Number of digits from 1 to n

Question

Let $n$ be a natural number $k$ digits. Show that the quantity $Q$ of digits required to write the natural numbers from $1$ to $n$ is:

$Q = k(n+1) - \underbrace{111\ldots11}_{k\textrm{ digits}}$

Answer 1

As this question is now 14 hours old, I will give an answer.

First sum the total number of digits of all the natural numbers $ \le k-1.$ Call this sum $S.$

There are $9 \times 10^{i-1}$ natural numbers of length $i,$ so the total number of digits of natural numbers of length $i$ is $9i \times 10^{i-1}.$ Hence

$$ S=9(1+2.10 + 3.10^2 + \cdots + (k-1)10^{k-2}) = (k-1)10^{k-1} - \frac{10^{k-1}-1}{9}.$$

Now the first natural number with $k$ digits is $10^{k-1}$ and so there are $n- 10^{k-1}+1$ natural numbers from $10^{k-1}$ to $n$ inclusive. Each of these numbers has $k$ digits and so the number of digits required to write the natural numbers from $1$ to $n$ is

$$ k( n- 10^{k-1}+1) + S = k( n- 10^{k-1}+1) + (k-1)10^{k-1} - \frac{10^{k-1}-1}{9}$$

$$ = k(n+1) - \frac{10^k-1}{9}.$$

Answer 2

Start with the one digit numbers. How many digits are needed for them. Do you include zero? How many digits to go from 10 through 99? And so on...

Answer 3

HINT $\ $ Chopping off the unit digit is a 10-to-1 map from k+1 digit onto k-digit naturals. Therefore, since there are 9 naturals with 1-digit, there are 90 with 2-digits, and 900 with 3-digits, etc.