2010 AMC 12A Problem 22

Question

What is the minimum value of

\begin{gathered} f(x) = |x-1|+|2x-1| \\ {}+|3x-1|+\cdots+|119x-1|? \end{gathered}

49

50

51

52

53

f

is piecewise linear with corners at

x=\tfrac1k;

the minimum is where its slope turns from negative to positive

On

\left[\tfrac1m,\tfrac1{m-1}\right]

the slope is

7140-(m-1)m;

find where it vanishes

The function

f

is piecewise linear with breakpoints at

x=\tfrac1k.

On the interval

\left[\tfrac1m,\tfrac1{m-1}\right]

its slope is

\begin{aligned} &\sum_{k=m}^{119}k \\ &\quad {}-\sum_{k=1}^{m-1}k=7140-(m-1)m, \end{aligned}

where

7140=\tfrac{119\cdot120}{2}.

This slope is zero when

(m-1)m=7140,

i.e.

m=85,

so the minimum occurs at the right endpoint

x=\tfrac1{84}.

There, terms with

k\le84

contribute

\tfrac{84-k}{84}

and terms with

k\ge85

contribute

\tfrac{k-84}{84},

so

\begin{aligned} f\left(\tfrac1{84}\right) &= \frac{3486}{84}+\frac{630}{84} \\ &= 41.5+7.5=49. \end{aligned}

Thus, A is the correct answer.

Problem 22 in other years