Difference between revisions of "1987 AJHSME Problems/Problem 6"

(New page: ==Problem== The smallest product one could obtain by multiplying two numbers in the set <math>\{ -7,-5,-1,1,3 \}</math> is <math>\text{(A)}\ -35 \qquad \text{(B)}\ -21 \qquad \text{(C)}\...)
 
 
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==See Also==
 
==See Also==
 
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{{AJHSME box|year=1987|num-b=5|num-a=7}}
[[1987 AJHSME Problems]]
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 22:52, 4 July 2013

Problem

The smallest product one could obtain by multiplying two numbers in the set $\{ -7,-5,-1,1,3 \}$ is

$\text{(A)}\ -35 \qquad \text{(B)}\ -21 \qquad \text{(C)}\ -15 \qquad \text{(D)}\ -1 \qquad \text{(E)}\ 3$

Solution

To get the smallest possible product, we want to multiply the smallest negative number by the largest positive number. These are $-7$ and $3$, respectively, and their product is $-21$, which is $\boxed{\text{B}}$

See Also

1987 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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

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