If x and y are integers and x^2 + y^2 is odd, which of the following : Problem Solving (PS)

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16 Oct 2019, 21:42

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x^2 + y^2 is odd only when one is definitely even and the other one definitely odd

(A) x --> Can be even or Odd --> NO

(B) y --> Can be even or Odd --> NO

(C) x+y --> Even + Odd or Odd + Even = Odd always --> NO

(D) xy+y --> x(y + 1)
--> Case 1: If x = odd, y = even ; x(y + 1) = odd*odd = Odd
--> Case 2: If x = even, y = odd ; x(y + 1) = even*even = Even
--> Two outcomes are possible --> NO

(E) xy = odd*even or even*odd = Even always --> YES

IMO Option E

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16 Oct 2019, 21:55

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Just the sum of two odd and even numbers will be odd.so, one of thr x² or y² are odd.
We know just the odd number to the power of 2 will be odd.
So, one of the x or y must be odd and their sum will be odd too. But their product will be surelly even.
Option E

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16 Oct 2019, 22:27

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If x and y are integers and x2+y2 is odd, which of the following must be even?

If x2 + y2 is odd, this can be inferred that one of x2 and y2 is odd, and another is even.
So, we'll get either x is odd, y is even OR x is even, y is odd.

Now, let's see each answer choice on which choice yield EVEN

(A) x - can be both even/odd
(B) y - can be both even/odd
(C) x+y - definitely odd: since if x is even, y is odd and x+y = odd OR if y is even, x is odd, x+y = odd
(D) xy+y can be both even/odd, if x is even, y is odd; xy+y = odd OR if y is even, x is odd; xy+y = even
(E) xy - one of x and y needs to be even, which even * any number = even - CORRECT answer choice

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17 Oct 2019, 02:22

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Given: x and y are integers => (x,y)= 0, +I, -I
x^2 + y^2 = odd => x and y have opposite even-odd nature. (e+ o => o)

Q Type: Must be true

1. x - can be even or odd
2. y - can be even or odd
3. x+ y => odd (given)
4. xy+ y ----- can be even or odd
if x=2, y =3 ,then xy+ y = 6+3 = 9 (odd)
if x=3, y=2, then xy+ y = 6+ 2 = 8 (even)

5. xy -- always even
Correct answer- E

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17 Oct 2019, 03:13

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Quote:

If x and y are integers and x^2+y^2 is odd, which of the following must be even?

(A) x
(B) y
(C) x+y
(D) xy+y
(E) xy

\(x^2+y^2=odd…e+o=odd\)

(A) x: x=(o,e)
(B) y: y=(o,e)
(C) x+y: o+e=o
(D) xy+y: y(x+1)=o(e+1)=o(o)=o
(E) xy: oe=e

Answer (E)

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17 Oct 2019, 04:56

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Imo. E

If x and y are integers and x^2+y^2 is odd, which of the following must be even?

x^2+y^2 = odd means one is even and another one is odd, but can't say which one is odd or even. Hence, A & B out.

(A) x - may be even or odd
(B) y - may be even or odd
(C) x+y - must be odd
(D) xy+y - xy = even & y = even or odd. So, xy +y = even or odd
(E) xy - must be even.

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17 Oct 2019, 06:12

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for the given condition to to be valid of x2+y2 is odd
either of them has to be even and odd
so IMO E ; xy will be even

If x and y are integers and x2+y2 is odd, which of the following must be even?

(A) x
(B) y
(C) x+y
(D) xy+y
(E) xy

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17 Oct 2019, 06:36

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For the x^2 + y^2 to be odd, x,y must be odd and even in either combination

So only xy values will be even

OA:E

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17 Oct 2019, 06:59

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Raising an integer by any power will not change its odd/even property.
Also note,
Odd + Odd = Even
Even + Even = Even
Even + Odd = Odd
So in order for x^2 + y^2 to be odd, one of the variables has to even and one has to be odd.
Using numbers we can start by x = 2 and y = 3 or x = 3 and y = 2.

A.) x = 2, yes if x = 2 - no if x = 3 (incorrect as it could or could not be even)
B.) same as above
C.) x + y, 2 + 3 = 5 which is odd (incorrect)
D.) xy + y - if x = 2/y = 3, 2(3) + 3 = 6 + 3 = 9 (ODD), if x = 3/y = 2, 3(2) + 2 = 6 + 2 = 8, (EVEN) (could be either so incorrect)
E.) xy - 2(3) = 6 (EVEN)

Note,
Odd * Odd = Odd
Even * Odd = Even
Even * Even = Even

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17 Oct 2019, 07:03

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x and y are integers
\(x^{2}+ y^{2}\)— odd number

—> In order \(x^{2}+ y^{2}\) to be odd number, one of x and y must be odd integer.

Which of the following must be even?

A) x could be odd

B) y could be odd

C) x+y — as above mentioned, odd+even=odd number

D) xy+y= y*(x+1) could be even or odd number
—> if x=0(even) and y=1
1*(0+1)=1 (odd number)

E) xy =even*odd= even ( always true)

The answer is E

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17 Oct 2019, 07:45

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Quote:

If x and y are integers and x^2+y^2 is odd, which of the following must be even?

(A) x
(B) y
(C) x+y
(D) xy+y
(E) xy

x^2+y^2= Odd, is possible in two case given below:
1. x is odd and y is even
2. x is even and y is odd.

A. Wrong. x can be even or odd.
B. Wrong. y can be even or odd.
C. Wrong. In both the cases, x+y is odd.
D. Wrong. xy+y=Even*odd+Odd=Even+odd=Odd or Odd*even+even=Even.
E. Correct.

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17 Oct 2019, 10:04

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If x and y are integers and \(x^2+y^2\) is odd, which of the following must be even?

Expression \(x^2+y^2\) is odd .e. o.o + e.e OR e.e or o.o. Hence if x is odd then y is even and vice-a-versa. Let's check

(A) x
It can be ether odd or even.

(B) y
Same as above

(C) x+y
o + e OR e + o. Always Odd.

(D) xy+y
o.e + e = e OR e.o + o = o. Either even or Odd case.

(E) xy

o.e OR e.o. Both cases are even.

Answer E.

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17 Oct 2019, 10:22

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X^2 + y^2 = odd
This implies => odd + even = odd.

Means one if them must be ODD and one of them must be EVEN. Since (odd)^2 = odd and (even)^2 = even.

Implies xy = even!

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17 Oct 2019, 21:06

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If x and y are integers and x^2+y^2 is odd, which of the following must be even?

If x and y are integers and x^2+y^2 is odd, meaning either one of them must be even and the other must be odd.
if x is odd, y is even
if x is even, y is odd

(A) x; can be either even or odd depends on y
(B) y; can be either even or odd depends on x
(C) x+y; odd plus even is odd therefore not correct
(D) xy+y; if x is odd the equation is even if x is even the equation is odd therefore not correct
(E) xy; odd multiply even is always even yay correct

therefore
E

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04 Dec 2024, 11:10

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solved in 45 seconds. test cases: 1 and 1, 2 and 2, 2 and 3, and 3 and 3.

E

If x and y are integers and x^2 + y^2 is odd, which of the following

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If x and y are integers and x^2 + y^2 is odd, which of the following

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