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Problem 483

AMC 12 late, AIME early

Algebra Difficulty 4.9 Find the answer

1. Let $n$ be a positive integer, then the minimum value of $|n-1|+|n-2|+\cdots \cdots+|n-100|$ is ).
A. 2500
B. 4950
C. 5050
D. 5150

Your answer

Official solution

1. A

When $n \geqslant 100$ , the sum mentioned is $100 n-5050$ . It reaches its minimum value of 4950 when $n=100$ . When $1 \leqslant n \leqslant 100$ , the sum is $(n-1)+\cdots+(n-n)+(n+1-n)+\cdots+(100 \quad n)=n^{2}-101 n+5050=\left(n-\frac{101}{2}\right)^{2}-\left(\frac{101}{2}\right)^{2}+5050$ .
Therefore, the minimum value of 2500 is reached when $n=50$ or 51.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.