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

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<math>\mathrm{(A)}\ 5\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 25\qquad \mathrm{(D)}\ 27\qquad \mathrm{(E)}\ 30</math>
 
<math>\mathrm{(A)}\ 5\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 25\qquad \mathrm{(D)}\ 27\qquad \mathrm{(E)}\ 30</math>
  
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== Solutions ==
+
== Solution 1 ==
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=== Solution 1 ===
 
 
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>\fbox{25 (C)}</math>
 
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>\fbox{25 (C)}</math>
  
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=== Solution 2 ===
+
== Solution 2 ==
 
The prime factorization of <math>45</math> is <math>3^2 * 5</math>. Therefore, the 5 distinct integer factors must have some negative numbers in them. Because there are two <math>3</math>'s in the prime factorization, one of them must be negative and the other positive. Because there is a <math>-3</math>, there must also be a <math>-1</math> to cancel the negatives out. The 5 distinct integer factors must be <math>-3, 3, 5, -1, 1</math>. The corresponding values of <math>a, b, c, d, </math> and <math>e</math> are <math>9, 3, 1, 7, 5</math>. and their sum is <math>\fbox{25 (C)}</math>
 
The prime factorization of <math>45</math> is <math>3^2 * 5</math>. Therefore, the 5 distinct integer factors must have some negative numbers in them. Because there are two <math>3</math>'s in the prime factorization, one of them must be negative and the other positive. Because there is a <math>-3</math>, there must also be a <math>-1</math> to cancel the negatives out. The 5 distinct integer factors must be <math>-3, 3, 5, -1, 1</math>. The corresponding values of <math>a, b, c, d, </math> and <math>e</math> are <math>9, 3, 1, 7, 5</math>. and their sum is <math>\fbox{25 (C)}</math>
  

Latest revision as of 01:10, 16 February 2021

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

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

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 $\fbox{25 (C)}$

See also

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

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