What Were the Individual Costs of the Three Items at 7-11?

[ceptimus]

I went into a 7-11 store and chose three items. I was charged $7.11

Intrigued that the cost was the same as the name of the store, I asked how the total was calculated. The checkout operator told me she simply multiplied together the price of the three items. Outraged, I insisted that she add the three prices rather than multiply them, but the total was still $7.11

What were the individual costs of the three items?

 

[BobG]

Solution:

This is a little ugly. I created a triangle with a base of 100 and modified the other two sides until the sum of the tangents was about 7.11 (length wound up being 145.7 and 156.3). For any triangle, the product of the tangents is equal to the sum of the tangents.

Unfortunately, this only got me close (I think arbitrarily setting the base hurt my accuracy). Then I just played around with each price until it worked out: $2.17, $4.15, and $0.79.
[/color]

 

[ceptimus]

Well that's very close, and I like the method.

Edit: I misremembered the puzzle. The original one had the same question, but with four items. There is an exact solution to that one. I suppose that Bob got as close as possible to an exact solution with the corrupted version I posted.

 

[dextercioby]

Let the 3 unknowns be:a,b,c.We know that:
$$abc=a+b+c=7.11$$
Solving for a,b,c assumes that one of the 3 variables is to be taken as a parameter.Let that be "a".We impose upon "a" to satisfy:
$$a>0$$.It should remain true for "c" and "b" as well.
Then:$$b=\frac{7}{ac}$$(*)
$$a+\frac{7.11}{ac}+c=7.11\Rightarrow a^{2}c+7.11+ac^{2}=7.11ac$$.
Solving for c yields 2 solutions:
$$c_{1,2}=\frac{7.11a-a^{2}\pm\sqrt{(a^{2}-7.11a)^{2}-28.44}}{2}$$(**)

The relation (**) yields conditions on "a",beside the one of being positive.
1)$$(a^{2}-7.11)^{2}\geq 28.44\Rightarrow i)a\geq \sqrt{7.11+\sqrt{28.44}}\sim 3.57;ii)a\leq \sqrt{7.11-\sqrt{28.44}}\sim 1.29$$
We conclude $$a\in (0,1.29]\cup [3.57,7.11)$$
Then it has to be solved this set of inequations:
$$0<{7.11a-a^{2}\pm\sqrt{(a^{2}-7.11a)^{2}-28.44}}<14.22$$

And then from these set of solutions another condition for "b":
$$0<\frac{7.11}{ac}<7.11$$

The trick,really brain teaser,is to ask of these solutions,algebraic numbers,to have maximum 2 decimals,as fractions of cents (0.01$) do not exist.

The trick is:apparently the number of triplets is infinite.For numbers with infinite number of decimals.

Daniel.

 

[dextercioby]

To prove my logic right,chose "b=1$".Then u can find two possible solutions:
$$a\sim 4.495,c\sim 1.605;a\sim 1.565,c\sim 4.535$$
I'll let u compute for $$b=1.01;b=1.02;b=1.03;...$$

Daniel.

 

[BobG]

There's more than one solution.

How about:

$3.21+$3.21+$0.69[/color]

I should have gotten this answer first, since it's the answer to a possible version of a completely different brain teaser ("Going off on a tangent..." back in July).

The trick,really brain teaser,is to ask of these solutions,algebraic numbers,to have maximum 2 decimals,as fractions of cents (0.01$) do not exist.

I'm doing these on a slide rule. I only have 3 significant digits, max.

 

[Gokul43201]

The problem isn't that nuch of a teaser unless you want exact solutions, in which case you have diophantine equations in the costs of the items measured in cents.

But my brain is on holiday right now, so... no solving these now.

 

[MattTuc13]

What about sales tax?!

 

[Brickster]

Matt, that made me laugh so hard. Must live in Alberta, Canada, no sales tax there.