I often wondered about the special properties of 9 in our number system. I wondered if there were similar special numbers in other systems. We count in base-10, which means there a 10 unique digits and we then use another place of digits once we get to 10. In other base systems, like base 16, and base 8, I wondered if there were similar numbers like 9 - or if 9 itself had these special properties.
As it turns out, it is not 9 itself that is special, but it is the number 1 less than the base system. So, for base 16, the magic number would be 15. Further, any factor of that number has similar special properties... so in this case 3 and 5 and in our base-10 system, that is why 3 has special properties as well.
In base-8, it would be 7 that has this special property.
So, I wrote a little proof that hopefully explains it. I wrote this years ago, and thought I blogged it, but I guess I didn't. So, here it is:
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.
QED
