Let f_0(x)=x+|x-100| -|x+100|, and for n≥1, let f_n(x)=|f_n-1(x)|-1. For how many values of x is f_100(x)=0?

2014 AMC 12A Problem 24

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24.

Let

f_0(x)=x+|x-100|

-|x+100|,

and for

n\ge1,

let

f_n(x)=|f_{n-1}(x)|-1.

For how many values of

x

f_{100}(x)=0?

299

300

301

302

303

Answer: C

Concepts:absolute value recursion counting intersections

Difficulty rating: 2520

Small hint:

f_0(x)=\pm k

for a nonnegative integer

k,

then

f_k(x)=0

and afterward the values cycle

0,-1,0,-1,\ldots

Big hint:

f_{100}(x)=0

needs

f_0(x)=2k

with

-50\le k\le50

; graph

f_0

and count intersections with each line

y=2k

Solution:

f_{n-1}(x)=\pm k,

then

f_n(x)=k-1.

So if

f_0(x)=\pm k

for a nonnegative integer

k,

then

f_k(x)=0,

after which the sequence alternates

0,-1,0,\ldots

Thus

f_{100}(x)=0

exactly when

f_0(x)=2k

for some integer

-50\le k\le50.

Now

f_0(x)=x+|x-100|

-|x+100|

equals

x+200

for

x\lt-100,

-x

for

-100\le x\lt100,

and

x-200

for

x\ge100.

Its graph is piecewise linear with turning points

(-100,100)

and

(100,-100).

A line

y=2k

meets this graph three times for

-49\le k\le49

and twice for

k=\pm50.

The total is

99\cdot3+2\cdot2=301.

Thus, the correct answer is C.

2014 AMC 12A Problem 24

Answer: C

Difficulty rating: 2520

Small hint:

Big hint:

Solution:

Problem 24 in other years