Math Is Fun Forum

MajikWaffle

Member

Registered: 2005-12-14

Posts: 11

An Algebraic Problem

The cost of four items was $7.11 whether you added or multiplied the individual items. What was the cost of each item.

Our problem of the week for precalc. Problem is whether there is a set number of solutions. So help me out if you can please

Last edited by MajikWaffle (2006-03-13 18:30:02)

MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: An Algebraic Problem

abcd = a+b+c+d = 7.11

We can also take advantage of the fact that (I assume) there are no items that have fractional cents.

In cents:

a+b+c+d = 711
abcd = 711 × 10[sup]6[/sup]

And the prime factors of 711 × 10[sup]6[/sup] are 2[sup]6[/sup]  × 3[sup]2[/sup]  × 5[sup]6[/sup]  × 79

Trial and error from here?

---

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,424

Re: An Algebraic Problem

I've heard of this problem before. All I did is used google and this is the page I got.
Majikwaffle, I think this solves your problem!

---

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: An Algebraic Problem

Oh, rats! Already solved!

---

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman