Problem 305Published on Sunday, 10th October 2010, 04:00 am; Solved by 417
Let's call S the (infinite) string that is made by concatenating the consecutive positive integers (starting from 1) written down in base 10.
Thus, S = 1234567891011121314151617181920212223242...
It's easy to see that any number will show up an infinite number of times in S.
Let's call f(n) the starting position of the nth occurrence of n in S.
For example, f(1)=1, f(5)=81, f(12)=271 and f(7780)=111111365.
Find f(3k) for 1k13.