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2000 AIME I Problems/Problem 12

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Problem

Given a function for which f(x) = f(398 - x) = f(2158 - x) = f(3214 - x) 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*}

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

352 - (46 - 24 + 1) - (351 - 200 + 1) = \boxed{ 177 }

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: f(x) = \cos \left(\frac{360}{352}(x-23)\right) (in degrees).

See also

2000 AIME I (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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