2003 AMC 12B Problems/Problem 24

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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

See also

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