TITLE: A Question on Digit Occurences
QUESTION [6 upvotes]: Here's a question I was thinking about:
For all positive integers n, list the decimal representation of the numbers 1, 2, 3, ..., n without any leading zeroes. Does there exist an n such that this list contains an equal number of each of the digits 0, 1, 2, 3, ..., 9? (For example, if n=15, the list is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, and it contains eight 1's, one 0, and so on.)
I thought about it for quite a while, and intuitively it seems very unlikely, but I couldn't formulate a rigorous proof. Could you guys help me in discovering one?
Thanks!

REPLY [3 votes]: no, the number of times zero appears is always less than the number of times one appears. 
proof:
set $k_i(n)$ equal to the number of times $1$ has appeared in the $10^i$'s place minus the number of times $0$ has appeared in the $10^i$'s place.
each $k_i(n)$ is non-negative and if the place of the first digit of n is $10^m$, $k_m(n)$ is positive, so the sum of all the $k_i(n)$ is positive.