2000 AMC 8 Problems/Problem 7

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
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All AJHSME/AMC 8 Problems and Solutions

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