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Is it possible for the number created by the consecutive numbers $1$ to $n$ where $n > 1$ be a palindrome eg. $1234567\ldots n$?

I believe this is a contest problem, but how would one solve this problem without looking up the hints?

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    Just for clarification: by "the number created by ordering 1 to $n$", do you mean $$123456789101112\cdots n?$$2011-08-16
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    @Zev: yeah that is what I mean2011-08-16
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    By "ordering" ... Are you allowed to specify the order of the numbers? That is, with $n=3$, would $132$ (though obviously not a palindrome) be a number under consideration? If so, then do we require that multi-digit blocks (for "ten", "eleven", "one hundred eighty seven", etc) remain in tact during the shuffling?2011-08-16
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    In Base Two, with $n = 3$: $11011$.2011-08-16
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    @Day: Hmmm... I don't think so. Perhaps I should have said consecutive numbers $1$ to $n$. I have trouble thinking up a solution without resorting to hint. I am just curious how to approach this problem with out hints.2011-08-16
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    The easiest way is to look at the hint and then intensively study how you could have done it without the hint..2011-08-16
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    @Listing: that's what I am doing.2011-08-16
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    What's the difference between looking up the hints and posting the problem to this website?2011-08-17
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    @Gerry: I want to see how to solve it in some other way.2011-08-17
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    In base 3, lots of solutions, including (1)(10)(2)(12)(22)(21)(20)(11).2011-08-17
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    How do we know what way is "another way" if you don't tell us what way you already know?2011-08-17
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    You want to look for distinguishing features of the string 123456...n which prevent it from being a palindrome. Once you think of searching for long runs of zeros, the solution should come quickly.2011-08-17
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    @Robert: Re-ordering the number blocks apparently wasn't intended to be allowed. That said, since the answer to the original question is "no", it seems fair to ask about the case where re-ordering *is* allowed.2011-08-17
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    @Day, 1 19 8 17 6 15 4 13 2 10 20 12 3 14 5 16 7 18 9 11.2011-08-17

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A solution can be found on page 43 of Andreescu and Andrica, Number Theory: Structures, Examples, and Problems, which page I was able to access on Google Books.

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    The [link](http://books.google.com/books?id=Mlt2O1rR9xIC&pg=PA43).2011-08-17
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    @J.M. You're the link man!2012-02-18
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    For those who do not want to read it thoroughly, if $10^5$ is the largest degree of $10$ before $n$, then there is only one piece $...99991000010001...$ with $0000$, which thus should be in the center of palindrome, but is not symmetric.2018-03-15