# 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?

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

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

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.

• This is very similar to your prior question.
– lulu
Commented Aug 30, 2022 at 16:05
• You can find the bounds for the perfect squares having $1,2,3,\cdots$ decimal digits as a shortcut. Commented Aug 30, 2022 at 17:10

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.