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Possible Duplicate:
Is it possible for the number created by ordering $1$ to $n$ where $n \geq 1$ be a palindrome?

I've been thinking about this problem for a while now, and have come up with nothing useful.

If anybody knows the solution, or the break-through idea, it'll be much appreciated. It's from a Russian 1996 Olympiad:

Can the number obtained by writing the numbers from $1$ to $n$ in order (for some $n>1$) be a palindrome?

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    Here's my two cents: study the distribution of zeroes (which is rather sparse). Notice that 123456789 is not a palindrome. Continuing, we get 12345678910. This is not a palindrome. The next candidate is 12345678910 followed by the next (nonzero) 10 digits. This is not a palindrome. Thus we continue until we hit the next zero, and so on. You'll end up having to investigate blocks in between zeroes, which must be symmetric about the zeroes (the zeroes, in a sense, act line a mirror). The answer should be: no, this isn't possible.2012-02-18
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    @anon: True, it is a duplicate. However, the question was not answered. I looked up the book, and it isn't there.2012-02-18
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    OP: The question was answered, and the link J.M. gave in the comment works fine for me, showing that indeed there is a proof at the top of page 43. If you don't have access for any reason I've [uploaded it for you](http://i.imgur.com/6v4kB.png).2012-02-18

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