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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 $31622776^2 = 9.9999996\cdot10^{14}$. 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 $\frac{10^{n}-1}{k} \leq 3.2$ otherwise it will have $2n-1$ digits

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    $\begingroup$ $\sqrt{10}\approx3.16$ $\endgroup$ Jul 18, 2019 at 22:08
  • $\begingroup$ @J.W.Tanner and what does it help me? $\endgroup$
    – ESCM
    Jul 18, 2019 at 22:17
  • $\begingroup$ see my answer below $\endgroup$ Jul 18, 2019 at 22:27
  • $\begingroup$ EduardoS: because that's what base-10 logarithm does. floor(10^log(X)) tells you the number of (decimal) digits of X; hence floor(10 ^ 2log(X)) for X^2. As you found empirically. $\endgroup$
    – smci
    Jul 19, 2019 at 19:21

3 Answers 3

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

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

So the cutoff you've observed is exactly at $\sqrt{10} \approx 3.162277660168379332$, times powers of 10.

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

Now if $10^{n-1}\le k \lt \sqrt{10} \times10^{n-1}$,

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

whereas if $\sqrt{10} \times10^{n-1} \lt k \lt 10^n$,

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

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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 $10^{m-1} \le k < \sqrt{10^{2m-1}}$

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

Now... thing for a moment. If $\sqrt{10} \approx 3.1622776601683793319988935444327....$, then $10^{m-1}*\sqrt{10}$ 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 $\sqrt{10}$.

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.

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