2007 AMC 12A Problems/Problem 14
Problem
Let a, b, c, d, and e be distinct integers such that
What is a+b+c+d+e?
- (A) 5
- (B) 17
- (C) 25
- (D) 27
- (E) 30
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 -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, , 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 | |
