Difference between revisions of "2007 AMC 12A Problems/Problem 14"

Revision as of 18:25, 24 October 2020

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

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 $25$ $(C).$

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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 8: / Line 8: @@
 ==Solutions==
 ===Solution 1===
-If 45 is expressed as a product of five distinct integer factors, the absolute value of the product of any four it as least <math> |(-3)(-1)(1)(3)|=9</math>, so no factor can have an absolute value greater than 5. Thus the factors of the given expression are five of the integers <math>\pm 3, \pm 1, \pm 5</math>. The product of all six of these is <math>-225=(-5)(45)</math>, 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 25 (C).
+If <math>45</math> is expressed as a product of five distinct integer factors, the absolute value of the product of any four is at least <math> |(-3)(-1)(1)(3)|=9</math>, so no factor can have an absolute value greater than <math>5</math>. Thus the factors of the given expression are five of the integers <math>\pm 3, \pm 1, \pm 5</math>. The product of all six of these is <math>-225=(-5)(45)</math>, so the factors are <math>-3, -1, 1, 3,</math> and <math>5.</math> The corresponding values of <math>a, b, c, d,</math> and <math>e</math> are <math>9, 7, 5, 3,</math> and <math>1,</math> and their sum is <math>25</math> <math>(C).</math>
 ===Solution 2===

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