Difference between revisions of "2014 AMC 12A Problems/Problem 24"
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==Solution 2== | ==Solution 2== | ||
− | First, notice that the | + | First, notice that the recursive rule moves the current value <math>1</math> closer to <math>0</math>. Upon reaching <math>0</math>, it alternates between <math>-1</math> and <math>0</math>. This means that <math>f_{100}(x) = 0</math> exactly when <math>|f_0(x)| \le 100</math> (to reach <math>0</math> in time) and <math>f_0(x)</math> is even (so <math>f_{100}(x) \ne -1</math>). |
+ | |||
+ | Casework each part of <math>f_0(x)</math> (where the expressions in the absolute values do not change sign): | ||
+ | <cmath>x \le -100 \implies f_0(x) = x-(x-100)+(x+100) = x+200</cmath> | ||
+ | so even <math>-300 \le x \le -100</math> work. | ||
+ | <cmath>-100 \le x \le 100 \implies f_0(x) = x-(x-100)-(x+100) = -x</cmath> | ||
+ | so even <math>-100 \le x \le 100</math> work. | ||
+ | <cmath>100 \le x \implies f_0(x) = x+(x-100)-(x+100) = x-200</cmath> | ||
+ | so even <math>100 \le x \le 300</math> work. | ||
+ | |||
+ | Putting these together, all even <math>x</math> where <math>-300 \le x \le 300</math> work. So, the answer is <math>2\cdot150+1 = \boxed{\textbf{(C)}\ 301}</math>. | ||
+ | ~revised by [[User:emerald_block|emerald_block]] | ||
== Video Solution by Richard Rusczyk == | == Video Solution by Richard Rusczyk == |
Revision as of 20:54, 4 November 2021
Problem
Let , and for
, let
. For how many values of
is
?
Solution 1
1. Draw the graph of by dividing the domain into three parts.
2. Apply the recursive rule a few times to find the pattern.
Note: is used to enlarge the difference, but the reasoning is the same.
3. Extrapolate to . Notice that the summits start
away from
and get
closer each iteration, so they reach
exactly at
.
reaches
at
, then zigzags between
and
, hitting
at every even
, before leaving
at
.
This means that at all even
where
. This is a
-integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or
.
(Revised by Flamedragon & Jason,C & emerald_block)
Solution 2
First, notice that the recursive rule moves the current value closer to
. Upon reaching
, it alternates between
and
. This means that
exactly when
(to reach
in time) and
is even (so
).
Casework each part of (where the expressions in the absolute values do not change sign):
so even
work.
so even
work.
so even
work.
Putting these together, all even where
work. So, the answer is
.
~revised by emerald_block
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2014amc12a/383
~ dolphin7
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.