True or Not?

Let the contribution of 1's to $f(k)$ from the last digit $(10^0)$ be $a_0$ ; second last, $a_1$ ; third last, $a_2$ and so on. For example: for $f(6)$ , $a_0=1$ from 1, while $a_d = 0$ for $d \ge 1$ , therefore, $f(6) = a_0 = 1$ ; and for $f(16)$ , $a_0 = 2$ from 1 and 11, and $a_1 = 7$ from 10 to 16 and $a_d = 0$ for $d \ge 2$ , therefore, $f(16) = a_0+a_1 = 2+7 = 9$ .

We note that for every 10, $a_0$ adds 1. Therefore, for $10^n$ , $a_0 = \dfrac {10^n}{10} = 10^{n-1}$ . Similarly, for every 100, $a_1$ adds 10 (from 10 to 19), and $a_1 = 10 \times \dfrac {10^n}{100} = 10^{n-1}$ ; for every 1000, $a_2$ adds 100 (from 100 to 199), and $a_2 = 100 \times \dfrac {10^n}{1000} = 10^{n-1}$ ...; for for every $10^n$ , $a_{n-1}$ adds $10^{n-1}$ , and $a_{n-1}= 10^{n-1} \times \dfrac {10^n}{10^n} = 10^{n-1}$ ; and from the $10^n$ digit, $a_n = 1$ . Therefore, we have:

$f(10^n) = \sum_{k=0}^n a_k = \sum_{k=0}^{n-1} 10^{n-1} + a_n = \boxed{n(10^{n-1}) + 1}$