Percent of perfect squares that have an odd digit for their hundreds place

[checkittwice]

Suppose you consider the set of all perfect squares (the squares of all of the integers).


What percent of them have an odd digit for their hundreds place?

 

[CaptainBlack]

suppose you consider the set of all perfect squares (the squares of all of the integers).


What percent of them have an odd digit for their hundreds place?

41%

cb

 

[checkittwice]

41%

cb

Edit: Some of my text was removed because I don't agree with it.


I really meant to ask (hopefully an easier question with less work):


"What percent of the perfect squares have an odd digit in their tens place?"


Edit:

I sent a PM to CB, because I did not want to have any post in
front of Opalg's solution to my amended question.

 

[CaptainBlack]

I would chop off that 1% and make it 40%.

However, I don't have the work for you at this
time to back it up.

When I get the chance I will check my reasoning and the counting program.

CB

 

[Opalg]

I really meant to ask (hopefully an easier question with less work):


"What percent of the perfect squares have an odd digit in their tens place?"

If you square the number $a+10b+100c+\ldots$ (where $a,\,b,\,c\ldots$ are digits from 0 to 9) then you get

$a^2 + 20ab + \ldots$ (everything else is a multiple of 100).

If you are only interested in whether the tens digit is even or odd, then a multiple of 20 makes no difference. So everything depends on whether the tens digit in $a^2$ is even or odd. The squares of the digits from 0 to 9 are

00, 01, 04, 09, 25, 49, 64, 81 (tens digit even)

and

16, 36 (tens digit odd).

Each of these is equally likely to occur, so I reckon that the proportion of perfect squares with an odd digit in their tens place is 20%.

 

[CaptainBlack]

If you square the number $a+10b+100c+\ldots$ (where $a,\,b,\,c\ldots$ are digits from 0 to 9) then you get

$a^2 + 20ab + \ldots$ (everything else is a multiple of 100).

If you are only interested in whether the tens digit is even or odd, then a multiple of 20 makes no difference. So everything depends on whether the tens digit in $a^2$ is even or odd. The squares of the digits from 0 to 9 are

00, 01, 04, 09, 25, 49, 64, 81 (tens digit even)

and

16, 36 (tens digit odd).

Each of these is equally likely to occur, so I reckon that the proportion of perfect squares with an odd digit in their tens place is 20%.

That is what I get.

CB

 

[Opalg]

That is what I get.

... and I agree with your 41% for those with an odd digit in the hundreds place. I simply counted the number of odd hundreds-place digits in the list of the first 100 squares (as listed here, for example). The proportion in any other sequence of 100 consecutive squares will be the same.