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?

2007 AMC 12A Problem 14

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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

The five factors

6-a,\ldots,6-e

are distinct integers whose product is

45

Check the positive and negative divisors of

45

; only

\{-3,-1,1,3,5\}

gives five distinct factors

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.