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A garden path is constructed with m rectangular stone pavers of equal length and a gravel strip between each pair of consecutive pavers. Each gravel strip is \frac{2}{3} as long as a paver. The total length of the path from the start of the first paver to the end of the last paver is 420 inches. If m is an integer and the length, in inches, of each gravel strip is an integer greater than 10, what is the value of m ?

Explanation	Calculations	Help
$We pick p as the length of each paver and g as the length of each gravel strip.$	$p = length of each paver (in inches) g = length of each gravel strip (in inches)$	Theory & Tactics Method Card TRANS1-A Click to view full details
$We translate "The total length of the path from the start of the first paver to the end of the last paver is 420 inches." into an equation.$	$m p + (m - 1) \times g = 420$	Theory & Tactics Method Card TRANS1-B Click to view full details Theory & Tactics Method Card TRANS1-C Click to view full details
$We translate "Each gravel strip is \frac{2}{3} as long as a paver." into an equation.$	$g = \frac{2}{3} p$	Theory & Tactics Method Card TRANS1-E Click to view full details
$We substitute g = \frac{2}{3} p into the total-length equation and simplify.$	$m p + (m - 1) \times \frac{2}{3} p = 420 p \times (m + \frac{2}{3} (m - 1)) = 420 p \times \frac{5 m - 2}{3} = 420 (5 m - 2) p = 1260$
$We need g to be an integer greater than 10, and since g = \frac{2}{3} p, we let p = 3 k so that g = 2 k .$	$p = 3 k g = 2 k$	Theory & Tactics Method Card TRANS1-A Click to view full details
$We replace p = 3 k in (5 m - 2) p = 1260 and simplify.$	$(5 m - 2) (3 k) = 1260 (5 m - 2) k = 420$
$We look for integer solutions to (5 m - 2) k = 420 with k \geq 6 so that g = 2 k > 10, and we find that m = 6 .$	$5 m - 2 = 28 k = \frac{420}{28} = 15 g = 2 \times 15 = 30$

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C

A garden path is constructed with m rectangular stone pavers of equal length and a gravel strip between each pair of consecutive pavers. Each gravel strip is \frac{2}{3} as long as a paver. The total length of the path from the start of the first paver to the end of the last paver is 420 inches. If m is an integer and the length, in inches, of each gravel strip is an integer greater than 10, what is the value of m ?

Explanation	Calculations	Help
$We pick p as the length of each paver and g as the length of each gravel strip.$	$p = length of each paver (in inches) g = length of each gravel strip (in inches)$	Theory & Tactics Method Card TRANS1-A Click to view full details
$We translate "The total length of the path from the start of the first paver to the end of the last paver is 420 inches." into an equation.$	$m p + (m - 1) \times g = 420$	Theory & Tactics Method Card TRANS1-B Click to view full details Theory & Tactics Method Card TRANS1-C Click to view full details
$We translate "Each gravel strip is \frac{2}{3} as long as a paver." into an equation.$	$g = \frac{2}{3} p$	Theory & Tactics Method Card TRANS1-E Click to view full details
$We substitute g = \frac{2}{3} p into the total-length equation and simplify.$	$m p + (m - 1) \times \frac{2}{3} p = 420 p \times (m + \frac{2}{3} (m - 1)) = 420 p \times \frac{5 m - 2}{3} = 420 (5 m - 2) p = 1260$
$We need g to be an integer greater than 10, and since g = \frac{2}{3} p, we let p = 3 k so that g = 2 k .$	$p = 3 k g = 2 k$	Theory & Tactics Method Card TRANS1-A Click to view full details
$We replace p = 3 k in (5 m - 2) p = 1260 and simplify.$	$(5 m - 2) (3 k) = 1260 (5 m - 2) k = 420$
$We look for integer solutions to (5 m - 2) k = 420 with k \geq 6 so that g = 2 k > 10, and we find that m = 6 .$	$5 m - 2 = 28 k = \frac{420}{28} = 15 g = 2 \times 15 = 30$

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C