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I read about this in Parc, S., “50 Visions of Mathematics”, Oxford University Press, (2014) and I wondered if this was true for any number base rather than just base 10.
On trying three digit numbers in octal and hexadecimal I found that they each had an equivalent kernel. This led me to solve the equations for finding the kernel, but using a variable, β, for the number base rather than restricting it to decimal numbers.

The results were as follows:
For two digit numbers there is a kernel if the base is of the form 3*(n – 1) + 5
For three digit numbers there is a kernel if the base is even i.e. of the form 2*n
For four digit numbers there is a kernel if the base is of the form 5*n, though there are kernels for base 2 and base 4.
For five digit numbers there are kernels for bases of the form 3*n

As I was using the computer algebra add-on for Microsoft Word I chose not to solve individually the hundreds of equations for six digit numbers. However, solving the equations based on the base-ten six digit numbers I did find that there are kernels for bases of the form 9*n + 8, 2*n, 15*n + 10, and 2*(n + 2).
In the above n takes the values 0, 1, 2, ...

I haven’t a proof, but it does seem likely that for any value with 3*n digits and an even base, there is a kernel which has n digits of value β – 1, β/2, and β/2 – 1. This is true for three digit numbers, six digit numbers, and some nine and twelve digit numerical examples I’ve tried.

I’ve not seen any references to this so I thought I would post these results as they may be of interest to other readers.

Philip Hickin

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