8
$\begingroup$

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

$\endgroup$
  • 13
    $\begingroup$ $\sqrt{10}\approx3.16$ $\endgroup$ – J. W. Tanner Jul 18 at 22:08
  • $\begingroup$ @J.W.Tanner and what does it help me? $\endgroup$ – Eduardo S. Jul 18 at 22:17
  • $\begingroup$ see my answer below $\endgroup$ – J. W. Tanner Jul 18 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 at 19:21
24
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.