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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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06 Jul 2013, 07:52

2

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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

(2) When x+y is divided by 5, the remainder is 2

hi,

let a and b and c are 3 arbitray integers.

and if ==>a/c==>remainder x and ===>b/c==>remainder is y

then ==>remainder of (a*b)/c=remainder of a*remainder of b ==>remainder of (a+b)/c=remainder of a+remainder of b ==>remainder of (a-b)/c=remainder of a-remainder of b

so using this clearly we can say that we need both the statement to solve this. hence C _________________

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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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28 Mar 2015, 10:04

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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, 15:11

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, 07:03

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

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16 Dec 2015, 12:46

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Expert's post

NoHalfMeasures wrote:

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

The best way would be to check them yourself by following the method shown below for (1).

You are given the following ,

a=nA+p and b=nB+q

Thus, (a+b) = n(A+B) + p+q , or in other words, you will get a remainder of p+q when you divide a+b by n.

You can come up with similar relations based on method above. Additionally, you should not be remembering these relations and should be applying them as and when needed from first principles. Learning these obscure and not that common relations will only end up confusing you. _________________

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