Prove that a two digit number is divisible by the sum of its digits if it is divisible by $9$. Prove that a two-digit number with non-zero digits (except $99$) is divisible by the sum of its digits if the number is divisible by $9$.
 A: It is easy to prove that if a number is divisible by $9$ then sum of its digits will also be divisible by $9$, you can do it that way:
Suppose the two digit number about which we are talking is of the form $10a+b$. As given $9|10a+b\implies 9|10a+b+9b\implies9|10(a+b)\implies9|a+b$.
It is clear that Sum of two digits can never be greater than $18$ (maximum being $9+9$ in case of $99$). So either sum of digits is $9$ or $18$, $18$ in the case of $99$ which is the only exception. And for all other pairs of digits $(a,b)$ the sum of digits $(a+b)$ is going to be $9$. 
As $9|10a+b \implies 9=a+b|10a+b$, and we are done.
A: Let the two-digit number be $10a+b$ for $a,b \in [0,9] \cap \Bbb{N}$. We know that $9 \mid (10a+b)$. We also know that $9 \mid 9a$, so we get:
$$9 \mid ((10a+b)-9a) \implies 9 \mid (a+b)$$
Now, since $a,b \in [0,9] \cap \Bbb{N}$, we have $0 \leq a+b \leq 18$. Now that we know $9 \mid (a+b)$, we know that $a+b \in \{0,9,18\}$.
Case 1: $a+b=0$: The only way this can happen is if $a=b=0$, meaning $10a+b=0$, which is not a two-digit number, so we don't need to prove this case.
Case 2: $a+b=9$: We have $9 \mid (10a+b) \implies (a+b) \mid (10a+b)$.
Case 3: $a+b=18$: This can only happen if $a=b=9$, which means $10a+b=99$. This is stated as an exception in the theorem, so we don't need to prove this case.
Thus, we have proven all of the cases we needed to prove for this theorem, so we are done.
A: $$(9)(1)=9$$
To get to $9(2)$ we add $9$ to $9(1)$, to get to $9k$ we add $9$ to $9(k-1)$. Or in other words we add $10-1$ subtracting $1$ from the units digit then adding $1$ to the tens digit. Subtracting $1$ then adding $1$ preserves the digit sum of $9$ because $x+1-1=x$, when assuming no carrying over complications happen. The only trouble we get to is when we run into a $0$ in the units place, then we have to do some carrying over. This occurs for the first time when we want to add $10-1$ to $9(10)$ (it's easy to check it does not occur at $9(5)$). However we are good before then, so it follows that the digit sum for $9(1),9(2),....9(10)$ are all $9$. And our numbers are all multiples of $9$, so they are divisible by their digit sum.
