Difference between revisions of "2003 AMC 12B Problems/Problem 24"

(hrmm.. my solution is actually completely incorrect, but I got the right answer)
 
m (Solution: typo)
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\qquad\mathrm{(E)}\ 2004</math>
 
\qquad\mathrm{(E)}\ 2004</math>
 
== Solution ==
 
== Solution ==
 +
{{solution}}
  
 
== See also ==
 
== See also ==

Revision as of 17:18, 5 February 2008

Problem

Positive integers $a,b,$ and $c$ are chosen so that $a<b<c$, and the system of equations

$2x + y = 2003 \quad$ and $\quad y = |x-a| + |x-b| + |x-c|$

has exactly one solution. What is the minimum value of $c$?

$\mathrm{(A)}\ 668 \qquad\mathrm{(B)}\ 669 \qquad\mathrm{(C)}\ 1002 \qquad\mathrm{(D)}\ 2003 \qquad\mathrm{(E)}\ 2004$

Solution

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See also

2003 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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 AMC 12 Problems and Solutions