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What is the remainder when x^2+y^2 is divided by 5, where x

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What is the remainder when x^2+y^2 is divided by 5, where x [#permalink]

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06 May 2006, 04:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the remainder when x^2+y^2 is divided by 5, where x and y are positive integer?

1). When x+y is divided by 5, the remainder is 1.
2). When x-y is divided by 5, the remainder is 2.

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07 May 2006, 10:40
(I) gives (x + y) = {1,11,21,...} or {6,16,26,...}
(II) gives (x - y) = {2,12,22,..} or {7,17,27,...}

(x^2 + y^2) = (x + y)^2 - 2xy = (x - y)^2 + 2xy

As we do not get any information about xy from either (I) or (II) we can rule out A and B.

Now combine them.

(x+y)^2(mod5) = 1 (as the last digit would be 1 or 6)
(x-y)^2(mod5) = 4 (as the last digit would be 4 or 9)

Add them and you would get 2(x^2 + y^2) on the LHS.

On the RHS you would get 1+4=5. So 2(x^2 + y^2)(mod 5) = 0 ==> (x^2 + y^2)(mod 5) = 0. So ans is C.
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Zooroopa

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08 May 2006, 08:52
Agree with C
here is brutal and unneccessary approach:

From st1 we learned that x+y is a number that ends to 6 or 1
From st2 we learned that x-y ends to 7 or 2
So lets assume, that x+y=16, x-y=12. x=14, y=2. the remainder for x^2+y^2 is 0.
Now, lets assume that x+y=21, x-y=17. x=19, y=2. the remainder is 0.

from these, we can see that one of them will always end to 4 or 9, and the other to 1 or 4 respectively, which gives us 5 at the end of X^2+Y^2
Thus the remainder is always 0, and the answer is C

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08 May 2006, 09:48
Its C. Each individual is INSUFF. Combining. Square and add.
which gives 2(x^2 + y^2) = multiple of 5.
Since x and y are integers, (x^2+y^2) is multiple of 5.

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08 May 2006, 09:48
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What is the remainder when x^2+y^2 is divided by 5, where x

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