Difference between revisions of "2014 AMC 12A Problems/Problem 24"
(→Solution 2) |
(→Solution 2) |
||
Line 20: | Line 20: | ||
==Solution 2== | ==Solution 2== | ||
− | First, notice that the recursion and the definition of <math>f_0(x)</math> require that for all <math>x</math> such that <math>-100 \le x \le 100</math>, if <math>f_{100}(x)=0</math>, then <math>f_0(x)</math> is even. Now, we can do case work on <math>x</math> to find which values of <math>x</math> (such that <math>-100 \le x \le 100</math>) make <math>f_0(x)</math> even. The answer comes out to be all the even values of <math>x</math> in the range <math>-100 \le x \le 100</math>. So, the answer is <math>\textbf{ | + | First, notice that the recursion and the definition of <math>f_0(x)</math> require that for all <math>x</math> such that <math>-100 \le x \le 100</math>, if <math>f_{100}(x)=0</math>, then <math>f_0(x)</math> is even. Now, we can do case work on <math>x</math> to find which values of <math>x</math> (such that <math>-100 \le x \le 100</math>) make <math>f_0(x)</math> even. The answer comes out to be all the even values of <math>x</math> in the range <math>-100 \le x \le 100</math>. So, the answer is <math>\boxed{\textbf{C} 301}</math>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2014|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2014|ab=A|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:05, 15 February 2014
Contents
Problem
Let , and for
, let
. For how many values of
is
?
Solution
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, and the answer is thus
.
(Revised by Flamedragon)
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
. So, the answer is
.
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.