that case its not divisible.
So (2) is not sufficient.
So my answer is A.



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Re: Divisibility
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20 Aug 2009, 21:25
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... So that answer is A.
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Re: Divisibility
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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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Re: Divisibility
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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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Re: Divisibility
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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, xy=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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Re: Divisibility
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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, xy=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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Re: Divisibility
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Updated on: 26 Aug 2009, 11:04
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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Re: Divisibility
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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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Re: Divisibility
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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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Re: Divisibility
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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 xy=3. The answer is still A though.Thanks for a correction.



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Re: Divisibility
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26 Aug 2009, 11:52
Given p,x,y>0; y is odd  of the form (2k1); p = \(x^2+y^2\) Is x divisible by 4? Stat1: p = 8n+5 => 8n+5 = \(x^2 + (2k1)^2\) for different values of n and k rearranging we get \(x^2 = 4(kk^2+1+2n)\) => \(x^2\) 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: (xy) = 3 or x(2k1)= 3 x=2(k+1) => x will be divisible by 4 depending on the value of k
Insuff.
IMO A



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Re: Divisibility
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25 Sep 2009, 23:19
bml wrote: Given p,x,y>0; y is odd  of the form (2k1); p = \(x^2+y^2\) Is x divisible by 4? Stat1: p = 8n+5 => 8n+5 = \(x^2 + (2k1)^2\) for different values of n and k rearranging we get \(x^2 = 4(kk^2+1+2n)\) => \(x^2\) 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: (xy) = 3 or x(2k1)= 3 x=2(k+1) => x will be divisible by 4 depending on the value of k
Insuff.
IMO A great explanation == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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Re: If p, x, and y are positive integers, y is odd, and p = x^2
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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
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