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If the smaller of 2 consecutive odd integers is a multiple of 5 , which of the following could NOT be the sum of these 2 integers?

Explanation	Calculations	Help
$We translate “2 consecutive odd integers” by letting the smaller be a and the next be a + 2 .$	$a, a + 2$	Theory & Tactics Method Card TRANS5-D Click to view full details
$We translate “the smaller is a multiple of 5” by writing a = 5 n, where n is an integer.$	$a = 5 n$	Theory & Tactics Method Card TRANS5-E Click to view full details
$Since a is odd, n must be odd; we translate “ n is odd” by letting n = 2 k + 1 .$	$n = 2 k + 1$	Theory & Tactics Method Card TRANS5-A Click to view full details
$We translate “the sum of these 2 integers” by forming a + (a + 2) .$	$Sum = a + (a + 2)$	Theory & Tactics Method Card TRANS1-C Click to view full details
$We substitute a = 5 (2 k + 1) into the sum and simplify.$	$5 (2 k + 1) + (5 (2 k + 1) + 2) = 10 (2 k + 1) + 2 = 20 k + 12$
$A valid sum must be of the form 20 k + 12 . Testing option C: 22 = 20 \times 1 + 2, which does not leave 12, so 22 cannot be the sum.$	$22 = 20 \times 1 + 2 22$

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If the smaller of 2 consecutive odd integers is a multiple of 5 , which of the following could NOT be the sum of these 2 integers?

Explanation	Calculations	Help
$We translate “2 consecutive odd integers” by letting the smaller be a and the next be a + 2 .$	$a, a + 2$	Theory & Tactics Method Card TRANS5-D Click to view full details
$We translate “the smaller is a multiple of 5” by writing a = 5 n, where n is an integer.$	$a = 5 n$	Theory & Tactics Method Card TRANS5-E Click to view full details
$Since a is odd, n must be odd; we translate “ n is odd” by letting n = 2 k + 1 .$	$n = 2 k + 1$	Theory & Tactics Method Card TRANS5-A Click to view full details
$We translate “the sum of these 2 integers” by forming a + (a + 2) .$	$Sum = a + (a + 2)$	Theory & Tactics Method Card TRANS1-C Click to view full details
$We substitute a = 5 (2 k + 1) into the sum and simplify.$	$5 (2 k + 1) + (5 (2 k + 1) + 2) = 10 (2 k + 1) + 2 = 20 k + 12$
$A valid sum must be of the form 20 k + 12 . Testing option C: 22 = 20 \times 1 + 2, which does not leave 12, so 22 cannot be the sum.$	$22 = 20 \times 1 + 2 22$

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22