To the nearest 10%, what percentage of all numbers contain a 9?

You probably need to clarify that you mean “what percentage of the set of all integers contains the digit ‘9’? Because there are an infinite number of “numbers” that don’t contain that digit (and another infinite set that does). So that’s kind of a non-starter.

However, the logic still holds.

There is an infinite set of integers that contain the digit 9. There is also an infinite set of digits that does not.

Maybe you need to restrict your range some more.

10% of all whole numbers contain the number 9.

It’s actually more than that, @zenvelo, because 100% of the numbers from 90 – 99 do, as well as 10% of all the numbers from 0 – 89. So 10% is an initial floor, but it’s a rising percentage.

@CWOTUS But 90% of all numbers do not have a 9.

This was a puzzler on CarTalk a few years ago.

How can you make that assertion, @zenvelo? In the numbers from 0 to 100 alone, 19% of all those integers contain the digit “9”.

9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99

Now, if you’re talking about integers divisible by 9 or integers that end in 9, then it’s a different story, but that’s not what the OP asks.

This is a really interesting question….

My first instinct was 10%. But I don’t EVER trust my instincts in math! They’re usually wrong.

So I thought about it further, breaking it down first into the numbers 0–99:
For 90% of those numbers (all of them except 90–99), 10% have a 9. 9, 19, 29, etc. For the other 10%, 100% of them have a 9 of course.
10% * 90 + 100% * 10 = .1 * 90 + 1 * 10 = 19%

Now I think about 100–999…
Again for the last 10%, 100% of them will contain a 9. So let’s think about the first 90%. For the numbers 100–899, you can ignore the first digit because you know it won’t be a 9. So then you’re basically just looking at the last two digits, which is 0–99 again, which we already determined to contain a 9 19% of the time. So now we have…
19% * 90 + 100% * 10 = .19 * 90 + 1 * 10 = 27.1%

27.1% of 3 digit numbers have a 9, and 19% of 2 digit numbers (I treated 0–9 as 2 digit numbers here, by calling them 00, 01, 02 etc), and there are 10 times as many 3 digit numbers as there are 2 digit numbers, so our average so far is:

(27.1% * 10 + 19% * 1) / (11) = 26.363636…%

I think the pattern will continue – as we keep adding digits, I think for each iteration we’re going to have the previous iteration’s percentage * 90 + 100% * 10. We’d have to add that sequence into infinity to get our answer, and at first I figured it would turn out to be a convergent sequence with a finite result. But when I started actually trying it, it soon became obvious that it wasn’t going to stop before 100. So I must have made a mistake or bad assumption somewhere. I’m really needing to get to bed right now, but I’d like to look at this more tomorrow if someone else hasn’t already solved it.

I’m happy with my wording @CWOTUS , and I’m happy with yours.

The question remains

To the nearest 10%, what percentage of all integers contain the digit 9?

lol

All numbers is an infinite set, including all numbers of all sizes. Most of the huge ones will have a 9, enough that it will be almost but not quite 100%. For example, the percentage of 100,000-digit numbers that does not have a 9 is very very small, and the set of all numbers includes numbers of any length, which only get less and less likely to not have a 9. So the approximation of the percentage would be 99.9999% with as many 9’s as you care to add, or some more accurate designation in calculus that takes into account that you’re talking about the proportion of an infinite set.

To the nearest 10%, 100% of all integers include a 9.

…..hhm…..let’s see now…....9, being the square root of 81, times 6,561, the square root of 81…...(OK, divide by 99…....carry the 9…....multiply that by…....9.9…...
.

(umm….can I get back to youz on this….??)

But…we’re talking about infinity….

OK, I am certain now that it is a convergent series that converges on 100%. @Zaku is correct.

The logic I used last night was pretty good, just ignore the paragraph where I separated out 3 and 2 digit numbers and averaged them. That was a mistake.

1. Start with 0–9. 10% have a 9. 0–8 (9 numbers) do not, 9 (1 number) does. 1/(1+9) = 10%

2. Then 0–99. 00–89, ignore the first digit which will never be a 9. You then have just the second digit, which is 0–9 repeated, which we already know from step 1 contains a 9 10% of the time. So in the numbers 00–89, which are 90% of the numbers 0–99, our percentage is still 10%. But for the last 10%, 90–99, it’s 100%. .9 * 10 + .1 * 100 = 19%.

3. Then 0–999. Same logic: look at 000–899. The first digit will never be a 9. So we just have 00–99 to consider, which we know from step 2 has a 9 at the rate of 19%. Again this chunk represents 90% of all the numbers we’re currently considering. The last 10% again has a 9 100% of the time. .9 * 19 + .1 * 100 = 27.1%

4. The pattern continues. At this point I’m going to get less formal because I’m not a mathematician. I repeated the pattern, which is .9 * (previous answer) + .1 * 100 many many times with the help of some tools on my calculator. It converged very slowly towards 100% (pic of calculator screen: http://imgur.com/a/UqXuT)

10%, my original instinct, is the answer to another similar but definitely distinct question, “What percent of digits in all whole numbers are a 9?” Picture both of these situations as a venn diagram – in that question, the circles don’t overlap, but in the OP’s question, they do because numbers can contain a 9 but that doesn’t preclude them from containing other digits too. Furthermore! I would argue that the answer to that question is not in fact 10% exactly but slightly higher than 10%! Zero doesn’t get its fair share of the pie because we don’t use leading zeros to represent whole numbers, so zero never gets to be the first digit of any number, whereas all the other digits do.

I agree with @Mariah and @Zaku in that, as you increase the range of numbers, the percentage which contain a 9 approaches a limit of 100%.

I would say that you do not need some kind of infinite series to figure this out. Take a list of all numbers of length N. For example, if N = 3 then you are dealing with 000 through 999. The proportion of numbers containing the digit 9 is then 1 – (9/10)^N.

This is derived by considering each digit independently, in which case a given digit has a 9/10 probability of not being 9. Then (9/10)^N is the probability of all digits not being 9. Subtracting that from 1 gives the probability that at least one digit is a 9.

Ah great simplification Phi! Only nitpick I have is that 0 is never the first digit, so being pedantic it’d be, I think, 1 – (8/9)(9/10)^(N-1). Obviously this doesn’t really matter for coming to the conclusion that the limit approaches 100%.