Difference between revisions of "2000 AMC 8 Problems/Problem 7"
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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 | 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: <math>(-8)\times(-6)\times(-4) = (-8)\times(24) = -192</math> and <math>(-8)\times5\times7 = (-8)\times35 = -280</math>. | exist: <math>(-8)\times(-6)\times(-4) = (-8)\times(24) = -192</math> and <math>(-8)\times5\times7 = (-8)\times35 = -280</math>. | ||
| − | The latter is smaller, so <math>\boxed{\text{(B) | + | The latter is smaller, so <math>\boxed{\text{(B)\ -280}}</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2000|num-b=6|num-a=8}} | {{AMC8 box|year=2000|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:35, 4 July 2013
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: 
and 
.
The latter is smaller, so 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
