# Difference between revisions of "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 $\fbox{25 (C)}$