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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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possible solution | Comment 3 of 7 |
Using the first equality only f(0) = f(398), f(1) = f(397), . . . f(198) = f(200), f(199) = f(199) then the sequence repeats in reverse order.  That's 200 values.

From the last equality f(-57) = f(999), f(-58) = f(998), . . . f(-398) = f(658).   But using the first equality, f(-57) = f(341), f(-58) = f(340), . . ., down to f(-398) = f(0) and we've counted those already.

Offhand I don't see how to reduce the range 399<=x<=657.   That's 259 more values for a total of 459 possible distinct values assuming I haven't miscounted.   

  Posted by xdog on 2014-09-15 13:04:35
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