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If x and y are integers, is x^2 + 11x + 13y an even number?

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Bunuel
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If x and y are integers, is x^2 + 11x + 13y an even number? [#permalink] New post   17 Nov 2023, 07:00
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If \(x\) and \(y\) are integers, is \(x^2+11x+13y\) an even number?

(1) \(x=11\)
(2) \(y=13\)
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B
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Its_me_aka_ak
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Re: If x and y are integers, is x^2 + 11x + 13y an even number? [#permalink] New post   27 Nov 2023, 03:53
Bunuel wrote:
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

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Re: If x and y are integers, is x^2 + 11x + 13y an even number? [#permalink] New post   27 Nov 2023, 04:53
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Its_me_aka_ak wrote:
Bunuel wrote:
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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Its_me_aka_ak
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Re: If x and y are integers, is x^2 + 11x + 13y an even number? [#permalink] New post   28 Nov 2023, 00:09
Bunuel wrote:
Its_me_aka_ak wrote:
Bunuel wrote:
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.

Oh, now it makes sense. Thank you
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Re: If x and y are integers, is x^2 + 11x + 13y an even number? [#permalink]
28 Nov 2023, 00:09
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Bunuel
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