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Problem 305


Problem 305


Reflexive Position

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 1≤k≤13.


自反位置

我们将十进制下的连续正整数(从1开始)拼接起来,构造出(无限长的)字符串S。
也即是说,S = 1234567891011121314151617181920212223242…

很容易看出,任何数都会在S中出现无限多次。

我们称f(n)是n在S中第n次出现时的起始位置。
例如,f(1)=1,f(5)=81,f(12)=271,以及f(7780)=111111365。

对于1≤k≤13,求∑f(3k)。