Its_me_aka_ak
Bunuel
If \(x\) and \(y\) are integers, is \(x^2+11x+13y\) an even number?
(1) \(x=11\)
(2) \(y=13\)
How is it B I don’t get it. Bunuel kindly help
Posted from my mobile device If \(x\) and \(y\) are integers, is \(x^2+11x+13y\) an even number?\(x^2+11x+13y=x(x+11)+13y\). Observe that the first term, x(x+11), is even regardless of x: if x is even, it's obviously even, and if x is odd, then x + 11 is even, making x(x+11) even again. Thus, \(x^2+11x+13y=even+13y\), which will be odd if y is odd and even if y is even. As we can see, we only need to know whether y is odd or even to answer the question.
(1) \(x=11\).
Not sufficient.
(2) \(y=13\).
As discussed above, knowing the even/odd parity of y is sufficient. To demonstrate: y is odd, therefore, \(x^2+11x+13y=even+13y=even+odd=odd\). Sufficient.
Answer: B.
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