2010 AMC 12A Problems/Problem 22

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Problem 22

What is the minimum value of $\left|x-1\right| + \left|2x-1\right| + \left|3x-1\right| + \cdots + \left|119x - 1 \right|$?

$\textbf{(A)}\ 49 \qquad \textbf{(B)}\ 50 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 52 \qquad \textbf{(E)}\ 53$

Solution

If we graph each term separately, we will notice that all of the zeros occur at $\frac{1}{m}$, where $m$ is any integer from $1$ to $119$, inclusive.

The minimum value occurs where the sum of the slopes is at a minimum, since it is easy to see that the value will be increasing on either side. That means the minimum must happen at some $\frac{1}{m}$.


The sum of the slope at $x = \frac{1}{m}$ is

$\sum_{i=m+1}^{119}i - \sum_{i=1}^{m}i$

$=\sum_{i=1}^{119}i - 2\sum_{i=1}^{m}i$

$=-m^2-m+7140$


Now we want to minimize $-m^2-m+7140$. The zeros occur at $-85$ and $84$, which means the slope is $0$ where $m = 84, 85$.

We can now verify that both $x=\frac{1}{84}$ and $x=\frac{1}{85}$ yield $\boxed{49\ \textbf{(A)}}$.