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Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

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31 Aug 2012, 03:17

My approach, combining both statements:

(5n+r)^2=25n^2+10nr+r^2 25n^2 is divisible by 5, 10nr too. So we only have to square the remainder to find the remainder of the square of n. The same is true for adding and subtracting numbers, which should be clear. Remainder of x is 1 greater than remainder of y (because of statement 1). --> Options for remainders of x and y: (1,0), (2,1), (3,2), (4,3), (5,4)-->(0,4) Statement 2 tells us to add x+y, so remainders of our options are: 1+0=1, 2+1=3, 3+2=5+0, 4+3=5+2, 0+4=4 Statement 2 says that the remainder is 2, when adding. Only (4,3) accomplishes this. So 4^2+3^2=16+9=25 The remainder will always be 0 and both statements combined are therefore sufficient.

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

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27 Jan 2013, 20:53

Bunuel wrote:

vikram4689 wrote:

Both are correct examples but the one i gave does not follow what you mentioned

You said since 2*P is divisible by 5, therefore P is divisible by 5... This is not correct P=5/2

Not so. Your example is NOT correct.

I'm saying that if p is an integer and 2p is divisible by 5 then p must be divisible by 5. Now, check your example, is it correct?

It should be: 2*5 is divisible by 5 (p=5) so 5 (p) is divisible by 5.

Hope it's clear.

I believe vikram4689 is correct. If we know that 2P is divisible by 5, then we only know that P is divisible by 2.5.

Let, P=x^2+y^2 and N=5a^2+2a+5b^2+4b+1

Therefore, we have already proven:

2P=5N P=(5/2)N

The only way for P to be divisible by 5 is if N is even (contains a factor of 2). Unfortunately, we can't factor out a 2 from N. Therefore, whether or not N is even depends on the values of A and B.

By looking at the equation of N, we can see that it will be even only if A and B are not both positive or both negative.

It works out that the equations (x-y)=5a+1 and (x+y)=5b+2 can both be satisfied only when A and B are not both positive or both negative. This is why Bunuel's incorrect assumption still works in this case.

There is no way I could have figured all of this out in 2 or 3 minutes though.

Both are correct examples but the one i gave does not follow what you mentioned

You said since 2*P is divisible by 5, therefore P is divisible by 5... This is not correct P=5/2

Not so. Your example is NOT correct.

I'm saying that if p is an integer and 2p is divisible by 5 then p must be divisible by 5. Now, check your example, is it correct?

It should be: 2*5 is divisible by 5 (p=5) so 5 (p) is divisible by 5.

Hope it's clear.

I believe vikram4689 is correct. If we know that 2P is divisible by 5, then we only know that P is divisible by 2.5.

Let, P=x^2+y^2 and N=5a^2+2a+5b^2+4b+1

Therefore, we have already proven:

2P=5N P=(5/2)N

The only way for P to be divisible by 5 is if N is even (contains a factor of 2). Unfortunately, we can't factor out a 2 from N. Therefore, whether or not N is even depends on the values of A and B.

By looking at the equation of N, we can see that it will be even only if A and B are not both positive or both negative.

It works out that the equations (x-y)=5a+1 and (x+y)=5b+2 can both be satisfied only when A and B are not both positive or both negative. This is why Bunuel's incorrect assumption still works in this case.

There is no way I could have figured all of this out in 2 or 3 minutes though.

Think again - It's not 'Bunuel's incorrect assumption'. It is given that x and y are integers so, given that P=x^2+y^2, P will be a positive integer. Now that we know that P is an integer, if 2P is divisible by 5, P must be divisible by 5.

Given that a and b are positive integers and that 'a' is divisible by 'b', a MUST have b as a factor. It's a basic mathematical fact. When I say, 'a' is divisible by 'b', I mean 'a' is a multiple of 'b' which means 'b' is a factor of 'a'. Try some examples to convince yourself.

This statement has far reaching implications. Think of questions like: Is \(2^{100}\) divisible by 3? I will say 'no' without a thought. Is \(2^{10}*3^{100}*5\) divisible by 7? Again, absolutely no!

To reiterate, if 'a' is to be divisible by 'b', a must have b as a factor. (talking about positive integers)
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Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

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30 Jan 2013, 21:48

Bunuel wrote:

If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?

(1) When x-y is divided by 5, the remainder is 1 --> \(x-y=5q+1\), so \(x-y\) can be 1, 6, 11, ... Now, \(x=2\) and \(y=1\) (\(x-y=1\)) then \(x^2+y^2=5\) and thus the remainder is 0, but if \(x=3\) and \(y=2\) (\(x-y=1\)) then \(x^2+y^2=13\) and thus the remainder is 3. Not sufficient.

(2) When x+y is divided by 5, the remainder is 2 --> \(x+y=5p+2\), so \(x+y\) can be 2, 7, 12, ... Now, \(x=1\) and \(y=1\) (\(x+y=2\)) then \(x^2+y^2=2\) and thus the remainder is 2, but if \(x=5\) and \(y=2\) (\(x+y=7\)) then \(x^2+y^2=29\) and thus the remainder is 4. Not sufficient.

(1)+(2) Square both expressions: \(x^2-2xy+y^2=25q^2+10q+1\) and \(x^2+2xy+y^2=25p^2+20p+4\) --> add them up: \(2(x^2+y^2)=5(5q^2+2q+5p^2+4p+1)\) --> so \(2(x^2+y^2)\) is divisible by 5 (remainder 0), which means that so is \(x^2+y^2\). Sufficient.

Answer: C.

Hope it's clear.

HI... y cant we just take examples...4 and 3 (14 and 13, 24 and 23)are the only nos. which will satisfy both the condition.. Hence, by checking cyclicity for 14 and 13 even...we can say that both the statements are reqd wot say?

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

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28 Mar 2015, 09:04

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Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

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13 Nov 2015, 14:11

Bunuel wrote:

If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?

(1) When x-y is divided by 5, the remainder is 1 --> \(x-y=5q+1\), so \(x-y\) can be 1, 6, 11, ... Now, \(x=2\) and \(y=1\) (\(x-y=1\)) then \(x^2+y^2=5\) and thus the remainder is 0, but if \(x=3\) and \(y=2\) (\(x-y=1\)) then \(x^2+y^2=13\) and thus the remainder is 3. Not sufficient.

(2) When x+y is divided by 5, the remainder is 2 --> \(x+y=5p+2\), so \(x+y\) can be 2, 7, 12, ... Now, \(x=1\) and \(y=1\) (\(x+y=2\)) then \(x^2+y^2=2\) and thus the remainder is 2, but if \(x=5\) and \(y=2\) (\(x+y=7\)) then \(x^2+y^2=29\) and thus the remainder is 4. Not sufficient.

(1)+(2) Square both expressions: \(x^2-2xy+y^2=25q^2+10q+1\) and \(x^2+2xy+y^2=25p^2+20p+4\) --> add them up: \(2(x^2+y^2)=5(5q^2+2q+5p^2+4p+1)\) --> so \(2(x^2+y^2)\) is divisible by 5 (remainder 0), which means that so is \(x^2+y^2\). Sufficient.

Answer: C.

Hope it's clear.

Question for Bunuel. Theoretically, could you have used only q or p to represent the qoutient in this example. Wouldn't they represent the same whole factor of times that 5 can go into each statement?

If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

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14 Dec 2015, 06:03

IMO this question is a walk in the park if we know the following rules:

You can always ignore the original number for remainder calculation. If you know a/n gives remainder p and b/n gives remainder of q then- i) remainder of (a+b)/n = remainder of (p+q)/n ii) remainder of (a-b)/n = remainder of (p-q)/n iii) remainder of (a^2+b^2)/n = remainder of (p^2+q^2)/n iv) remainder of (a^2-b^2)/n = remainder of (p^2-q^2)/n v) remainder of (a*b)/n = remainder of (p*q)/n vi) remainder of (a/b)/n = remainder of (p/q)/n . . . etc

But Im not quite sure about the accuracy of these rules.

So Moderators and Math Experts:

How do the above rules look neat or preposterous ?;) Any corrections, additions, exceptions? Thanks
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If x and y are integer, what is the remainder when x^2 + y^2
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14 Dec 2015, 06:03

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