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22 Jan 2023, 02:16
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C
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53% (02:13) correct
47%
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wrong
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If x and y are positive integers, what is the remainder of x^2 + y^2, when it is divided by 4?
(1) x and y are different prime numbers.
(2) x – y = 2
(1) x and y are different prime numbers.
(2) x – y = 2
22 Jan 2023, 13:59
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Bunuel
Statement 1
(1) x and y are different prime numbers.
Case 1:
x = 2 & y = 3
x^2 + y^2 = 13
Remainder(13/4) = 1
Case 2:
x = 3 & y = 5
x^2 + y^2 = 34
Remainder(13/4) = 2
The statement is not sufficient, and we can eliminate A and D.
Statement 2
(2) x – y = 2
x = 2 + y
\(x^2 = y^2 + 4 + 4y\)
Adding \(y^2\) on both sides
\(x^2 + y^2= 2y^2 + 4 + 4y\)
The statement doesn't provide any further information, hence not sufficient.
Eliminate B.
Combined
From 2, we know that \(x^2 + y^2= 2y^2 + 4 + 4y\)
And y is a prime number. As y > 3, we can represent y as \(6n \pm 1\)
\(2y^2 + 4 + 2y = 2(6n \pm 1)^2 + 4 + 4(6n \pm 1)\)
Note, that the last two terms are divisible by 4, hence the remainder will depend only on the first term
\( 2(6n \pm 1)^2 = 2(36n^2 + 1 \pm 12n)\)
The first and the third term is divisible by 4, hence the remainder depends on the middle term.
Remainder of (x^2 + y^2) = Remainder (2/4) = 2.
The statements combined are sufficient.
Option C
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