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Remainder problem

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Remainder problem [#permalink] New post 28 Jun 2009, 06:56
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Sorry if the question has been posted before.

Is x^2+y^2 divisible 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 3
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Re: Remainder problem [#permalink] New post 28 Jun 2009, 10:22
Man this problem will eat time on the GMAT, if we dont know of any easy way.

The only solution I found out was by trying numbers which eats time.....

(1) x=7 and y=1 satisfy this stmt. Also 7^2 + 1^2 = 50 divisible by 5.
x=10, y=-1 also satisfy this stmt. But 10^2 + -1^2 = 101 is not divisible by 5.
INSUFF.

(2) x=5, y=3 satisfy this stmt. Also 5^2+3^2 = 31 not divisble by 5
x=6, y=2 satisfy this stmt. But 6^2 + 2^2 =40 is divisble by 5.
INSUFF.

Taking both stmt 1 and 2, the only numbers that can satisfy this condition are 7 and 1.
So x^2 + y^2 is divisible by 5.

Answer is C. What is the OA???
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Re: Remainder problem [#permalink] New post 28 Jun 2009, 12:00
x^2 + y^2 = (x+y)^2-2xy = (x-y)^2 + 2xy
If (x+y) = 1 (mod 5) then (x+y)^2 = 4 (mod 5)
since we do not know what is 2xy equal in (mod 5) so we can not know whether (x+y)^2-2xy is equal to 0 or not. INSUFF

If (x-y) = 3 (mod 5) then (x-y)^2 = 9 = 4 (mod 5)
since we do not know what is 2xy equal in (mod 5) so we can not know whether (x-y)^2+2xy is equal to 0 or not. INSUFF

Gathering both datas together =>
(x+y) = 1 mod 5
(x-y) = 3 mod 5
(x+y)^2 + (x-y)^2 = 2(x^2+y^2)
So
(x+y)^2 + (x-y)^2 = 1 + 9 = 10 = 0 (mod 5)
since 2(x^2+y^2) = 10 = 0 (mod 5)
x^2 + y^2 = 0 in module 5.
That is to say, x^2 + y^2 is divisible by 5 ;)
Both are needed
Answer is C
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Re: Remainder problem [#permalink] New post 28 Jun 2009, 17:38
I got C by using the numbers, but here is what tried.

Is x^2+y^2 divisible 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 3

Stmt 1.
X - Y = 5D + 1
square both sides
X^2 + y^2 = 25D^2 + 10D + (1 + 2XY)..............1
Now we can say that 25D^2 + 10D + (2xy+1) can be divisible by 5 if (2xy + 1) is divisible by 5. But we can not say anything abt. this. So insuff.

Stmt. 2
X + Y = 5D + 3
square both sides
X^2 + Y^2 = 25D^2 + 9 + 15D -2XY....................2
This will be divisible by 5 if (9-2XY) is divisible by 5. We can not say this so insuff.

Now combine both.......eq1 = eq2 (as both equals X^2 + y^2), but i m not able to get any useful information from this.
can someone take this approach further or else let me know in this approach itself is incorrect.
Re: Remainder problem   [#permalink] 28 Jun 2009, 17:38
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