If x and y are positive integers, what is the remainder when (x^2 + y^

To know this, we need to know the remainder of x^2 and y^2 when divided by 4.
As equations look a bit complex and our statements are very simple, we'll first try a few numbers.
This is an Alternative approach.

(1) As there are only 4 possible classes of integers with respect to 'remainder when divided by 4' (remainder = 0,1,2,3), checking the first 4 pairs is enough.
Say x = 1, y = 2. Then 1^2+2^2 = 5 which has remainder 1. Say x = 2, y = 3. Then 4 + 9 = 13 which has remainder 1. Say x = 3, y = 4. Then 9+16=25 which also has remainder 1. Last x = 4, y = 5 Which gives 16+25=41, remainder 1.
Then (1) is sufficient.

(2) Once again, there are only 4 options, depending on the remainder classes when divided by 4: x has remainder 0 and y remainder 1, x has remainder 0 and y remainder 3, x remainder 2 and y remainder 1 and x remainder 2 and y remainder 3. Looking at (1), we'll see what we've already checked: (4,5) is the first option, (4,3) is the second option, (2,1) is the third option and (2,3) the fourth option. So we'll get the same answer.
Sufficient.

(D) is our answer.

\(\frac{x^2 + y^2}{4}\)= Remainder?

Statement 1: x and y are consecutive integers

Case 1 : \(\frac{2^2+3^2}{4}\)= \(\frac{13}{4}\){Remainder is 1}
Case 2: \(\frac{3^2+4^2}{4}\)= \(\frac{25}{4}\){Remainder is 1}

Helen Statment 1 is sufficient.

Statement 2: x = even & y = odd
Same as Statement 1. Hence it is sufficient too.

Answer IMO is D

The remainder of (x^2 + y^2) divided by 4, is the remainder of x^2 divided by 4 + y^2 divided by 4.

St1:
-x or y must be an even number and a square of an even number divided by 4 have a remainder of 0.
-The other number must be odd, so square of an positive odd number should be: 1, 9, 25, 49.... all of them divided by 4 have a remainder of 1.

0+1 = 1

Statment 1 is sufficient.

Stat 2 = Sta 1 is sufficient.

D is the answer

Par of GMAT CLUB'S New Year's Quantitative Challenge Set

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