Overpaying your bill

Question

(Please tell me exactly how you found the answer and what you did because I'm a Year 7 and this is a Year 9 question.)

Mr. Rich decides to pay a bill using internet banking. The bill is a two-digit whole number $ab$. He accidentally inserts an extra digit after the b. Later he finds out he overpaid his bill by $647. What was his original bill?

I started working it out like this

$(100a+10b+c)-(10a+b)=9(10a+b)+c$

But then I got stuck. What I don't understand is how you're meant to proceed. Are you meant to do anything with the $c$? Plus, are the digits different by placing two different pro numerals or are they the same? Can anyone help me?

Answer 1

You are already on the right way:

Actual bill: $10a+b$.

Actually Paid: $100a + 10b+c$

Overpaid: $647=(100a+10b+c)-(10a+b)=90a+9b+c=647$

With the last equation and with the knowledge that $a,b,c\in\{0,1,\ldots,9\}$ (i.e. they can be only integers from 0 to 9) you can solve this.

Hint: Try e.g. $a=2$ and see that in this case either $b$ or $c$ need to be greater than 9.

Answer 2

$(100a+10b+c)-(10a+b)=647$.

$90a+9b+c=647$.

There are three unknowns and one equation. We also require integer solutions from the field $\Bbb Z_{10}$.