Difference between revisions of "2007 AMC 12A Problems/Problem 14"
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| − | ==Problem== | + | == Problem == |
| − | + | Let <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> be distinct integers such that | |
| − | Let a, b, c, d, and e be distinct | + | <math>(6-a)(6-b)(6-c)(6-d)(6-e)=45</math> |
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What is <math>a+b+c+d+e</math>? | What is <math>a+b+c+d+e</math>? | ||
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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> | ||
| − | ==Solution== | + | == Solution 1 == |
| − | If 45 is expressed as a product of five distinct integer factors, the absolute value of the product of any four | + | 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 == | ||
| + | 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> | ||
==See also== | ==See also== | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 01:10, 16 February 2021
Contents
Problem
Let 
, 
, 
, 
, and 
be distinct integers such that
What is 
?
Solution 1
If 
is expressed as a product of five distinct integer factors, the absolute value of the product of any four is at least 
, so no factor can have an absolute value greater than 
. Thus the factors of the given expression are five of the integers 
. The product of all six of these is 
, so the factors are 
and 
The corresponding values of 
and are 
and 
and their sum is 
Solution 2
The prime factorization of is 
. Therefore, the 5 distinct integer factors must have some negative numbers in them. Because there are two 
's in the prime factorization, one of them must be negative and the other positive. Because there is a 
, there must also be a 
to cancel the negatives out. The 5 distinct integer factors must be 
. The corresponding values of and are 
. and their sum is
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
