2007 AMC 12A Problem 14

Below is the professionally curated solution for Problem 14 of the 2007 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

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Concepts:factoring bounding to limit cases

Difficulty rating: 1440

14.

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

$5$

$17$

$25$

$27$

$30$

Solution:

The five factors are distinct integers multiplying to $45.$ If any factor had absolute value more than $5,$ the remaining four (distinct) would have product at least $|(-3)(-1)(1)(3)|=9,$ forcing the total above $45.$

So the factors come from $\pm 1,\pm 3,\pm 5.$ The product of all six of these is $-225=(-5)(45),$ so the five factors are $-3,-1,1,3,5.$

Then $a,b,c,d,e$ are $9,7,5,3,1$ in some order, and their sum is $25.$

Thus, the correct answer is C.

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