# A Question on Digit Occurences

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!

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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.

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Is there a simple proof of this claim? – coffeemath Nov 11 '13 at 0:22
Here is the proof, as requested. – Zackkenyon Nov 11 '13 at 0:40
It looks like this shows the number of $1$'s appearing for $1\le k \le n$ exceeds the number of $0$'s, which is enough. I'd suggest adjusting the notation to $k_i(n)$ for the difference between the number of $1$'s minus the number of $0$'s which appear for $1 \le k \le n$ and maybe elaborating on sub-proofs of $k_i(n)\ge 0$ and the statement re. numbers whose most significant digit is in the $10^m$ position. (and +1 -- that wasn't my downvote.) – coffeemath Nov 11 '13 at 1:21
Sorry, this proof seemed instantly clear to me, and I thought maybe user107905 maybe just wanted a hint. But then downvotes and challenges to my honor. – Zackkenyon Nov 11 '13 at 1:39
I didn't see it immediately, even when looking at it. Had to do some paperwork to get the proof. I hope you didn't view my "is there a simple proof" as a challenge of your answer, it was just a request for a fill-in, as you placed in the answer afterwards. Thanks. – coffeemath Nov 11 '13 at 2:24