Thread: Problem 50
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Old November 22nd, 2008, 02:22 AM
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NonCommAlg NonCommAlg is offline
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this problem is for the moderators only! lol (just kidding!)

suppose a_1 \leq a_2 \leq \cdots \leq a_n, and f(x)=\sum_{i=1}^n |x-a_i|. put: m=\left \lfloor \frac{n}{2} \right \rfloor. prove that: \min f(x)=\sum_{k=1}^m (a_{n+1-k} - a_k).
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