Data Sufficiency

If x and y are integers, what is the remainder when x² + y² 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

If x and y are integers, what is the remainder when (x^2 + y^2) divided by 5?
A) When (x - y) is divided by 5, the remainder is 1.
B) When (x + y) is divided by 5, the remainder is 2.

Statement 1:
x-y = 5a + 1 = 1, 6, 11, 36...
Since x and y can take on many different integer values, INSUFFICIENT.

Statement 2:
x+y = 5b + 2 = 2, 7, 12, 17...
Since x and y can take on many different integer values, INSUFFICIENT.

Statements combined:
x-y = 1, 6, 11, 21...
x+y = 2, 7, 12, 17...
(x-y) + (x+y) = 2x.
Since x is an integer, (x-y) + (x+y) must be even.
Thus, x-y and x+y are either BOTH EVEN or BOTH ODD.

Note the following:
(x-y)² + (x+y)² = (x² - 2xy + y²) + (x² + 2xy + y²) = 2(x² + y²).

(x-y)² = 1, 36, 121, 256...
(x+y)² = 4, 49, 144, 289...

If (x-y)² and (x+y)² are BOTH EVEN, their sum will have a units digit of 0:
36+4=40.
256+144=400.
A units digit of 0 implies a multiple of 10.

If (x-y)² and (x+y)² are BOTH ODD, their sum will have a units digit of 0:
1+49=50.
121+289=410.
A units digit of 0 implies a multiple of 10.

Thus, whether (x-y)² and (x+y)² are BOTH EVEN or BOTH ODD:
(x-y)² + (x+y)² = multiple of 10.
2(x² + y²) = multiple of 10
x² + y² = (multiple of 10)/2 = multiple of 5.

Thus, when x²+y² is divided by 5, the remainder will be 0.
SUFFICIENT.

The correct answer is C.

richachampion wrote:If x and y are integers, what is the remainder when x² + y² 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

Statement 1 - Potential values for x - y: 1, 6, 11, 16

Case 1: x - y = 1; x = 1 and y = 0. x² + y² = 1 + 0 = 1. Remainder of 1 when divided by 5.
Case 2: x - y = 1; x = 2 and y = 1. x² + y² = 4 + 1 = 5. Remainder of 0 when divided by 5.
Not Sufficient.

Statement 2 - Potential values for x + y: 2, 7, 12, 17

Case 1: x + y = 2; x = 1 and y = 1. x² + y² = 1 + 1 = 1. Remainder of 2 when divided by 5.
Case 2: x + y = 7; x = 4 and y = 3. x² + y² = 16 + 9 = 25. Remainder of 0 when divided by 5.
Not Sufficient.

Together: Case 1: x - y = 1 and x + y = 7; x = 4 and y = 3
x² + y² = 16 + 9 = 25. Remainder of 0 when divided by 5.

Case 2: Case 1: x - y = 6 and x + y = 12; x = 9 and y = 3
x² + y² = 81 + 9 = 90. Remainder of 0 when divided by 5.

No matter what we pick, we'll get a reminder of 0. Together the statements are sufficient. Answer is C

richachampion wrote:If x and y are integers, what is the remainder when x² + y² 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

Fun algebraic solution for testing together:

x - y = 5t + 1
(x - y)² = (5t + 1)²
x² - 2xy + y² = 25t² + 10t + 1

x + y = 5z + 2
(x - y)² = (5z + 2)²
x² + 2xy + y² = 25z² + 20z + 4


Sum the equations in red to get

2x² + 2y² = 25t² + 25z² + 10t + 20z + 5
x² + y² = 5(5t² + 5z² + 2t + 4z+ 1)/2
x² + y² = (Multiple of 5)/2
If we're dealing with integers, x² + y² = will have to be a multiple of 5.