Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫  π  -¹ ² ³ °

You are not logged in.

## #1 2006-03-13 18:26:01

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)

Offline

## #2 2006-03-13 20:00:49

MathsIsFun
Registered: 2005-01-21
Posts: 7,631

### 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

Offline

## #3 2006-03-13 20:13:54

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,812

### 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 is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

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

Offline

## #4 2006-03-13 20:17:28

MathsIsFun
Registered: 2005-01-21
Posts: 7,631