TITLE: $f(n)=$ how many times number $2$ appears in numbers from $1$ to $n$
QUESTION [6 upvotes]: $f(n)$ calculates how many times $2$ appears in numbers from $1$ to $n$. For example, $f(1)=0$, $f(2)=1$ and $f(12)=2$. 
What is the first $n$ for which $f(n)=n$?
I would need first of all to figure out the function itself, if you can help me. And also the result, so I can compare after I solve the function myself.
Please help me with this one, I am stuck. I tried to do it through MS Excel, but the number is too large for what Excel can handle. Any help is highly appreciated. 
Thanks!
p.s: thanks to Sidd, I now have an answer. However, if someone can also figure out the mathematical function for this, it would be very helpful.

REPLY [4 votes]: If you have a number like $6387$, then instead of doing $f(6387)$ you can do $f(6000) + f(300) + f(80) + f(7)$. For any $n$ in the form of $a*10^b$ where $a,b \in \mathbb{Z}$ ,
$f(n) = \begin{cases}
  ab*10^{b-1} + 10^b & \text{:$b\ge1, a\gt2$} \\
  ab*10^{b-1} + (n \bmod 10^b) + 1 & \text{:$b\ge1, a=2$} \\
  ab*10^{b-1} & \text{:$b\ge1, a=1$} \\
  1 & \text{:$b=0, a\ge2$} \\
  0 & \text{:$b=0, a\lt2$} \\
\end{cases}$
For an explanation on how that equation works, take the number $6000$. If you put a $2$ in the units place (_ _ _ 2), then for all $4$ digit numbers, $6$ numbers can occupy the first spot, $10$ the second, and $10$ the third. $6*10*10 = 600$. The $2$ can also be placed in the tens and hundreds place, so make that $600*3$. Now a $2$ can also be placed in the thousands place since 2 <= 6, but there can be 10 other digits everywhere else, so it would be $10*10*10$, leaving you with $1800+1000=2800$. (That is the basic logic behind what I did).
I do not know if you are looking for a function, or just an answer. But I know the answer is $28263827$. 
import java.io.File;

import java.io.PrintWriter;

public class test {
    public static long count = 0;

    public static void main(String[] args) {
        long i = 1;
        find(i);
        while (i != count) {
            if(i%100000==0){
                System.out.println(i + ":" + count);
            }

            i++;
            find(i);
        }

        System.out.println(i);
    }

    public static void find(long s) {
        String s2 = Long.toString(s);
        char[] chars = s2.toCharArray();

        for (char c : chars) {
            if (c == '2')
                count++;
        }
    }
}