2000 AMC 8 Problems/Problem 7
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Problem
What is the minimum possible product of three different numbers of the set 
?
Solution
The only way to get a negative product using three numbers is to multiply one negative number and two positives or three negatives. Only two reasonable choices
exist: 
= (-8)\times(24) = -192
(-8)\times5\times7 = (-8)\times35 = -280
{\text{(B) 7}}$.
See Also
| 2000 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
