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#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 big_smile

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

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

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

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

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

ganesh
Moderator
Registered: 2005-06-28
Posts: 13,490

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! up smile


Character is who you are when no one is looking.

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

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

Re: An Algebraic Problem

Oh, rats! Already solved!


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

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