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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 further reduction (spoiler?) | Comment 4 of 7 |
f(1) = f(397).
But f(x) = f(2158-x), so f(397) = f(1761)
In general, f(x) = f(1760 + x), so the values repeat with a cycle of 1760

Also, f(x) = f(3214-x), so f(397) = f(2817)
In general, f(x) = f(2816 + x), so the values also repeat with a cycle of 2816

If 1760 and 2816 were relatively prime, then there would only be one possible value for integer values of x, but this is not the case.
Their greatest common divisor is 352, so the values repeat with a cycle of 352.

Therefore, there are at most 352 possible function values for integer values of x.

Further improvement might be possible.  I am still looking.

Edited on September 16, 2014, 2:40 pm
  Posted by Steve Herman on 2014-09-16 09:10:10

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