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

(Solution)
 
Line 12: Line 12:
  
 
{{AMC8 box|year=1999|num-b=2|num-a=4}}
 
{{AMC8 box|year=1999|num-b=2|num-a=4}}
 +
{{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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png