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

#### checkittwice

##### Member
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

##### Well-known member
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

##### Member
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.

Last edited:

#### CaptainBlack

##### Well-known member
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

##### MHB Oldtimer
Staff member
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%.

• Sudharaka

#### CaptainBlack

##### Well-known member
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

##### MHB Oldtimer
Staff member
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.