Let a, b, c, d, and e be distinct integers such that \beginaligned &(6-a)(6-b)(6-c) \ &\quad ×(6-d)(6-e) \ &=45. \endaligned What is a+b+c+d+e?

2007 AMC 12A Problem 14 — Solution

2007 AMC 12A Problem 14

Attempt Problem 14 of the 2007 AMC 12A below, then check your answer against the professionally curated solution 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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14.

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

$\begin{aligned} &(6-a)(6-b)(6-c) \\ &\quad {}\cdot(6-d)(6-e) \\ &=45. \end{aligned}$

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

$5$

$17$

$25$

$27$

$30$

Answer: C

Concepts:factoring bounding to limit cases

Difficulty rating: 1440

Solution:

The five factors are distinct nonzero integers with product $45,$ so each is a positive or negative divisor of $45.$ Checking $\pm1,\pm3,\pm5,\pm9,\pm15,\pm45,$ only the five-element set $\{-3,-1,1,3,5\}$ has product $45.$

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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