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

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

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New post 27 Nov 2018, 01:04
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If x and y are positive integers, what is the remainder when (x^2 + y^  [#permalink]

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New post 27 Nov 2018, 01:32
Bunuel wrote:
If x and y are positive integers, what is the remainder when (x^2 + y^2) is divided by 4 ?


(1) x and y are consecutive integers

(2) x is even and y is odd


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.
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Re: If x and y are positive integers, what is the remainder when (x^2 + y^  [#permalink]

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New post 27 Nov 2018, 05:17
\(\frac{x^2 + y^2}{4}\)= 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
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Re: If x and y are positive integers, what is the remainder when (x^2 + y^  [#permalink]

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New post 27 Nov 2018, 07:24
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
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Re: If x and y are positive integers, what is the remainder when (x^2 + y^  [#permalink]

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Re: If x and y are positive integers, what is the remainder when (x^2 + y^  [#permalink]

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New post 29 Dec 2018, 04:18
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Statement 1: if x & y are consecutive integers, \((x^2 + y^2)\) will be sum of square of one even positive integer and square of one odd positive integer. Now, square of even integer is always divisible by 4 since even integer will have 2 as a factor and since 2 is squared, it will be divisible by 4. Coming to the square of the odd integer, this number when divided by 4 will always give 1 as remainder. Let's understand why:

Odd integer can be written as 2n+1. Now \((2n+1)^2\)=4\(n^2\)+1+4n. When this expression is divided by 4, we get that 4\(n^2\)+4n to be divisible by 4 and 1/4 leaving a remainder of 1.

Thus, \((x^2 + y^2)\) when divided by 4, will always leave a remainder of 1. Hence, sufficient

Statement 2: This statement is a repeat of the same concept used in statement 1. Hence, by default this statement is also sufficient

Answer is D
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Re: If x and y are positive integers, what is the remainder when (x^2 + y^   [#permalink] 29 Dec 2018, 04:18
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