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Can you concatenate the numbers from I to XVI in some order, such that the resulting 38-letter string is a palindrome?

This puzzle was inspired by a recent puzzle by Dmitry Kamenetsky.

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    $\begingroup$ Thanks for posting this. Here is an extension for you: for which $n$ can you concatenate the Roman numbers from 1 to $n$, such that the resulting string is a palindrome? $\endgroup$ Mar 24, 2022 at 2:00
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    $\begingroup$ Good question! I originally wanted to ask this for numbers I to XX. But that doesn't work so I changed it to 16. n > 16 is not possible. One constraint is that the number of I, V, X required must be even except at most one. But n=20 should be ok in that respect. What happens is that from n=17 you have 3 numbers containing "XV". The string must contain these in reverse. But "VX" is invalid in a roman number. So there must be a split after each "V". Now there are only 2 numbers ending in "V", so... $\endgroup$
    – Florian F
    Mar 24, 2022 at 7:19

1 Answer 1

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One such palindrome:

XIIIVIIIVIIXIIVXVIXXIVXVIIXIIVIIIVIIIX

Puzzle: stop here and decipher that before continuing.

XIII V III VII XI IV XV IX XIV XVI I XII VI II VIII X
13 5 3 7 11 4 15 9 14 16 1 12 6 2 8 10

I obtained this solution by finding shorter palindromes using subsets of the numbers.

Specifically, I started with 14 and 16:
XIV XVI <=> IV XV IX
Then worked with 8 and 13:
XIII V <=> VIII X
And resolved the rest, using 7 and 12 in a similar manner:
III VII XI <=> I XII VI II

Taking one half of each these to form the first half of the full palindrome yields the final result.

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