Difference between revisions of "2000 AIME I Problems/Problem 12"
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== Problem == | == Problem == | ||
− | + | Given a [[function]] <math>f</math> for which | |
+ | <cmath> | ||
+ | f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)</cmath> | ||
+ | holds for all real <math>x,</math> what is the largest number of different values that can appear in the list <math>f(0),f(1),f(2),\ldots,f(999)?</math> | ||
== Solution == | == Solution == | ||
− | {{ | + | <cmath>\begin{align*}f(2158 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\ |
+ | f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{align*}</cmath> | ||
+ | |||
+ | Since <math>\mathrm{gcd}(1056, 1760) = 352</math> we can conclude that (by the [[Euclidean algorithm]]) | ||
+ | |||
+ | <cmath>f(x) = f(352 + x)</cmath> | ||
+ | |||
+ | So we need only to consider one period <math>f(0), f(1), ... f(351)</math>, which can have at most <math>352</math> distinct values which determine the value of <math>f(x)</math> at all other integers. | ||
+ | |||
+ | But we also know that <math>f(x) = f(46 - x) = f(398 - x)</math>, so the values <math>x = 24, 25, ... 46</math> and <math>x = 200, 201, ... 351</math> are repeated. This gives a total of | ||
+ | |||
+ | <cmath>352 - (46 - 24 + 1) - (351 - 200 + 1) = \boxed{ 177 }</cmath> | ||
+ | |||
+ | distinct values. | ||
+ | |||
+ | To show that it is possible to have <math>f(23), f(24), \ldots, f(199)</math> distinct, we try to find a function which fulfills the given conditions. A bit of trial and error would lead to the [[cosine]] function: <math>f(x) = \cos \left(\frac{360}{352}(x-23)\right)</math> (in degrees). | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=2000|n=I|num-b=11|num-a=13}} | |
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:50, 10 March 2015
Problem
Given a function for which holds for all real what is the largest number of different values that can appear in the list
Solution
Since we can conclude that (by the Euclidean algorithm)
So we need only to consider one period , which can have at most distinct values which determine the value of at all other integers.
But we also know that , so the values and are repeated. This gives a total of
distinct values.
To show that it is possible to have distinct, we try to find a function which fulfills the given conditions. A bit of trial and error would lead to the cosine function: (in degrees).
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.