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**MajikWaffle****Member**- Registered: 2005-12-14
- Posts: 11

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)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,591

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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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,171

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.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,591

Oh, rats! Already solved!

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

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