Odd Tally Counter

A normal tally counter has 4 4 digits and counts 1 1 by 1 1 , hence there are 10000 10000 different numbers can be shown on it (in range 0000...9999 0000 ... 9999 ).

Suppose you have an odd tally counter. Instead 1 1 by 1 1 , it counts x x by x x . In case of overflow, the counter only shows the last 4 4 digits of the number.

For example, if x=1001 x = 1001 the counter will show these numbers : 0000,1001,2002,...,9009,0010,1011,... 0000, 1001, 2002, ..., 9009, 0010, 1011, ...

Let F(n) F(n) be the function returning the number of different numbers can be shown on odd tally counter with x=n x = n , find the last 3 3 digits of i=19999F(i) \sum_{i=1}^{9999} F(i).

As an explicit example : F(1)=10000 F(1) = 10000

This problem is taken from TOKI Open Contest January 2013 (Problemsetter : Me)

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