AoPSWiki
Visit the AoPS Book Store.

2000 AIME I Problems/Problem 12

From AoPSWiki

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(2518 - x) = f(x) &= f(3214 - (2158 - x)) &= f(1056 + x)\\f(398 - x) = f(x) &= f(2158 - (398 - x))...

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 (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Do you have what it takes to be the next brilliant trader, researcher, or developer at Jane Street Capital? Find out in the Careers in Mathematics Forum.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us