If k is a positive even integer, which of the following must be an odd integer? I....

Question

If k is a positive even integer, which of the following must be an odd integer? I. k^2-3 k+4 II. k^5+3 III. 7 k-7

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D We start by expressing k as twice an integer to reflect that it is even. k = 2n for some integer n We substitute k = 2n into k^2 - 3k + 4 and factor out 2 to test its parity. (2n)^2 - 3(2n) + 4 = 4n^2 - 6n + 4 =2(2n^2 - 3n + 2) =2(integer) 4n^2 - 6n + 4 is even We substitute k = 2n into k^5 + 3 and evaluate its parity. (2n)^5 + 3 = 32n^5 + 3 32n^5 is even 32n^5 + 3 is odd We substitute k = 2n into 7k - 7 and evaluate its parity. 7(2n) - 7 = 14n - 7 14n is even 7 is odd 14n - 7 is odd We see that only the second and third expressions must be odd, so the correct answer is choice D. Answer: D

Explanation	Calculations	Help
$We start by expressing k as twice an integer to reflect that it is even.$	$k = 2 n for some integer n$
$We substitute k = 2 n into k^{2} - 3 k + 4 and factor out 2 to test its parity.$	$(2 n)^{2} - 3 (2 n) + 4 = 4 n^{2} - 6 n + 4 = 2 (2 n^{2} - 3 n + 2) = 2 \times (integer) 4 n^{2} - 6 n + 4 is even$
$We substitute k = 2 n into k^{5} + 3 and evaluate its parity.$	$(2 n)^{5} + 3 = 32 n^{5} + 3 32 n^{5} is even 32 n^{5} + 3 is odd$
$We substitute k = 2 n into 7 k - 7 and evaluate its parity.$	$7 (2 n) - 7 = 14 n - 7 14 n is even 7 is odd 14 n - 7 is odd$
$We see that only the second and third expressions must be odd, so the correct answer is choice D.$	$Answer: D$

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