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Is x^2+y^2 divisible by 5? 1). When x-y is divided by 5, the

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Manager
Joined: 19 Aug 2006
Posts: 222
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Is x^2+y^2 divisible by 5? 1). When x-y is divided by 5, the [#permalink]  20 Nov 2006, 05:53
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99. 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
Director
Joined: 28 Dec 2005
Posts: 922
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Kudos [?]: 37 [0], given: 0

I get C. Together they are sufficient.

From A: 5p+1=x-y
From B: 5q+3=x+y

Adding the 2 equations: x = (5(p+q)+4)/2
Subtracting the 2 equations B-A: y = (5(q-p)+2)/2

Then x^2+y^2 = ((5(p+q)+4)/2)^2 + ((5(q-p)+2)/2)^2

=25(p+q)^2+4 + 25(q-p)^2+1

=25((p+q)^2 + (q-p)^2) + 5

Hence divisible by 5.
VP
Joined: 25 Jun 2006
Posts: 1175
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for me it is E.

i get this:

x^2+y^2 = [25(m^2+n^2)]/2 + 5m + 15n + 5.

we can't be sure m^2 + n^2 is an even number, so we can't guarantee the first term divided by 5 is an integer.
Senior Manager
Joined: 08 Jun 2006
Posts: 342
Location: Washington DC
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Kudos [?]: 23 [0], given: 0

Yes, got the point of tennis_ball
It should be E

In fact, x^2 + y^2 is not divisible by 5 when x = 4.5 and y = 3.5, whereas these values satisfy other conditions.
Senior Manager
Joined: 23 Jun 2006
Posts: 387
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tennis ball is right...

x=2 y=1 satisfy (1) and (2) and x^2+y2 is divisible by 5.
however -
x=4.5 y=3.5 satisfy (1) and (2) and x^2+y^2 is not even an integer... not to say that it is not divisible by 5.

nothing said in the question that x and y should be integers.

hence E.
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 181 [0], given: 2

this is a very simple problem...

I get E..

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

1) ok so we are told (x-y)=5m+1; but we dont know anything about (x+Y)..insuff

2) same thing

now C) lets look

(x-y)(x+y), we have remainders 3 and 1...pick a number

x-y=6; x+y=8; 8*6=42/5 remainder is 2

pick another number x-y =11; x+y=8; 8*11=88/5 remainder is 3...

Insuff...E it is..
Director
Joined: 28 Dec 2005
Posts: 922
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Kudos [?]: 37 [0], given: 0

Thanks everyone, yes, I assumed x and y are integers
Intern
Joined: 08 Jul 2006
Posts: 11
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Kudos [?]: 0 [0], given: 0

Just for the record

x^2+y^2 is not equal (x-y)(x+y)

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