2007 AMC 12A Problems/Problem 14

Revision as of 09:56, 29 September 2007 by Identity (talk | contribs) (Solution)

Problem

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

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

What is $a+b+c+d+e$?

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

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).

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions