Difference between revisions of "2000 AIME I Problems/Problem 12"

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== Solution ==
 
== Solution ==
<cmath>\begin{align*}f(2518 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\
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<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>
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f(398 - x) = f(x) &= f(2158 - (398 - x)) &= f(1760 + x)\end{align*}</cmath>
  
 
Since <math>\mathrm{gcd}(1056, 1760) = 352</math> we can conclude that (by the [[Euclidean algorithm]])
 
Since <math>\mathrm{gcd}(1056, 1760) = 352</math> we can conclude that (by the [[Euclidean algorithm]])
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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 ==

Latest revision as of 17:50, 10 March 2015

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 (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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