Even-Odd number of digits in Perfect Squares

[billn]

Not a formal student, so this IS NOT homework. I'm just a long time fan of math.

Trying to pass along some tips I've used throughout my life, as an aid I generated a list of 1,000 consecutive integers and their squares. I became intrigued with the change in the number of digits in the squares from Odd to Even. In particular, the last root before the change and its relationship to the square root of 10. The attached PDF contains my work.

I do not understand the relationship between those roots and the square root of 10. I see it, but don't understand it. I've walked away from it several times in the hope that when I returned, the 'trees' would disappear and I could see the "forest". Hasn't happened!

Thanks!!

 

[billn]

Thanks anyway.

 

[Dr.Peterson]

Trying to pass along some tips I've used throughout my life, as an aid I generated a list of 1,000 consecutive integers and their squares. I became intrigued with the change in the number of digits in the squares from Odd to Even. In particular, the last root before the change and its relationship to the square root of 10. The attached PDF contains my work.

I do not understand the relationship between those roots and the square root of 10. I see it, but don't understand it. I've walked away from it several times in the hope that when I returned, the 'trees' would disappear and I could see the "forest". Hasn't happened!

Perhaps no one answered you because you seemed to have answered your own question (or else because they didn't want to open a pdf). When you didn't get an answer, you should have tried clarifying your question: What is it about this that you don't understand? What are the "trees" you are seeing?

Let's take one example, the step from 31 to 32.

Since [imath]31<10\sqrt{10} \approx 31.6[/imath], we know that [imath]31^2<(10\sqrt{10})^2 = 1000[/imath], so it has only 3 digits.

Adding 1 to that, we find that [imath]32>10\sqrt{10} \approx 31.6[/imath], so [imath]32^2>(10\sqrt{10})^2 = 1000[/imath], so it has 4 digits.

That should be enough to explain your observation, unless I'm missing your point.

Here is a more advanced version:

The number of digits in a number [imath]N[/imath] is [imath]\left\lfloor\log_{10}N\right\rfloor + 1[/imath]. For example, 12345 has [imath]\left\lfloor\log_{10}12345\right\rfloor + 1 =\left\lfloor 4.09\right\rfloor + 1 = 4 + 1 = 5[/imath] digits.

So the number of digits in [imath]N^2[/imath] is [imath]\left\lfloor\log_{10}N^2\right\rfloor + 1 = \left\lfloor 2\log_{10}N\right\rfloor + 1[/imath], which increases each time [imath]2\log_{10}N[/imath] reaches an integer. It reaches an even integer (leading to an odd number of digits) when N reaches a power of 10. It reaches an odd integer (leading to an even number of digits) when [imath]\log_{10}N[/imath] equals an odd multiple of 1/2, meaning that N is 10 raised to an odd multiple of 1/2 -- that is, a power of 10 times [imath]\sqrt{10}[/imath]. And that's what you observed.