"xy + z is odd"
two numbers can add to give an odd sum only if they have opposite parity. hence:
case 1: xy is odd, z is even
there's only one way this can happen:
x = odd, y = odd, z = even. ............(1)
case 2: xy is even, z is odd
there are 3 ways in which this can happen:
x = even, y = even, z = odd .........(2a)
x = odd, y = even, z = odd ..........(2b)
x = even, y = odd, z = odd .........(2c)
statement (1)
pull out x:
xy+xz = x(y + z) is even.
this means that at least one of x and (y + z) is even.
--> if x is even, regardless of the parity of (y + z), then the answer to the prompt question is "yes" and we're done.
--> the other possibility would be x = odd and (y + z) = even. this is impossible, though, as it doesn't satisfy any of the cases above.
therefore, the answer must be "yes".
sufficient.
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statement (2)
this means that y and xz have opposite parity.
--> y = even, xz = odd --> this means x = odd, y = even, z = odd. that's case (2b), which gives "no" to the question.
at this point you're done, because STATEMENTS CAN'T CONTRADICT EACH OTHER, se you know that "yes" MUST be a possibility with this statement (as statement #1 gives exclusively "yes" answers).
if you use this statement first, you'll have to keep going through the cases.
insufficient.
ans = A..
