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Among the numbers 1, 2, . . . , $10^{10}$, are there more of those containing the digit 9 in their decimal notation, or of those with no 9?

Please give me some hint how to prove this.

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    Hint: Easy to count those with no $9$...basically there are nine candidates for each slot! – 2017-01-16
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    @ lulu what do you mean by 9 candidates? – 2017-01-16
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    Well, largest place can only be $0,1$, but if it's $0$ then the next largest can be any of $0-8$, so there are nine candidates for that slot. And nine for the next, and so on. Makes life a bit simpler if you add $0$ to your list (but it's no big deal to remove it). – 2017-01-16
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    To be clear, ignoring the case of $n=10^{10}$ there are ten available slots. Each of those slots can be filled with any of the digits from $0-8$ so there are nine choices for each of those ten slots. Now, if you choose all $0's$ you get the number $0$ which isn't in your list, so subtract $1$ for that. Then add $1$ for $n=10^{10}$. – 2017-01-16
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    It would be interesting to see when the change occurs. Clearly for single-digit numbers there are more without "9" than with "9." As the answer shows, this changes by the time we get to 9 digits. Where is the bifurcation point? And as the number of digits rises further, the proportion of those not containing a "9" must drop ever closer to 0. – 2017-01-16

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The amount of numbers not containing the digit $9$ is:

$\left(\sum\limits_{n=1}^{10}8\cdot9^{n-1}\right)+1=$

$8\cdot\left(\sum\limits_{n=1}^{10}9^{n-1}\right)+1=$

$8\cdot\left(\sum\limits_{n=0}^{9}9^{n}\right)+1=$

$8\cdot\left(\frac{9^{10}-1}{9-1}\right)+1=$

$3486784401$

The amount of numbers containing the digit $9$ is therefore:

$10^{10}-3486784401=6513215599$

As $6513215599>3486784401$, there are more numbers containing the digit $9$ than not.

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    sorry, I am new to this. would you explain a little bit more on how did you construct the summation? and why do you have to add 1? – 2017-01-16
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    @Tmm: The summation is for each number of digits (between $1$ digit and $10$ digits). The "$+1$" is for the only $11$-digit number ($10^{10}$). – 2017-01-16
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    what about the number 8? – 2017-01-16
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    @Tmm: The first digit is $\in[1,8]$ (hence $8$ options). Each one of the other digits is $\in[0,8]$ (hence $9$ options). – 2017-01-16
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    @ barak manos thanks! – 2017-01-16
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    @Tmm: You're welcome. Do you understand the complete answer then? – 2017-01-16
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    @ barack manos Yes, I do now. It's just a basic counting problem. I have been reading textbook while thinking of your answer. It all makes sense to me now! – 2017-01-16
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    @Tmm: Great. Feel free to accept the answer by clicking on the V next to it :) – 2017-01-16
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Since the sum of the reciprocals of the integers lacking the digit 9 converges, there are many more integers with 9 than without it.

The same for any base and any digit.

Another way.

There are $10^n$ n-digit numbers, and there are $9^n$ n-digit numbers without the digit 9.

The ratio is $(9/10)^n$.

If $n=10$, this is $(9/10)^{10} =(1-1/10)^{10} \approx 1/e $ since $(1-1/n)^n \approx 1/e$.

Therefore, about $1-1/e \approx 0.63$ of these numbers have a 9 and $1/e \approx 0.37$ do not have a 9.

Therefore more have them have a 9.

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    @ marty cohen Would you expand a little bit more on your reasoning? – 2017-01-16
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    I did that from my phone. When I get to my computer, I'll expand. – 2017-01-16
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    @ thanks! can't wait. – 2017-01-16
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    @martycohen +1. Interesting new approach I've never seen. – 2018-01-21