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MathRevolution
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If x and y are positive integers, is xy even? 1) x+y is odd 2) xy is e 27 Jul 2017, 01:10
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77% (00:53) correct 23% (00:52) wrong based on 141 sessions

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If x and y are positive integers, is xy even?

1) \(x+y\) is odd
2) \(x^y\) is even
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D
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sashiim20
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If x and y are positive integers, is xy even? 1) x+y is odd 2) xy is e 27 Jul 2017, 01:16
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MathRevolution
If x and y are positive integers, is xy even?

1) \(x+y\) is odd
2) \(x^y\) is even
Show more

1) \(x+y\) is odd

\(Even + Odd = Odd\)

Therefore either \(x\) or \(y\) has to be even.

Hence \(xy\) will be Even.

I is Sufficient.

2) \(x^y\) is even

Even raised to any integer is even.

Odd raised to any integer will be odd.

\(x\) is even. Therefore \(xy\) will be even.

II is Sufficient.

Answer (D)...
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If x and y are positive integers, is xy even? 1) x+y is odd 2) xy is e 30 Jul 2017, 18:56
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==> If you modify the original condition and the question, in order to get xy=even, it can either be x=even? or y=even?. For con 1), from (x,y)=(odd,even),(even,odd), you always get yes, hence it is sufficient. For con 2), you always get x=even, hence it is yes and sufficient.

The answer is D.
Answer: D
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If x and y are positive integers, is xy even? 1) x+y is odd 2) xy is e 02 Apr 2019, 10:00
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sashiim20
MathRevolution
If x and y are positive integers, is xy even?

1) \(x+y\) is odd
2) \(x^y\) is even

1) \(x+y\) is odd

\(Even + Odd = Odd\)

Therefore either \(x\) or \(y\) has to be even.

Hence \(xy\) will be Even.

I is Sufficient.

2) \(x^y\) is even

Even raised to any integer is even.

Odd raised to any integer will be odd.

\(x\) is even. Therefore \(xy\) will be even.

II is Sufficient.

Answer (D)...
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When u r saying that x^y , for any number the answer will be even if x is even and odd if x is odd.

So in this case y can be odd or even .


hence the product xy can be even or odd .
So how come the condition is sufficint
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If x and y are positive integers, is xy even? 1) x+y is odd 2) xy is e 02 Apr 2019, 10:16
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aaggarwal191
sashiim20
MathRevolution
If x and y are positive integers, is xy even?

1) \(x+y\) is odd
2) \(x^y\) is even

1) \(x+y\) is odd

\(Even + Odd = Odd\)

Therefore either \(x\) or \(y\) has to be even.

Hence \(xy\) will be Even.

I is Sufficient.

2) \(x^y\) is even

Even raised to any integer is even.

Odd raised to any integer will be odd.

\(x\) is even. Therefore \(xy\) will be even.

II is Sufficient.

Answer (D)...
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When u r saying that x^y , for any number the answer will be even if x is even and odd if x is odd.

So in this case y can be odd or even .


hence the product xy can be even or odd .
So how come the condition is sufficint
Show more

Even * Even = Even
Even * Odd = Even
Odd * Even = Even
Odd * Odd = Odd

Given: \(x^y\) even. Hence, \(x\) is even. If it was odd, it would be: \(odd * odd * odd\) y times, which would give us an odd number.
Regardless of \(y\) nature (even or odd), \(xy\) will be always even.

Hope it's clear.
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