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Hey I don’t know what to call the post I think the name doesn’t reflect the question very well (considering limits are a separate thing in maths), but the question, I think, is very intriguing.

When a 1-digit number is squared the number of digits that could be in the square is 1 or 2. In other words the first odd number and first even number. (Yes, I am aware 0 is an even number as well but I am going to ignore that fact because a number can’t have 0 digits. If you want to you can say “The number of digits in the square of a 1-digit number is the first odd number or the second even number”)

And this is true in general, the number of digits in square of a number of n digits can have either the nth odd number of digits or nth even number of digits.

Now here comes the interesting part in the domain of 1 digit number, 3 is the last number which gets me an odd number of digits when squared. 31 is the last number which gives an odd number of digits, when considering only two digit numbers and 316 when squared gives me an odd number of digits and it is the last number in the three digit realm that does this. And I found that this pattern continues for at least seven digits. The pattern here is - 3, 31, 316, 3162, 31622, etc... The numbers that come onwards are just the previous number with a digit attached to the back of it.

Why is this happening?

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    $\begingroup$ Note: $\sqrt{10}=3.1622...$ $\endgroup$ Commented Dec 11, 2020 at 2:46
  • $\begingroup$ $n^2<10^{2n+1}\iff n<10^n\sqrt{10}$ $\endgroup$ Commented Dec 11, 2020 at 2:52
  • $\begingroup$ Oh mt god thanks for the help. I think I understand what you mean to say, but could you explain it more clearly, so that I can check if what I understood is correct $\endgroup$ Commented Dec 11, 2020 at 8:38
  • $\begingroup$ I should have used a different letter (such as $m^2$ and $m$ instead of $n^2$ and $n$) in my comment above; see my answer $\endgroup$ Commented Dec 11, 2020 at 14:26

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This is happening

because the largest number which gives an odd number ($2n+1$) of digits when squared

is the largest number $m$ such that $m^2<10^{2n+1}$,

which is the largest number $m$ such that $m<10^n\sqrt{10}$.

So $m$ has the first $n$ digits of $\sqrt{10}=3.1622...$

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    $\begingroup$ Thanks a lot! I was thinking along with the same lines. Thanks a ton.!! Good day! $\endgroup$ Commented Dec 12, 2020 at 9:07

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