Difference between revisions of "1999 AMC 8 Problems/Problem 3"

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==Solution==
 
==Solution==
  
By adding each pair, we can see that <math>\boxed{(D)}</math> gives us <math>0</math>, not <math>1</math>, as our sum.
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By adding each triplet, we can see that <math>\boxed{(D)}</math> gives us <math>0</math>, not <math>1</math>, as our sum.
  
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=1999|num-b=2|num-a=4}}
 
{{AMC8 box|year=1999|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 23:33, 4 July 2013

Problem

Which triplet of numbers has a sum NOT equal to 1?

$\text{(A)}\ (1/2,1/3,1/6) \qquad \text{(B)}\ (2,-2,1) \qquad \text{(C)}\ (0.1,0.3,0.6) \qquad \text{(D)}\ (1.1,-2.1,1.0) \qquad \text{(E)}\ (-3/2,-5/2,5)$

Solution

By adding each triplet, we can see that $\boxed{(D)}$ gives us $0$, not $1$, as our sum.

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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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