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Bunuel
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Bunuel
If x and y are positive integers, then xy + x is

A. Always even
B. Always odd
C. Even whenever x is even
D. Even whenever x + y is odd
E. It cannot be determined from the information given

x and y are positive integers, then xy + x..

Let's take even as 2 and odd as 3.


IMO option A is correct answer..

If you consider your fourth scenario, that turns out to be odd.
Hence option C should be correct.

Ya..you are correct..missed to check that...
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Bunuel
If x and y are positive integers, then xy + x is

A. Always even
B. Always odd
C. Even whenever x is even
D. Even whenever x + y is odd
E. It cannot be determined from the information given

I think the correct answer here is E.
See the following table
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nayanparikh

In your table if x is even, then xy+x is even, so C.
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xy + x can be factorized as x(y + 1) by taking out x common.

Now, anything multiplied by even is always even.

Answer (C).
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Bunuel
If x and y are positive integers, then xy + x is

A. Always even
B. Always odd
C. Even whenever x is even
D. Even whenever x + y is odd
E. It cannot be determined from the information given


Case x y xy + x
Case1 E E E*E + E = E + E = Even always
Case2 E O E*O + O = E + O = Odd always
Case3 O E O*E + E = E + E = Even always
Case4 O O O*O + O = O + O = Even always

A. Always even - NO
B. Always odd - NO
C. Even whenever x is even - YES
D. Even whenever x + y is odd - NO [One even & one odd, Case2 —> Odd, Case3 —> Even]
E. It cannot be determined from the information given - NO

IMO Option C

Pls Hit kudos if you like the solution

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Bunuel
If x and y are positive integers, then xy + x is

A. Always even
B. Always odd
C. Even whenever x is even
D. Even whenever x + y is odd
E. It cannot be determined from the information given

Why are we considering that x and y are distinct positive integers? If x and y are the same, then the result can be even in case x is odd.
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