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 == | ||
− | {{solution}} | + | <cmath>\begin{align*}f(2518 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\ |
+ | f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{eqnarray*}</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). | ||
+ | |||
+ | {{incomplete|solution}} | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=I|num-b=11|num-a=13}} | {{AIME box|year=2000|n=I|num-b=11|num-a=13}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Revision as of 12:05, 1 January 2008
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
\begin{align*}f(2518 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\ f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{eqnarray*} (Error compiling LaTeX. )
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 |