Figuring out a digit knowing that the number is a multiple of 9 I am trying to help my kid with a homework problem, and he's insisting that the only way to solve is trial and error. I want to know if there is a more systematic way - it sounds as though the teacher may have taught them to do it by guessing and verifying, but it has stoked my curiosity.

The cost of each item is 9 dollars. We know that some customer paid $18C43$,
  but the figure C is blurred out. How much did we pay? and How many
  items were purchased?

We can unfold $18C43 = 18043 + 100C$. And we know that $9n = 18043 + 100C$. But we end with have two unknowns in one equation. Of course we know that both C and n are natural numbers.
 A: Hint: The amount paid is a multiple of $9$. Recall casting out nines...
Solution:
Casting out nines in $18C43$ leaves $C+7$. This must be a multiple of $9$. Since $0 \le C \le 9$, we have $2 \le C+7 \le 16$, and so the only solution is $C+7=9$, that is, $C=2$.
A: $9n=18043+100C$
divide through by $9.$  
We really only care about the remainders.
$18043/9 = (2004) r 7$
$100/ 9 = (11) r 1$
9 must divide $(7 + C)$
This approach will work for any divisor.  But 9's are special.
If the sum of the digits is divisible by 9, then the number is divisible by 9.
A: keep casting out nines until you can't any more...
$18C43 \to 1+8+C+4+3 \to 16+C \to 7+C$
Then you need to have $7+C = 9$. 
So $C = 2$
Why does this work?
$$18C43 =
\left\{ \begin{array}{rlo}
   10000 &= 1 + 1 \cdot 9999\\
   +8000 &= 8 + 8 \cdot 999\\
    +C00 &= C + C \cdot 99 \\
     +40 &= 4 + 4 \cdot 9 \\
      +3 &= 3
\end{array} \right.
$$
 If you examine this, you see that 
\begin{align}
    18C43 
    &= 1+8+C+4+3 + (\text{a multiple of 9}) \\
    &= 16+C + (\text{a multiple of 9}) \\
    &= 1+6+C + (\text{another multiple of 9}) \\
    &= 7+C + (\text{another multiple of 9}) \\
\end{align}
So, if $18C43 $ has to be a multiple of $9$, then $7+C$ has to be a multiple of $9$.
The only choice you have is $7 + C = 9$
A: $9n =18043 + 99C +C $
$n=2004 \frac 79 + 11C + \frac C9$
So $\frac 79+\frac C9=\frac {7+C}9$ is a whole number.  $C = 0....9$ so $C+7=7,....16$.  So the only possible answer is $C+7= 9$ and $C=2$
So $9n=18243$
$n=2027$
(And $2027 =2004 \frac 79+22+\frac 29$)
===
But there's also $18C43\implies 1+8+C+4+3=16+C\implies 1+6+C=7+C\implies 7+C=9$ so $C=2$ because, as everyone knows, "a number is divisible by 9 if and only if the sum of ots digits are divisible by 9".
But I have mixed feelings about teaching that to kids.  It'It's trick that kids like.  And a really useful one.  But I don't like teaching "magic".
