a short mathematical proof

I wrote this up a while ago and thought I'd post it for someone to stumble across...

I think it is valid.

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proof of base-1 properties

Let n be the base of the numbering system being used.

Theorem:
Any number is divisible by (n-1) if its digits sum up to a number divisible by (n-1). This property can be recursively applied to the sum of the digits of the result until that result is simple enough to determine whether it is divisible by (n-1).

Proof:
Starting with 0, add (n-1). You get (n-1). The sum of the digit is also (n-1) which is certainly divisible by (n-1). From that point, every time you add another (n-1) the "tens" place will increment by 1 and you would subtract 1 from the "ones" place. This keeps the sum total of the digits equal and therefore still divisble by (n-1).
When a zero is once again in the "ones" place (occurs every n times), the addition of another (n-1) will not increment the "tens" place, however, it has just added (n-1) to the sum total of the digits. Therefore the sum of the digits continues to be divisible by (n-1).
Therefore, if the sum of the digits of any number is divisible by (n-1), the number itself is divisible by (n-1). In this proof we covered all possible multipliers of (n-1), so we can also say the converse is true. For any number that is divisible by (n-1), the sum of its digits is also divisible by (n-1).

Corollary:
Any factor of (n-1) inherits this property.

Proof:
Let f be a factor of (n-1) and m by the multiplier of that factor (for example, if n=15 and f=3 then m would equal 5). Again, start with 0, every time f is added to a number one of two cases exist:

If the "ones" place of the number is < (n-f) then
we will not be incrementing the "tens" place and the sum of the digits will be increased by f. Therefore, the sum of the digits remains divisible by f.

If the "ones" place of the number is >= (n-f) then
1 will be added to the "tens" place and this place will be decremented by (n-f). The sum of the digits is -(n-f) + 1 which equals -( (n-1) - f). Since f is a factor of (n-1) with m as its multiplier, this is the same as -((m*f) - f) or - ((m-1)*f). We are subtracting multiples of f from the sum of the digits. Therefore, the sum of the digits remains divisible by f.