<s>[INST] Sum of digits when $99$ divides $n$ Let $n$ be a natural number such that $99\mid n$. Show that $S(n)\ge 18$.
It is clear that $9 \mid S(n)$, but I cannot apply the divisibility rule of $11$. Please help!
 [/INST] If $99|n$ then $9|n$ and $11|n$.
Two well known rules.  
1) If $9|n$ then $9|S(n)$.  
So $S(n) = 9, 18,27,.... $ etc.  So the the only way $S(n) \ge 18$ can be false is if $S(n) = 9$.
2) If $11|S(n)$ then the sum of the even positioned digits plus the sum of the odd positions digits are either equal, or off by a multiple of $11$.
So if $S_e$ the sum of the even positioned digits and $S_o$ is the sum of the odd positioned digits.  
Then $S(n) = S_o + S_e = \begin{cases}2S_o &\text{if } S_o = S_e \\ 2S_o +11k &\text{if } S_e = S_o + 11k\text{ for some }k \ge 1 \\ 2S_e + 11k &\text{if } S_o = S_e + 11k\text{ for some }k \ge 1\end{cases}$
So either $S(n)$ is even. Or $S(n) \ge 11$  And $S(n) = 9,18,27,....$ so $S(n) \ge 18$.
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Remains to prove those two basic rules.
You've seen proofs a million times, haven't you?
Well, here they are for the million + 1 time:
If $n = \sum\limits_{k=0}^m d_k 10^k$ where $d_k$ are the digits of $n$ then
1) $n = \sum\limits_{k=0}^m d_k (10^k - 1) + \sum\limits_{k=0}^m d_k$
$= \sum\limits_{k=0}^m d_k (10^k - 1) + S(n)$.
Now $10^k -1 = 9999....9$ and is divisible by $9$ so $\sum\limits_{k=0}^m d_k (10^k - 1)$ is divisible by $9$.
[Actually $(a^{k-1} + a^{k-2} + .... + a + 1)(a-1) = a^k - 1$ means $a-1$ always divides $a^k -1$ so $10-1=9$ always divides $10^k -1$; is a more sophisticated proof.]
So $9|n$ if and only if $9|S(n)$.
2) $n = \sum\limits_{k=0}^m d_k 10^k = \sum\limits_{k=0; k\text{ even}}^m d_k 10^k + \sum\limits_{k=0; k\text{ odd}}^m d_k 10^k$
$\sum\limits_{k=0; k\text{ even}}^m d_k (10^k -1) + \sum\limits_{k=0; k\text{ even}}^m d_k + \sum\limits_{k=0; k\text{ odd}}^m d_k (10^k + 1) -\sum\limits_{k=0; k\text{ odd}}^m d_k$
$= \sum\limits_{k=0; k\text{ even}}^m d_k (10^k -1)+ \sum\limits_{k=0; k\text{ odd}}^m d_k (10^k + 1)+S_e - S_o$.
Claim: $11|10^k - 1$ if $k$ is even and $11|10^k + 1$ if $k$ is odd.
If $k$ is even  then $(a^{k-1} - a^{k-2} + a^{k-3} -....+a^3-a^2 +a -1)(a + 1) = a^k - 1$ so $a+1|a^k - 1$ if $k$ is even.  So $10 + 1 = 11$ divides $10^k-1$ if $k$ is even.
If $k$ is odd then $(a^{k-1} - a^{k-2} + a^{k-3} -.... - a^3 +a^2 -a + 1)(a +1)=a^k + 1$ so $a+1|a^k + 1$ if $k$ is odd. So $10 + 1 = 11$ divides $10^k +1$ if $k$ is odd.
So $11|n$ if and only if $11$ divides $S_e - S_o$.  Or in other words if $S_o$ and $S_e$ have a difference of a multiple of $11$.
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