2007 AMC 12A Problems/Problem 14

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Problem

Let a, b, c, d, and e be distinct integers such that

$(6-a)(6-b)(6-c)(6-d)(6-e)=45$

$\mathrm{(A)}\ 5\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 25\qquad \mathrm{(D)}\ 27\qquad \mathrm{(E)}\ 30$

Solution

If 45 is expressed as a product of five distinct integer factors, the absolute value of the product of any four it as least $|(-3)(-1)(1)(3)|=9$ , so no factor can have an absolute value greater than 5. Thus the factors of the given expression are five of the integers $\pm 3, \pm 1, \pm 5$ . The product of all six of these is $-225=(-5)(45)$ , so the factors are -3, -1, 1, 3, and 5. The corresponding values of a, b, c, d, and e are 9, 7, 5, 3, and 1, and their sum is 25 (C).

2007 AMC 12A (Problems • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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2007 AMC 12A Problems/Problem 14

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Problem

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