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A palindrome is a number or word that is the same when read forward and backward, for example, “176671” and “civic.” Can the number obtained by writing the numbers from 1 to n in order (n > 1) be a palindrome?

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That's why he defined a palindrome as "number or word" – Hagen von Eitzen Dec 27 '12 at 21:28
I covered all bases :) – fosho Dec 27 '12 at 21:28
Maybe I misread. Thought the word "word" was not there initially. – Thomas Andrews Dec 27 '12 at 21:29
up vote 5 down vote accepted

No. With $n=1$, we do have a palindrome of course. But for $n>1$ we can clearly exclude the case $n\le 10$. In fact, we need $n\equiv 1\pmod {10}$, as the palindromic string ms must end in "$\ldots 1$". Let $k\ge1$ with $10^k<n<10^{k+1}$. Then there is exactly one position in the assumed palindromic string "$12345\ldots54321$" where "$1\underbrace{0\ldots0}_k1$" occurs: In the absence of leading zeroes, only the two $1$s in this block can be leading digits of some numbers, and since no number has more than $k+1$ digits, indeed both $1$s must be leading digits, i.e. the only position where this pattern occurs is at the number $10^k$ (together with the leading $1$ of $10^k+1$). For a palindrome, such a uniquely occuring subpalindrome must be in the very middle. But it is preceeded by "$\underbrace{9\ldots9}_k$" from $10^k-1$ and followed by "$\underbrace{0\ldots0}_{k-1}1$" as the rest of $10^k+1$, contradicting the palindromic symmetry.

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Here's a beginning of the task where the numbers do not have to be in order. Note that a palindrome can have at most one digit which occurs an odd number of times - the centre digit if the number of digits is odd.

Now after 1, you have to have all digits 1-9 - nine digits. If you stop below 100 you will always have an odd number of digits (1-9 plus pairs from the two digit numbers). So you can work on digit parities to reduce the number of cases you have to consider.

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The answer is no - there's a variety of ways to prove this. For example, consider the number $k$ in the list with the most 0's, say $m$ of them. Clearly, $k$ must consist of a single digit followed by $m$ 0's, otherwise there would be a number before it with more 0's. Now we have two cases:

  • Case 1: $k$ is the only number with $m$ 0's. Then $k$ must be the middle number in the palindrome, otherwise $k$ does not have a unique number of 0's. However, the resulting concatenation of integers is not a palindrome since the number before $k$ , $(\ldots999)$, is not symmetric with the number after it: $(\ldots00001)$.
  • Case 2: There are multiple numbers with $m$ 0's. Consider the rightmost one, say $k=n\cdot10^m$ for some $n>1$. Then, $k$ is embedded in the palindrome between $((n-1)999\ldots)$ and $((n)00\ldots01)$. Since $k$ is the rightmost appearance of $m$ 0's, the leftmost appearance is thus $(10\ldots0(n))(00\ldots0(n))(99\ldots9(n-1))$. However, this is absurd since the left most appearance is naturally $10^m$.

Therefore, it's impossible for the concatenation of the $n$ integers $1$ through $n$ to be a palindrome.

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