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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

(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...

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.

Last edited by LenaA on 26 Aug 2009, 10:04, edited 1 time in total.

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

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.

(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

(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.

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

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.

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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