ok so this is what I came up with:
is even.
assume that it's true for k=a, we wish to prove it is true for a+1:
is a multiple of
because
is even.
That means our number can be written as:
As required.
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This was inspired from the proof that
always ends in
for natural
in base 10. (the case for n=10):
First term is an even multiple of half of ten, meaning its a multiple of 10, hence the addition of 6 makes it the last digit. Generalizing this:
if
The first and third term obviously end in 0. The middle term is an even multiple of half of ten, meaning that it's a multiple of 10 so it too ends in 0, hence the addition of 6 makes the last digit 6.
This is analogous to the more generalized version posted originally, which is just an extension of this proof to bases other than 10.(that are only divisible by 2 once)