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Distinct Value Deduction (Posted on 2014-09-14) Difficulty: 3 of 5
A function F is such that this relationship holds for all real x.

F(x) = F(398-x) = F(2158-x) = F(3214-x)

What is the maximum number of distinct values that can appear in the list F(0), F(1), F(2), ..., F(999).

No Solution Yet Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Some Thoughts The Final reduction (spoiler) | Comment 6 of 7 |
(See two previous posts, for the start of this solution)

f(1) = f(397) = f(45) because the function has a cycle of 352
f(2) = f(396) = f(44)
f(3) = f(43)
   .
   .
   .
f(22) = f(24)
f(23) = f(23)

So f(1) through f(22) are duplicates.
Our previous 199 are now reduced to 177 possibly distinct values.
177 is half of the 352 cycle, plus 1, which makes some sense.

Are there any more necessarily duplicate values between f(23) and f(199)?  
It appears not.

One function that meets all requirements, and has 177 distinct integer values, is a triangular wave function which takes its minimum at f(23) = 23, its maximum at f(199) = 199, descends again to 23 at f(375), and repeats in both directions with a cycle of 352.  
Between 23 and 199, f(x) = x. 
f(1079) is also a minimum, and f(1607) is also a maximum.
The function is symmetric around x = 199, so f(x) = f(398-x).
The function is symmetric around x = 1079, so f(x) = f(2158 - x).
The function is symmetric around x = 1607, so f(x) = f(3214 - x).

  Posted by Steve Herman on 2014-09-16 16:50:03
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