Sum of digits of numbers from $1$ to $10^n$ 
Without using a calculator, what is the sum of digits of the numbers from $1$ to $10^n$?

Now, I'm familiar with the idea of pairing the numbers as follows:
$$\langle 0, 10^n -1 \rangle,\, \langle 1, 10^n - 2\rangle , \dotsc$$
The sum of digits of each pair is $9n$. How can I prove this property retains for all pairs?
 A: Well, it's not too hard to figure out that the sum of digits of 0 or 1 to 9 is 45. That's very helpful, since that sequence will appear a bunch. Just add 1, and 46 is the sum of digits from 1 to 10 (or 0 to 10).
In a way, that and the base shifts are all you need I believe. Let's do 100. We're going to get 10 times 45 in the ones place, for all of the numbers. As for the tens place, it moves ten times slower - but they all still pair up ten times to make 45 (try doing 10 with 20 with 30, 11 with 21 with 31, etc). so the sum from 1 to 99 is 450 + 450 = 900, and then add 1 + 0 + 0 to get 901.
In general, it should be $ 45n 10^{n - 1} + 1 $, by the reasoning I have described.
A: There is a constant pattern of nos. I will demonstrate it for 2 digit nos.
00 01 02 03 04 05 06 07 08 09
10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29 and it goes so on.  
You can make a pattern of numbers for any digit nos.
Just add $0s$ before a number to make it of n-digit. Now due to this symmetrical pattern, all nos. 0,1,2,3,4. . . will be in equal amount.  
A: From $0000$ till $9999$ we have $10\,000$ four digit words. The $40\,000$ digits have an average value of $4.5$ each. Add $1$ for the extra number $10\,000$ and obtain $180\,001$.
A: Okay so the truly simple explanation is merely the fact that all those pairs sum to $999$ (i.e. in the case 0f $n=3$) . There's nothing more than that. In more details, one can see that this property is true for the first pair. Inductively, the next pair is made by increasing the first item by $1$ and decreasing the second by $1$, and so maintaining this property.
