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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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yet further reduction (spoiler?) | Comment 5 of 7 |
Why yes, of course, further improvement is possible.

I have previously shown that the function values for the integers repeat with a cycle of 352, so the only values that need to be considered are f(1), f(2) ... f(352)

But we know that

f(352) = f(46)
f(351) = f(47)
f(200) = f(198)
f(199) = f(199)

So values f(200) through f(352) are not unique, and there are 153 of those.

Maximum distinct values = 352 - 153 = 199.

Further reduction might be possible, in that there might be some duplication among f(1), f(2), ...f(199).  I will return to this later.

  Posted by Steve Herman on 2014-09-16 09:27:29
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