Find the 1000th digit of $N=61218243036\ldots$ If all the multiples of 6 are written side-by-side, then a large integer $N$ is generated as follows:
$$N=61218243036\ldots$$
The question is to find the $1000$th digit of $N$. Please simply give a hint as to how I can proceed. I have calculated the answer through brute force counting (with the help of a computer) but I am looking for something more elegant.
 A: There are $10^{k}-10^{k-1}=3(3 \times 10^{k-1})=6 (15 \times 10^{k-2})$ numbers with $k$ digits. Therefore, for $k \geq 2$, there are $15 \times 10^{k-2}$ multiples of $6$ with $k$ digits.
So the bit of $N$ that uses all the multiples of $6$ with at most $m$ digits will have length $1 + \sum_{k=2}^m k(15 \times 10^{k-2})=1 + 15(\sum_{k=0}^{m-2} (k+2) 10^k)$, which, after some algebra, is equal to $\frac{10^m(9m-1) - 26}{54}$.
Now, you can explicitly find the largest $m$ such that this expression is less than $1000$, and work from there...
A: There is only 1 multiple of 6 with one digit. 
There are [100/6 - 1] = 15 multiples of 6 with two digits: then 30 digits. 
There are [1000/6 - 15 - 1] = 150 multiples of 6 with three digits: then 450 digits. 
At this point, we have counted the first 1 + 30 + 450 = 481 digits of N. There remain other 1000 - 481 = 519 digits. The first 516 of these are included in the next 516/4 = 129 multiples of 6, which have all four digits. 
Thus, we have counted a total of 1 + 15 + 150 + 129 = 295 multiples of 6, and there remains three digits to arrive to the 1000th digit of N. Thus, the 1000th digit of N is the third digit of the 296° multiple of 6. Since 296 * 6 = 1776, the digit is 7.  
