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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.

Bumblebee
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Antoni Parellada
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4 Answers4

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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, $C=2$.

lhf
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$9n=18043+100C$ 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.

Doug M
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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 $7 + C = 9$

Steven Alexis Gregory
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$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".

fleablood
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