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

I'm getting answer E

1. x-y/5 gives remainder 1

therefore x-y=1, 6, 11, 16, 21
insufficient because we don't know x or y

2. x+y/5 gives remainder 3
therefore x+y = 3, 8, 13, 18, 23
insufficient

the formula is y = qx + r

equation 1: x-y = 5q + 1
equation 2: x+y = 5q + 3

combine equations and solve
x+y = x - y - 1 + 3
2y = 2
y = 1

if y = 1 x can be 7
7+1 = 8 and divided by 5 gives remainder 3
7-1 = 6 and divided by 5 gives remainder 1

however

17 + 1 = 18 and divided by 5 gives remainder 3
17-1 = 16 and divided by 5 gives remainder 1

so answer E

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

Answer:

Statement 1:
This tells us two things-
1) Either one of x and y is a multiple of 5
or
2) Both x and y are non -multiples of 5-( x-y=can result in multiple of 5 or non multiple of 5 )
Both x and y CANNOT be multiples of 5 -Not sufficient

Statement 2:
This tells us the same thing as 1
1) Either one of x and y is a multiple of 5
or
2) Both x and y are non -multiples of 5( x+y=can result in multiple of 5 or non multiple of 5 )
Both x and y CANNOT be multiples of 5 -Not sufficient

Combining (1) and (2)
Either one of x and y is a multiple of 5 or Both x and y are non -multiples of 5
this means
either x2 or y2 is a multiple of 5 or both are non multiples of 5
x2+y2- this can result in a multiple of 5 or a non multiple - so Not sufficient

Answer -(E)

First I must say this is not GMAT question. As every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

But let's deal with question as it's written. x^2+y^2 to be divisible by 5 it should be an integer first and then should have the last digit 0 or 5:

(1) x-y=5n+1 --> (x-y)^2=25n^2+10n+1 --> x^2+y^2=25n^2+10n+1+2xy. Now 25n^2+10n is divisible by 5, but what about 1-2xy? Not sufficient.

(2) x+y=5m+3 --> (x+y)^2=25m^2+30m+9 --> x^2+y^2=25m^2+30m+9-2xy. The same here. Not sufficient.

(1)+(2) Add the equations:
2(x^2+y^2)=25n^2+10n+25m^2+30m+10=25(n^2+m^2)+10(n+3m)+10.

x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+3m)+5. Well if n^2+m^2 is even, x^2+y^2 is an integer and it's divisible by 5. But what if n^2+m^2 is odd, then x^2+y^2 is not an integer at all, hence not divisible by 5.

Can we somehow conclude that n^2+m^2 is even? If we are not told that x and y are integers then not.

Answer: E.

To check this consider following pairs of x an y: x=4.5, y=3.5 and x=7, y=1,
x=4.5, y=3.5
x-y=4.5-3.5=1 first statement holds true, 1 divided by 5 remainder 1.
x+y=4.5+3.5=8 second statement holds true, 8 divided by 5 remainder 3.
x^2+y^2=4.5^2+3.5^2=32.5 not divisible by 5.

x=7, y=1
x-y=7-1=6 first statement holds true, 6 divided by 5 remainder 1.
x+y=7+1=8 second statement holds true, 8 divided by 5 remainder 3.
x^2+y^2=7^2+1^2=50 is divisible by 5.

Two different answers. Not sufficient. Answer: E.


Now let's see if we were told that x and y are positive integers, as GMAT does. Still would be quite hard 700+ question.

Last step:
x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+3m)+5. Can we now conclude that that n^2+m^2 is even?

x-y=5n+1 and x+y=5m+3. Add them up: 2x=5(n+m)+4 --> x=\frac{5(n+m)}{2}+2. As x is an integer n+m must be divisible by 2, hence either both are even or both are odd, in any case n^2+m^2 is even and divisible by 2. Hence x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+15m)+5 is an integer and divisible by 5.

In this case answer C.

Hope it's clear.
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That was tough, you are scaring me.
I just joined yesterday with a hope that i will get good DS questions.
This is too tough

1- case
(x-y)/5 --> 1 (Remainder)
(x-y)^2/5--> 1(Remainder)
nothing can be deduced for remainder of x^2+y^2 divided by 5
insufficient

2- case

(x+y)/5-->3 (remainder)
(x+y)^2/5-->4(remainder)

insufficient as nothing can be deduced for remainder of x^2+y^2 divided by 5

however adding two cases
2(x^2+y^2)/5--> remainder 0 (1+4=5 divisible by 5) hence divisible by 5

hence both the cases are required

alternatively take x=12, y=6
and cross check

hence answer is c

what is OA