Sum of the Digits Used

Sum of digits from $1$ to $9$ is $45$

This will come $10$ times ( sum of all the units digits)

Sum of the tens digits from $10$ to $19=10$

Similarly we would get the sum of tens digits as $20, 30,40.....90$

The total sum would be $(45\times 10) + (10 +20 +30..........90)$

$= 450 +450 =900$

The digit $k$ appears 10 times in the units place, and 10 times in the tens place. Hence, it is used $20$ times in the positive integers from 1 to 99. Since the sum of the digits from 1 to 9 is 45, the total sum of all digits is $20 \times 45 = 900$ .