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I am trying to group numbers based on the count of digits in the square they produce, and below is the list containing the grouping that I've done so far.

1 - 3     // Numbers producing 1 digit when squared ...
4 - 9     // Numbers producing 2 digits ...
10 - 31   // Numbers producing 3 digits ...
32 - 99   // and so on ...
100 - 316  
317 - 999  
1000 - 3162
3163 - 9999
10000 - 31622
31623 - 99999
100000 - 316227
316228 - 999999

As one can observe, every range that produces odd digit count has a lower bound of $10^{i-1}$, while its upper bound has an observable pattern. I want to know if this can be represented by a formula or if the upper bound is a known constant such that we can use it as $C \cdot 10^{i-1}$.

JD Hrnnts
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  • Have a look at the decimal expantion of $\sqrt{10}$, and see whether you recognize it. – Arthur Sep 20 '17 at 07:18
  • Is there a rule/theorem that is surrounding this? Also I want to know where did $\sqrt{10^{2k+1}}$ came from. – JD Hrnnts Sep 20 '17 at 07:30

1 Answers1

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The numbers producing $1$ digit when squared has squares between $1$ and $10$, which means that the numbers themselves lie between $\sqrt1$ and $\sqrt{10}$. The numbers producing two digits when squared has squares between $10$ and $100$, so they lie between $\sqrt{10}$ and $\sqrt{100}$.

Continuing this way, we see that numbers that leave $n$ digits when squared has squares between $10^{n-1}$ and $10^{n}$, so they must lie between $\sqrt{10^{n-1}}$ and $\sqrt{10^{n}}$. This is the pattern that you have picked up.

Arthur
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