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If p and r are integers and p + r is odd, which of the following must be true? I. p r is even. II. p^{2} + r^{2} is odd. III. (p - r)^{2} is even.

Explanation	Calculations	Help
$We first note that p + r being odd means one of p and r is odd and the other is even.$	$p is odd and r is even p is even and r is odd$	Theory & Tactics Method Card OE1-C Click to view full details
$For statement I, since one factor is odd and the other is even, their product must be even.$	$p r = odd \times even = even p r = even \times odd = even$	Theory & Tactics Method Card OE2-C Click to view full details Theory & Tactics Method Card OE2-D Click to view full details
$For statement II, squaring preserves parity, and then adding an odd square and an even square yields an odd result.$	$p^{2} = odd \times odd = odd r^{2} = even \times even = even p^{2} + r^{2} = odd + even = odd$	Theory & Tactics Method Card OE2-A Click to view full details Theory & Tactics Method Card OE2-B Click to view full details Theory & Tactics Method Card OE1-C Click to view full details
$For statement III, subtracting an even from an odd or an odd from an even gives an odd, and squaring an odd gives an odd, so (p - r)^{2} is odd, not even.$	$p - r = odd - even = odd p - r = even - odd = odd (p - r)^{2} = odd \times odd = odd$	Theory & Tactics Method Card OE3-C Click to view full details Theory & Tactics Method Card OE3-D Click to view full details Theory & Tactics Method Card OE2-A Click to view full details

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If p and r are integers and p + r is odd, which of the following must be true? I. p r is even. II. p^{2} + r^{2} is odd. III. (p - r)^{2} is even.

Explanation	Calculations	Help
$We first note that p + r being odd means one of p and r is odd and the other is even.$	$p is odd and r is even p is even and r is odd$	Theory & Tactics Method Card OE1-C Click to view full details
$For statement I, since one factor is odd and the other is even, their product must be even.$	$p r = odd \times even = even p r = even \times odd = even$	Theory & Tactics Method Card OE2-C Click to view full details Theory & Tactics Method Card OE2-D Click to view full details
$For statement II, squaring preserves parity, and then adding an odd square and an even square yields an odd result.$	$p^{2} = odd \times odd = odd r^{2} = even \times even = even p^{2} + r^{2} = odd + even = odd$	Theory & Tactics Method Card OE2-A Click to view full details Theory & Tactics Method Card OE2-B Click to view full details Theory & Tactics Method Card OE1-C Click to view full details
$For statement III, subtracting an even from an odd or an odd from an even gives an odd, and squaring an odd gives an odd, so (p - r)^{2} is odd, not even.$	$p - r = odd - even = odd p - r = even - odd = odd (p - r)^{2} = odd \times odd = odd$	Theory & Tactics Method Card OE3-C Click to view full details Theory & Tactics Method Card OE3-D Click to view full details Theory & Tactics Method Card OE2-A Click to view full details

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D