Largest Digit Sum in bases besides base 10 So I know that in base 10, the largest digit sum a number N can have will occur when all the digits are equal to 9. 
For example, 999 adds to 27, which adds to 9. 9999 adds to 36, which adds to 9 as well. 
My question is, does this same logic apply to other bases, such as base 2 or base 3? My initial thought is that it does not because if you look at 1111 in base 2, this sums to 4? 
 A: I think the term you're looking for is "digital root." Here's the Mathworld page about it: http://mathworld.wolfram.com/DigitalRoot.html On the one hand, the content is strictly limited to base $10$. But I don't have to worry about some unaccountable pedant or prankster messing up that page.
Call the base of numeration $b$. Then the smallest $n$-digit number in base $b$ is $b^{n - 1}$, for $n \geq 1$. The largest number with $n - 1$ digits is then of course $b^{n - 1} - 1$, and its base $b$ representation consists entirely of whatever digit corresponds to $b - 1$.
The digital roots repeat in predictably periodic cycles. The digital root of $1$ is obviously $1$. The digital root of $b - 1$ is obviously $b - 1$. Then, since $b$ is represented as $10$, its digital root is $1$, and the digital root of $b + 1$ is $2$ (as long as $b > 2$).
From this it follows that the digital root of $b^n$ is $1$, the digital root of $b^n + 1$ is $2$ and the digital root of $b^n - 1$ is $b - 1$.
Here's a concrete example in octal that you can probably check on your computer's calculator: $7$ is $7$, $63$ is $77$, $511$ is $777$, $4095$ is $7777$, etc.
A: Well in base $b$ the highest digit is $b-1$ so the highest digit sum will occur when all the digits are $b-1$.  So the highest sum you can have for an $n$ digit number is $(b-1)(b-1)...(b-1)$ and the sum is $(b-1) + .... = n(b-1)$. 
You have a second statement that the sum of those digits is $9$.  That's mostly true.  But consider $99,999,999,999$.  The sum of those digits is $99$.  And the sum of those is $18$ and not $9$.  But the sum of the digits of $18$ is $1+8=9$ so we do get to $9$ eventually after many steps.
And this is true of base $b$.
A number $N = \sum_{i=0} a_i b^i$ and the sum of the digits of $N$ is $\sum_{i=0} a_i$.  So if you subtract the two you get:
$\sum_{i=1} a_i (b^i - 1)$.  And $b^i - 1 = (b-1)\times(b^{i-1} + b^{i-2} + ... + b + 1)$.  So the difference between the number $N$ and sum of the digits is:
$\sum_{i=1} a_i(b-1)(\sum_{k=0}^{i-1}b^k) = (b-1)[\sum_{i=1} a_i(\sum_{k=0}^{i-1}b^k)]$ which is a multiple of $b-1$.
So $N$ is a multiple of $b-1$ if and only if the sum of the digits are.
Now it might not happen on your first step that you get $b-1$ but it will happen eventually.
Consider $1111_2 = 15_{10}$.  The sum of those digits is $4$.  And $4$ is a multiple of $2-1$.  In base $2$, we have $4_{10} = 100$ and then sum of those is $1 + 0 + 0 =1$.  So we are done.  But it took us $2$ steps.
Consider base $4$ as a better example:  $33333$ has the sum of the digitis is $3+3+3+3+3 = 15_{10} = 3*4 + 3 = 33_{4}$.  And the sum of the digits of $33$ is $3+3 = 6_{10} = 4 + 2 = 12_4$.  ANd the sum of the digits of $12_4$ is $1+2 = 3$.
Or base $3$.  $222222_3$ will have sum $2+2+2+2+2+2 = 12_{10} = 9 + 3 + 1 = 110_3$ and then sum of the digits of $110$ are $1+1+0 = 2$
This WILL always work.
