2007 AMC 12A Problems/Problem 14

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

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

Solutions

Solution 1

If $45$ is expressed as a product of five distinct integer factors, the absolute value of the product of any four is at 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 $\fbox{25 (C)}$

Solution 2

The prime factorization of $45$ is $3^2 * 5$ . Therefore, the 5 distinct integer factors must have some negative numbers in them. Because there are two $3$ 's in the prime factorization, one of them must be negative and the other positive. Because there is a $-3$ , there must also be a $-1$ to cancel the negatives out. The 5 distinct integer factors must be $-3, 3, 5, -1, 1$ . The corresponding values of $a, b, c, d,$ and $e$ are $9, 3, 1, 7, 5$ . and their sum is $\fbox{25 (C)}$

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

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2007 AMC 12A Problems/Problem 14

Contents

Problem

Solutions

Solution 1

Solution 2

See also