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I realized that, any number $k$ with $n$ digits, and when $k$ is squared, i.e $k^2$ will have $2n-1$ or $2n$ digits.

Per example, let $k = 583$, thus $n = 3$ and the digits of $583^2$ are $2n$.

But I started thinking, how can I narrow the result to be more precise?

But, I asked myself. How can I determine when I will have $2n-1$ or $2n$ digits?

What I did was the following:

$1)$ The last number of $1$ digit whose square has $2n-1$ digits, is $3$ since $3^2 = 9$, and the next number $4^2 = 16$ have $2n$ digits. And the quotient between the max numbers of $1$ digit($9$) and the last number of $1$ digit to the pow of $2$ with $2n-1$ digits(3) is: $9/3 = 3$

$2)$ The last number of $2$ digits whose square has $2n-1$ digits, is $31$, since $31^2 = 961$. The quotient here is $3.19354839$

$3)$ The last number of $3$ digits whose square has $2n-1$ digits, is $316$ since $316^2 = 99856$. The quotient here is $3.16139241$

$4)$ The last number of $8$ digits whose square has $2n-1$ digits, is $31622776$, since $31622776^2 = 9.9999996cdot10^14$. The quotient here is $3.16227768871$

The quotient is each time smaller and closer to $3.16$

$i)$ Why does the quotient between the largest number with $n$ digits and the last number with $n$ digits squared that have $2n-1$ follow this "pattern" closer and closer to $3.16$?

$i)$ With this I can assure for all numbers that: If i have a number, per example $k = 7558$ and $k$ have $4$ digits and the quotient between $9999/7558 < 3.2$, then $k$ have $2n = 8$ digits?

That more generally, I can assure you this?:

If i have a number $k$ with $n$ digits, this number have
$2n$ digits if $frac10^n-1k leq 3.2$ otherwise it will have $2n-1$
digits

If $k^2$ has $2n$ digits, then it is true that $10^2n-1 leq k^2 < 10^2n$, so we have $10^n-1sqrt10 = sqrt10^2n-1 leq k < sqrt10^2n = 10^n$.

If $k^2$ has $2n-1$ digits, then it is true that $10^2n-2 leq k^2 < 10^2n-1$, so we have $10^n-1 = sqrt10^2n-2 leq k < sqrt10^2n-1 = 10^n-1sqrt10$.

So the cutoff you've observed is exactly at $sqrt10 approx 3.162277660168379332$, times powers of 10.

First note this: $k$ has $n$ digits means $10^n-1le k lt 10^n.$

Now if $10^n-1le k lt sqrt10 times10^n-1$,

then $10^2n-2le k^2 < 10^2n-1$, so $k^2$ has $2n-1$ digits,

whereas if $sqrt10 times10^n-1 lt k lt 10^n$,

then $10^2n-1le k^2 < 10^2n$, so $k^2$ has $2n$ digits.

A (natural) number $k$ has $m$ digits if an only if $10^m-1 le k < 10^m$

So $10^2m-2 le k^2 < 10^2m$.

If $10^2m-2 le k^2 < 10^2m-1$ then $k^2$ will have $2m-1$ digits.

Taking the square roots we see This happens when $10^m-1 le k < sqrt10^2m-1$

$sqrt10^2m-1 = sqrt10*10^2m-2 = 10^m-1*sqrt 10$ which is never an integer.

Now... thing for a moment. If $sqrt10 approx 3.1622776601683793319988935444327....$, then $10^m-1*sqrt10$ will be $3.1622776601683793319988935444327...$ "shifted over $m$ decimal places".

So the largest possible natural number less than $10^m-1*sqrt 10$ will be the first $m$ digits of $sqrt10$.

So a $m$ digit number when square will have $2m-1$ digits if $k < 10^m-1*sqrt 10$ and will have $2m$ digits if $k > 10^m-1*sqrt 10$. And the way can tell if $k <$ or $k > 10^m-1*sqrt 10$ is by checking if the digits match the digits of $31622776601683793319988935444327...$ up to $m$ plaes.