How many zeros occur in all the natural numbers from 1 to 10^n inclusive where n is a natural number?

Question

10^n is 10 to the power of n.

note to firefly:
what about 101, 102,..., 201, 202,... etc?

Answer 1

Thanks for a very nice counting problem.
I think I have it:
You can check that the number of zeros between 10^(m-1)+1 and 10^m is: z[m] = 1+9(m-1)10^(m-2)
I have checked the following numbers:
m z[m]
1 1
2 10
3 181
4 2701
5 36001
6 450001
7 5400001
8 63000001
We are looking for the sum of these numbers:
Z[p]=Sum(m:1 to p) { z[m] } = p(10^(p-1) + 1) - (10^p - 1)/9
So for example inclusively from 1 to10^10 we find 8888888899 zeros.
To find the frequency of zeros we divide Z[p] by 10^p to get the asymptotic (1/10)p which we expect.
Please correct me if I am wrong.

Answer 2

Well, if n is 1, then the answer is 1
If n is 2, then the numbers with zeros are:
10,20,30,... 90, 100 which is 11 (two zeros in 100).
If n is 3, then for each "hundred", we have 11, and then at 1000 we have an extra, or 11*10+1

So a pattern emerges:
1
10+1
10*(10+1)+1 = 111
10*(10*(10+1)+1)+1 is the next, I presume, or 1111 or
the sum from 0 to n-1 of 10^(i-1)