2007 AMC 12A Problem 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?

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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.

Tagged: factoring · bounding to limit cases