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#1 2006-03-14 17:26:01

MajikWaffle
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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 (2006-03-14 17:30:02)

#2 2006-03-14 19:00:49

MathsIsFun
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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 × 106

And the prime factors of 711 × 106 are 26  32  56  79

Trial and error from here?


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

#3 2006-03-14 19:13:54

ganesh
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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.

#4 2006-03-14 19:17:28

MathsIsFun
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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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