Difference between revisions of "2014 AMC 12A Problems/Problem 24"
m (→Solution 2) |
(→Problem) |
||
| Line 2: | Line 2: | ||
Let <math>f_0(x)=x+|x-100|-|x+100|</math>, and for <math>n\geq 1</math>, let <math>f_n(x)=|f_{n-1}(x)|-1</math>. For how many values of <math>x</math> is <math>f_{100}(x)=0</math>? | Let <math>f_0(x)=x+|x-100|-|x+100|</math>, and for <math>n\geq 1</math>, let <math>f_n(x)=|f_{n-1}(x)|-1</math>. For how many values of <math>x</math> is <math>f_{100}(x)=0</math>? | ||
| − | <math>\ | + | <math>\textf{(A) }299\qquad |
\textbf{(B) }300\qquad | \textbf{(B) }300\qquad | ||
\textbf{(C) }301\qquad | \textbf{(C) }301\qquad | ||
\textbf{(D) }302\qquad | \textbf{(D) }302\qquad | ||
\textbf{(E) }303\qquad</math> | \textbf{(E) }303\qquad</math> | ||
| + | |||
==Solution 1== | ==Solution 1== | ||
Revision as of 13:30, 12 October 2018
Contents
Problem
Let 
, and for 
, let 
. For how many values of 
is 
?
$\textf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$ (Error compiling LaTeX. Unknown error_msg)
Solution 1
1. Draw the graph of 
by dividing the domain into three parts.
2. Look at the recursive rule. Take absolute of the previous function and down by 1 to get the next function.
3. Count the x intercepts of the each function and find the pattern.
The pattern turns out to be 
solutions,for x interval:[1,99], the function gain only one extra solution after 
because there is no summit on the graph any more, and the answer is thus 
.
(Revised by Flamedragon & Jason,C)
Solution 2
First, notice that the recursion and the definition of require that for all such that 
, if , then is even. Now, we can do case work on to find which values of (such that ) make even. The answer comes out to be all the even values of in the range , in the domain 
. So, the answer is 
or 
.
See Also
| 2014 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
