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Difficulty:
5%
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Question Stats:
82% (00:52) correct
18%
(00:51)
wrong
based on 210
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If n is an integer, which of the following must be an even number?
I. \(n^2+n\)
II. \(n^2-n\)
III. \(2n\)
A. I only
B. II only
C. III only
D. I and II
E. I , II and III
I. \(n^2+n\)
II. \(n^2-n\)
III. \(2n\)
A. I only
B. II only
C. III only
D. I and II
E. I , II and III
23 Jan 2024, 12:47
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On these problems, I like to distill the algebra into real, plain-English numbers. The benefit to doing this is coming up with answers that work in all scenarios without having to plug in numbers.
So the question at hand is, "which of the following must be an even number?" Recall our multiplication rule: a number is always even if multiplied by an even number. And our addition rule: a number is even if both numbers in the sum are odd or if both numbers in the sum are even. Onto the answer choices:
I) While plugging in numbers will work here, anytime you see \(n^2 + n\), know that we are multiplying two consecutive numbers, because it is simply \((n) * (n + 1)\) when factored. This alone is enough to tell us it is even because remember, for any set of integers, we're always alternating between even and odd. So, by definition, we're triggering our multiplication rule. ACs B and C are out.
II) This is the mirror image of I). Factored out, we get \(n * (n - 1)\), or the multiplication of two consecutive numbers. The multiplication rule is triggered again, so AC A is out.
III) Once again, our multiplication rule is triggered here because we're multiplying by an even number. AC D is out, leaving AC E as the correct answer.
So the question at hand is, "which of the following must be an even number?" Recall our multiplication rule: a number is always even if multiplied by an even number. And our addition rule: a number is even if both numbers in the sum are odd or if both numbers in the sum are even. Onto the answer choices:
I) While plugging in numbers will work here, anytime you see \(n^2 + n\), know that we are multiplying two consecutive numbers, because it is simply \((n) * (n + 1)\) when factored. This alone is enough to tell us it is even because remember, for any set of integers, we're always alternating between even and odd. So, by definition, we're triggering our multiplication rule. ACs B and C are out.
II) This is the mirror image of I). Factored out, we get \(n * (n - 1)\), or the multiplication of two consecutive numbers. The multiplication rule is triggered again, so AC A is out.
III) Once again, our multiplication rule is triggered here because we're multiplying by an even number. AC D is out, leaving AC E as the correct answer.
