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),…,f(999)?
Solution:
From the given identities f(x)=f(398−x),f(x)=f(2158−x), and f(x)=f(3214−x), derive the following identities:
f(x)f(x)f(x)f(x)=f(2158−x)=f(3214−(2158−x))=f(1056+x),=f(1056+x)=f(2158−(1056+x))=f(1102−x),=f(1056+x)=f(1102−(1056+x))=f(46−x), and =f(46−x)=f(398−(46−x))=f(352+x).
It follows from the last identity that f is periodic, and that the period of f divides 352. Thus every value in the list is found among f(0),f(1),…,f(351). The identity f(x)= f(398−x) implies that f(200),f(201),…,f(351) are found among f(0),f(1),…. f(199), and the identity f(x)=f(46−x) implies that f(0),f(1),…,f(22) are found among f(23),f(24),….f(199). Thus there can be at most 177 different values in the list. To see that the values f(23).f(24)…,f(199) can be distinct, consider the function f(x)=cos(352360(x−23)), whose argument is interpreted in degrees. It is routine to verify the required identities f(x)=f(398−x),f(x)=f(2158−x), and f(x)=f(3214−x), and to see that 1=f(23)>f(24)>f(25)>⋯>f(199)=−1.
The problems on this page are the property of the MAA's American Mathematics Competitions