# 2000 AIME I Problems/Problem 12

## Problem

Given a function $f$ for which $$f(x) = f(398 - x) = f(2158 - x) = f(3214 - x)$$ holds for all real $x,$ what is the largest number of different values that can appear in the list $f(0),f(1),f(2),\ldots,f(999)?$

## 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{align*}

Since $\mathrm{gcd}(1056, 1760) = 352$ we can conclude that (by the Euclidean algorithm) $$f(x) = f(352 + x)$$

So we need only to consider one period $f(0), f(1), ... f(351)$, which can have at most $352$ distinct values which determine the value of $f(x)$ at all other integers.

But we also know that $f(x) = f(46 - x) = f(398 - x)$, so the values $x = 24, 25, ... 46$ and $x = 200, 201, ... 351$ 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 $f(23), f(24), \ldots, f(199)$ 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 (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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