Odd Long Tally Counter

This problem is the more-evil-version of this problem.

A normal long tally counter has 7 7 digits and counts 1 1 by 1 1 , hence there are 107 10^{7} different numbers can be shown on it (in range 0000000...9999999 0000000 ... 9999999 ).

Suppose you have an odd long 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 7 7 digits of the number.

For example, if x=1000001 x = 1000001 the counter will show these numbers : 0000000,1000001,2000002,...,9000009,0000010,1000011,... 0000000, 1000001, 2000002, ..., 9000009, 0000010, 1000011, ...

Let F(n) F(n) be the function returning the number of different numbers can be shown on odd long tally counter with x=n x = n , find the last 3 3 digits of i=11071F(i) \sum_{i=1}^{10^7-1} F(i).

As an explicit example : F(1)=107 F(1) = 10^7

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

×

Problem Loading...

Note Loading...

Set Loading...