Difference between revisions of "2000 AMC 8 Problems/Problem 7"

m (Solution: wrong number in solution box)
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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) -280}}</math>.
+
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}}

Revision as of 14:15, 23 December 2012

Problem

What is the minimum possible product of three different numbers of the set $\{-8,-6,-4,0,3,5,7\}$?

$\text{(A)}\ -336 \qquad \text{(B)}\ -280 \qquad \text{(C)}\ -210 \qquad \text{(D)}\ -192 \qquad \text{(E)}\ 0$

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: $(-8)\times(-6)\times(-4) = (-8)\times(24) = -192$ and $(-8)\times5\times7 = (-8)\times35 = -280$. The latter is smaller, so $\boxed{\text{(B)\ -280}}$.

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
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