Difference between revisions of "2000 AIME I Problems/Problem 12"
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== Solution == | == Solution == | ||
− | <cmath>\begin{align*}f( | + | <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{eqnarray*}</cmath> | f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{eqnarray*}</cmath> | ||
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distinct values. | 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). | + | 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 == |
Revision as of 11:43, 29 November 2014
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(2158 - 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. ! LaTeX Error: \begin{align*} on input line 20 ended by \end{eqnarray*}.)
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.