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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?

Vepir
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Mark
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  • Just for clarification: by "the number created by ordering 1 to $n$", do you mean $$123456789101112\cdots n?$$ – Zev Chonoles Aug 16 '11 at 22:58
  • @Zev: yeah that is what I mean – Mark Aug 16 '11 at 23:06
  • 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? – Blue Aug 16 '11 at 23:19
  • In Base Two, with $n = 3$: $11011$. – Blue Aug 16 '11 at 23:26
  • @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. – Mark Aug 16 '11 at 23:26
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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.. – Listing Aug 16 '11 at 23:35
  • @Listing: that's what I am doing. – Mark Aug 16 '11 at 23:41
  • What's the difference between looking up the hints and posting the problem to this website? – Gerry Myerson Aug 17 '11 at 00:02
  • @Gerry: I want to see how to solve it in some other way. – Mark Aug 17 '11 at 00:15
  • In base 3, lots of solutions, including (1)(10)(2)(12)(22)(21)(20)(11). – Robert Israel Aug 17 '11 at 00:17
  • How do we know what way is "another way" if you don't tell us what way you already know? – Gerry Myerson Aug 17 '11 at 01:29
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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. – Dave Radcliffe Aug 17 '11 at 02:24
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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. – Blue Aug 17 '11 at 03:24
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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. – Gerry Myerson Aug 17 '11 at 04:39

1 Answers1

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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.

Gerry Myerson
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