Digit Sums

The problem is equivalent to sum of all digits in hundreds place + sum of all digits in tens place + sum of all digits in units place.

$x=H+T+U$

Now the numbers 1 to 9 appear 100 times each in 'hundreds place' (1 appears in 100 to 199, 2 appears in 200 to 299 and so on) Thus, $H=100 \times (1+2+\cdots+9)=100 \times 45 = 4500$ .

Similarly each numbers appears 10 times in the 'tens place' in each of the sets (100 to 199, 200 to 299, ..., 900 to 999)

Giving $T=9 \times 10 \times 45 = 4050$ .

Using a similar logic $U = 90 \times 45 = 4050$

Giving the solution as

$\huge{x=H+T+U = 12600}$

on noticing the pattern of occurrence,each number 1 to 9 appears exactly 280 times.the zeroes do not make a difference to the sum as 0+0=zero(lol) thus,you get 280(1+2+3+4+5+6+7+8+9)=280(45)=12600

We can form $9\times10\times10=900$ three digit numbers

Each number will be in this place:

$\frac{900}{9}=100$ in hundreds

$\frac{900}{10}=90$ in tens

$\frac{900}{10}=90$ in unities

Then

Hundred place sum will be $100(1+2+3..+9)=100(45)=4500$

Ten's place sum will be $90(0+1+2+3..+9)=90(45)=4050$

Unities place sum will be $90(0+1+2+3..+9)=90(45)=4050$

Total sum is $4500+4050+4050=\boxed{12600}$