A normal tally counter has \( 4 \) digits and counts \( 1 \) by \( 1 \), hence there are \( 10000 \) different numbers can be shown on it (in range \( 0000 ... 9999 \)).
Suppose you have an odd tally counter. Instead \( 1 \) by \( 1 \), it counts \( x \) by \( x \). In case of overflow, the counter only shows the last \( 4 \) digits of the number.
For example, if \( x = 1001 \) the counter will show these numbers : \( 0000, 1001, 2002, ..., 9009, 0010, 1011, ... \)
Let \( F(n) \) be the function returning the number of different numbers can be shown on odd tally counter with \( x = n \), find the last \( 3 \) digits of \( \sum_{i=1}^{9999} F(i)\).
As an explicit example : \( F(1) = 10000 \)
This problem is taken from TOKI Open Contest January 2013 (Problemsetter : Me)
by
Ammar Fathin Sabili
