Palindromic Repetitions

Arithmetic, algebra, number theory, sequence and series, analysis, ...
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Palindromic Repetitions

Post by jnmcd »

A palindrome is a number that is the same way forward as it is backwards.
For example, 22 is a palindrome.
Clearly, however, 23 is not.
But by adding 23 reversed on to itself (32), you get 55, which is a palindrome.
That took 1 repetition.
With 64, 1 repetition gives you 110, and since that isn't a palindrome, you must do another repetition. You'll end up with 121, which is a palindrome.
That took 2 repetitions.
So my question is this:
What is the lowest number to require more than 25 repetitions to become a palindrome? (Yes, brute force will work)

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Re: Palindromic Repetitions

Post by pj6444 »

Should we assume that all numbers will create a palindrome eventually? What if the number is already a palindrome to begin with?

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Re: Palindromic Repetitions

Post by NChaloult »

In the case of this problem, it would not matter if the theory that all numbers eventually arrive at a palindrome is later proven to be false, for we are only looking for the first one that takes more than 25 repetitions. Also, if a number already is a palindrome, then did it not take 0 repetitions to arrive at a palindrome? This is simply my thought process, at least...

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Re: Palindromic Repetitions

Post by jaap »

See also: ... cture.html
(Note: contains spoiler for the problem in the OP)

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Re: Palindromic Repetitions

Post by TripleM »

jaap wrote:(Note: contains spoiler for the problem in the OP)
I'm not sure it does; it contains a number which isn't known to ever become a palindrome, but we're looking for a number which does become a palindrome, but in more than 25 steps.

There is a smallest initial value *known* to take more than 25 steps, though it can't be proven it is the smallest. (I believe that's what pj6444 was getting at.)

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