7-11

by

Shanti Howard & Audrea Bankston

7-11 Problem*

A guy walks into a 7-11 store and selects four items to buy. The clerk at the counter informs the gentleman that the total cost of the four items is \$7.11. He was completely surprised that the cost was the same as the name of the store. The clerk informed the man that he simply multiplied the cost of each item and arrived at the total. The customer calmly informed the clerk that the items should be added and not multiplied. The clerk then added the items together and informed the customer that the total was still exactly \$7.11.

What are the exact costs of each item?

Note: I interpret the "exactly" and "exact" to mean just that -- not "to the nearest penny." Is there a solution with this strict
interpretatin of "exactly?"

Solution:

The easiest thing that came of this problem was the fact that I realized I really could add and subtract, multiply and divide. Doing both adding and subtracting within the same realm of a number besides 2+2 and 2*2 both being equal to 4...no one else has ever given me this challenge...and I still do not know if our answer is correct...

First:

We decided to find the factors of 711 (keep in mind that we are not working with decimals when we begin this....I think they would be an impediment. The factors of 711 are:

711 = (3)(3)(79)

Second:

I had to imagine myself actually going into this 7-11, picking up some items, going to the counter, and then having this weird clerk tell me he multiplied my items....how could I not go crazy and tell him off? How could I keep my sanity? Then, to top it all off, I would find out that my purchase, even after adding, was still \$7.11!!! I would be nuts!!!! How could this be? Where would I or could I begin?

This problem was driving me crazy!!! Shae said there was some sort of answer to this problem, but we had to probe into it in order to get the answer.

Shae also said that it (the product of the numbers to be multiplied) must have to do with the divisibility of the numbers 3 and 79. She also added that we might want to play around with the multiples of our factors (3 and 79). For instance, the easiest combination of numbers that we came up with are \$7.11 = (3)(3)(.79)(.32). I got the numbers from the factors, however, the last number was the difference between the sum of the factors and \$7.11.

Third:

Now, when we multiplied these (the above) numbers together, the product was 2.2752. This could not be right...the first three, remember, were the factors, so there's no way! We were not thinking straight here. (This is also a good time to bring up to my kids that they need to think before they say, think before they write, and think before they do). I decided to take the long route and start with the above numbers to see where it would take us. Here is what my list looks like:
 Product Sum Product Total .79 3 3 .32 7.11 2.2752 .79 2 2 2.32 7.11 7.3312 .79 2 2.16 2.16 7.11 7.371648 .79 2.08 2.08 2.16 7.11 7.38256896 .79 2.08 2.12 2.12 7.11 7.38519808 .79 1.50 2.70 2.12 7.11 6.78294 .79 1.50 2.82 2 7.11 6.6834 .79 1.60 2.72 2 7.11 6.87616 .79 1.70 2.62 2 7.11 7.03732 .79 1.80 2.52 2 7.11 7.16688 .79 1.71 2.61 2 7.11 7.051698 .79 1.79 2.53 2 7.11 7.155346 .79 1.77 2.55 2 7.11 7.13133 .79 1.76 2.56 2 7.11 7.118848 .79 1.75 2.57 2 7.11 7.10605

Also, we had to reread the problem more than one time because something stood out that we just could not shake, the problem stated:

Note: I interpret the "exactly" and "exact" to mean just that -- not "to the nearest penny." Is there a solution with this strict
interpretation of "exactly?"

Fourth:

If the problem says "exact" and "exactly"...and he means just that...not to the nearest cent, then this throws out our guesses of :

(.79)(1.76)(2.56)(2) = 7.118848

and

(.79)(1.75)(2.57)(2) = 7.10605

I do not see how it could not mean to the nearest cent. In the first and second case, maybe the clerk was not all that smart when he was doing the multiplication, since he did not realize he had to add to get the total price. Maybe, he "thought" it was okay to to say \$7.11 in either case - maybe he knows how to round up the hundredths place when there is a zero in there, but he does not know how if there is another number there. I am not sure...I must come back to this problem...maybe I will go back and look at the multiples of the factors 3 and 79...hopefully it will lead me down another shakey path.

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