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I recently came across a problem in a maths contest that stated the following: "The center of a natural number is the sum of four times its units digit and the number you get when you remove that digit. For each natural number n there's a series that starts with n and each element is the center of the previous one. For example, the series that starts on 2018 goes like this: 2018, 233, 35, 23, 14, 17... A number n is said to be centered if the series that begins with n has any element that has itself as its center. Determine every centered number. Justify your answer"

After calculating the center of the first few numbers, I realized that the only numbers that where their center were 13, 26 and 39 (1 + 3*4 = 13; 2 + 6*4 = 26; 3 + 9*4 = 39). I then noticed that the center of any multiple of 13 was another multiple of 13, and that if the number was larger than 39, the resulting number would be smaller than itself. That made me conclude that that all multiples of 13 are centered numbers.

My hypothesis is that the set of all centered numbers is the set of all multiples of 13. I would like to know how to analytically prove this.

ArmindoFlores
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The center of the number $n=10a+b$, $a$ a positive integer, $0\le b\le9$, is $m=a+4b$ (by my reverse engineering of your example sequence). We have that $n+3m=13a+13b$ is always a multiple of $13$. This implies that $m$ is a multiple of $13$ if and only if $n$ is. The claim follows from this and your list of self-centered numbers. After all, we always have $m\le n$ and $m<n$ unless $n$ is self-centered.

Jyrki Lahtonen
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  • The reason I got this one pretty quickly is that the transformation $10a+b\mapsto a+4b$ is a known "rule for divisibility by $13$". When facing the task of determining whether a given integer is divisible by $13$ you can perform this transformation as many times as you can (always replacing the integer with a smaller one). Then the number you started with is divisible by thirteen if and only if the last number is. – Jyrki Lahtonen Jan 11 '18 at 22:38
  • There is a slew of divisibility tests like this. See this thread for the most general version I'm aware of. – Jyrki Lahtonen Jan 11 '18 at 22:40
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You are converting the number $10a+b$ to $a+4b$. Let's see what happens to their remainder $\pmod {13}$: $a+4b=4(10a+b)-39a\equiv 4(10a+b)\pmod{13}$. This means, every time we make one step, we multiply the remainder (in division by $13$) by $4$. This means:

  • If the remainder is $0$, the remainder stays $0$. Thus a number already divisible by $13$ stays divisible by $13$. Such a number has a chance to have itself as a center and you have already figured out which of those numbers do.
  • If the remainder is $\ne 0$, the remainder stays $\ne 0$, but otherwise changes, because $4x\equiv x\pmod{13}$, i.e. $3x\equiv 0\pmod{13}$ implies $x\equiv 0\pmod{13}$ due to $13\not\mid 3$. Such a number obviously cannot have itself as a center.