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

now as per equation either x and y will be odd or both will be even .


I. xy + 1
if both are even then even plus odd can be odd
this equation can be correct
II. 3x + 5y
if x and y are even then above equation will even and if both are odd then odd+ odd is again even
so this equation cannot be odd
incorrect
III. xy + y
if xand y are odd then above equation will be odd +odd =even
if both are even then also even + even again even
This cannot be odd

so only equation 1 is correct

Answer is Option A
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Given x^2 + y^2 is even which means x,y should both be either odd or even
Taking first choice
i) xy + 1 If x, y is even then it might be odd
ii) 3x + 5y If x,y both are odd or even it'll be even
iii) xy + y => x(y+1) Here it'll always be even.

Hence Option A is correct answer
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If x and y are integers and x^2 + y^2 is even, which of the following may be odd?

I. xy + 1
II. 3x + 5y
III. xy + y

A. I only
B. III only
C. I and II only
D. I and III only
E. I, II, and III


­
X and Y are integers, means they can be both positive or negative.

x^2 + y^2 is even.

Sum of squares of two integers is Even, only when two cases happen.

Case 1: Odd + Odd = Even

Case 2: Even + Even = Even

which of the following may be odd?

I. xy + 1
II. 3x + 5y
III. xy + y

I. xy + 1

Applying case 1 and Case 2, we get

odd * odd + 1 = odd + 1 = Even

even * even + 1 = even + 1 = Odd

I may be ODD.

II. 3x + 5y

Applying case 1 and Case 2, we get

3*odd + 5* Odd = Odd + Odd = Even

3* Even + 5* Even = Even + Even = Even

II cannot be odd.

III. xy + y

Applying case 1 and Case 2, we get

Odd * Odd + Odd = Odd + Odd = Even

Even * Even + Even = Even + Even = Even

III cannot be odd.

Option A
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This is an interesting question here is the question asking which must be ODD.
The fastest way to solve it is the application of the formula of Even and Odd but also beware of the trap.

Lets breakdown the question stem-

If x and y are integers and x^2 + y^2 is even, which means the x and y both are same parity either both of them are Odd or Even. So the best is to consider the lowest positive integer so in that case We'll take the pair of E + E = E and the smallest positive number of even is 0,So let the value of X and Y both 0. Now lets break down the options.

Option -1
XY + 1
=(0)(0) + 1 = 1 so it is definitely an Odd

Option - 2
3X + 5Y
=3*0 + 5*0 = 0 + 0 = 0, which is an Even and question is asking about Odd so We'll eliminate it

Option - 3
XY + Y
=(0)(0) + 0 = 0, which is also an Even so We'll eliminate it.

The Correct Ans - A (Option- 1 Only)
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Bunuel
If x and y are integers and x^2 + y^2 is even, which of the following may be odd?

I. xy + 1
II. 3x + 5y
III. xy + y

A. I only
B. III only
C. I and II only
D. I and III only
E. I, II, and III
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