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The follow-up article appears to give some hints though: > A number of readers emailed to say they had discovered that repeatedly adding up the digits of any o

The follow-up article appears to give some hints though: > A number of readers emailed to say they had discovered that repeatedly adding up the digits of any of the kernels of Kaprekar's operation always equalled 9 (...) Professor Nishiyama has provided an explanation why this happens: it is because the result of performing Kaprekar's operation on any number is a multiple of 9. https://plus.maths.org/content/pluschat-15 https://plus.maths.org/content/pluschat-15 This is starting to look very similar to "if you repeatedly add all the digits of an integer represented in base 10 representation until you have a single digit, and the result is 3, 6 or 9, then it is divisible by 3". I forgot the exact explanation for that one, but IIRC has to do with the implicit calculation that is embedded in base 10 positional notation, other bases have a different number you can quickly verify the divisibility of this way. So maybe that (the "implicit calculation in base 10 representation" thing) is one part of the explanation. I mean whatever it is, it feels like a mix of all these operations imposing constraints upon each other and interacting with the recursive feedback loop to result in the convergence as an emergent property.

Arnavion 3y ago
>This is starting to look very similar to "if you repeatedly add all the digits of an integer represented in base 10 representation until you have a single digit, and the result is 3, 6 or 9, then it is divisible by 3". I forgot the exact explanation for that one, but IIRC has to do with the implicit calculation that is embedded in base 10 positional notation Yes. The number "xyz" is 100x + 10y + z. Each power of 10 can be split into 1 plus a multiple of 9, ie (x + y + z) + (99x + 9y). The second group where all the components are a multiple of 9 is of course divisible by 9 and by 3. The first group is the sum of the digits. So if the sum of the digits is divisible by 3, then the original number was divisible by 3. If the sum of the digits is also divisible by 9, then the original number was also divisible by 9. If the sum of the digits is not divisible by 3, the original number was not divisible by 3 either. This generalizes to any number of digits, and to any base N for testing divisibility by N-1 or factors of N-1.

Arnavion 3y ago

>This is starting to look very similar to "if you repeatedly add all the digits of an integer represented in base 10 representation until you have a single digit, and the result is 3, 6 or 9, then it is divisible by 3". I forgot the exact explanation for that one, but IIRC has to do with the implicit calculation that is embedded in base 10 positional notation Yes. The number "xyz" is 100x + 10y + z. Each power of 10 can be split into 1 plus a multiple of 9, ie (x + y + z) + (99x + 9y). The second group where all the components are a multiple of 9 is of course divisible by 9 and by 3. The first group is the sum of the digits. So if the sum of the digits is divisible by 3, then the original number was divisible by 3. If the sum of the digits is also divisible by 9, then the original number was also divisible by 9. If the sum of the digits is not divisible by 3, the original number was not divisible by 3 either. This generalizes to any number of digits, and to any base N for testing divisibility by N-1 or factors of N-1.