2000 AIME I Problems/Problem 12

Revision as of 10:43, 29 November 2014 by Jabberwock2 (talk | contribs) (Solution)

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{eqnarray*} (Error compiling LaTeX. ! LaTeX Error: \begin{align*} on input line 20 ended by \end{eqnarray*}.)

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 (ProblemsAnswer KeyResources)
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

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