# Word problem involving equality of multiplication and addition

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.

• 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 $c$. – Hetebrij Sep 11 '15 at 21:27
• @Hetebrij - I tried and updated the OP to reflect my attempt. Still getting nowhere. – Zain Patel Sep 11 '15 at 21:38

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.