Difference between revisions of "2007 AMC 12A Problems/Problem 14"
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==Solution== | ==Solution== | ||
| − | 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> | + | 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). |
==See also== | ==See also== | ||
Revision as of 09:56, 29 September 2007
Problem
Let a, b, c, d, and e be distinct integers such that
What is 
?
Solution
If 45 is expressed as a product of five distinct integer factors, the absolute value of the product of any four it as least 
, so no factor can have an absolute value greater than 5. Thus the factors of the given expression are five of the integers 
. The product of all six of these is 
, 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).
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 | |
