1
$\begingroup$

The problem is:

Johnny is at his local news-stand, looking at the various magazines on display. He picked up three magazines, $A$, $B$ and $C$. He passes the magazines to the shopkeeper, who enters the amounts of each magazine into the cash register.

'Hang on a minute!' says Johnny. 'You just pressed the multiplication button each time between amounts instead of the addition button.' The shopkeeper smiles and replies 'It doesn't matter. Either way, it comes to $£5.70$.'

What were the prices of the magazines?

Letting $a, b$ and $c$ be the prices of each magazine respectively, we obviously have $$abc = a + b + c = 5.70$$

So we have, $abc = a + b +c$, $abc = 5.70$ and $a + b + c = 5.70$ - but I'm not quite sure what to do with these equations. I did try:

$$a = \frac{b+c}{bc-1} \implies \frac{bc(b+c)}{bc-1} = 5.70$$

So $$bc(b+c) = 5.70bc-5.70 \implies b(b+c) = 5.70b - \frac{5.70}{c} \implies b^2+c(c - 5.70) + \frac{5.70}{c} = 0$$

This gets horrid though, so I don't think I'm on the right track.

$\endgroup$
  • $\begingroup$ Using the first one, you can write $a = \frac{b+c}{bc-1}$, which you can subsitute in the next equation, and thus you can write $b$ as function of $c$, which you can use to write $a$ as function of only $c$. Subsituting in the last equation yields a result for $c$. $\endgroup$ – Hetebrij Sep 11 '15 at 21:27
  • $\begingroup$ @Hetebrij - I tried and updated the OP to reflect my attempt. Still getting nowhere. $\endgroup$ – Zain Patel Sep 11 '15 at 21:38
1
$\begingroup$

Let's express all prices in cents, so that we have to deal with positive integer numbers only. Then your equations become: $$ \cases{ a+b+c=570 & \cr abc = 5\,700\,000 } $$ Decomposing $5\,700\,000$ into prime factors gets $2^5\cdot3\cdot5^5\cdot19$, so that now it is only a matter of trying to combine these factors into three numbers so that their sum be $570$. After some trial and error one ends up with $125$, $160$ and $285$, which is then the solution.

$\endgroup$
  • $\begingroup$ Beautiful and elegant! $\endgroup$ – Zain Patel Sep 11 '15 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.