Is it possible to calculate how many digit number I have to write if I write $n^2$ sequence from 1 to $1000^2$ without using calculator?

Question

From the fact that, for any positive integer n, it will require $1+\lfloor\log_{10}(n)\rfloor$ digits to write. When $\lfloor\ \rfloor$ is floor function.

If I write "1,4,9,16,25,...,1000000", the total digit number I have to write is

\begin{equation} \sum_{n=1}^{1000} ( 1+\lfloor\log_{10}(n^2)\rfloor ) = 1001 + \sum_{n=1}^{1000} (\left\lfloor \frac{2\ln(n)}{\ln(10)} \right\rfloor ) \end{equation}

The problem is that is it possible to calculate $\sum_{n=1}^{1000} \left\lfloor \frac{2\ln(n)}{\ln(10)} \right\rfloor$ without using calculator or not.

I think no but I am not sure. It might be possible in the way that I don't know.

Answer 1

If $1 \leq n \leq 3$ then $n^2$ has one digit.

If $4 \leq n \leq 9$ then $n^2$ has two digits.

If $10 \leq n \leq 31$ then $n^2$ has three digits.

If $32 \leq n \leq 99$ then $n^2$ has four digits.

If $100 \leq n \leq 316$ then $n^2$ has five digits.

If $317 \leq n \leq 999$ then $n^2$ has six digits.

And of course $1000^2$ has seven digits.

From this you can add up the total number of digits. You can easily confirm each of the numbers in the inequalities $a \leq n \leq b$ above just using pencil and paper if you know how to do long multiplication of three-digit numbers.