Ring Me Up!

September 5, 2014 at 7:11 am 2 comments

Cash RegisterWhen my college roommate contracted crabs, he went to CVS to buy some lice cream. As you can imagine, he didn’t want to announce to the world what he was buying or why, so he put the box on the counter with a notepad, a bottle of aspirin, a pack of cigarettes, a bag of M&M’s, and a tube of toothpaste — hoping the cream would blend in. The attractive co-ed clerk at the register rang him up without a second look.

As he walked out of the drug store thinking he had gotten away with it, he opened the cigarettes, put one to his lips, and realized he had nothing with which to light it. He returned to the checkout and asked the clerk for a pack of matches.

“Why?” she asked. “If the cream doesn’t work, you gonna burn ’em off?”

Ouch.

My luck with clerks wasn’t much better. At a grocery store, I placed a bar of soap, a container of milk, two boxes of cereal, and a frozen dinner on the check-out counter. The girl at the cash register asked, “Are you single?”

I looked at my items-to-be-purchased. “Pretty obvious, huh?”

“Sure is,” she replied. “You’re a very unattractive man.”

I did, however, have an exceptional experience at a convenience store. This is what happened.

I walked into a 7-11 and took four items to the cash register. The clerk informed me that the register was broken, but she said she could figure the total using her calculator. The clerk then proceeded to multiply the prices together and declared that the total was $7.11. Although I knew the prices should have been added, not multiplied, I said nothing — as it turns out, the result would have been $7.11 whether the four prices were added or multiplied.

There was no sales tax. What was the cost of each item?

As you might have guessed, that story is completely false. (The one about me being called ‘unattractive’ is a slight exaggeration. The one about my roommate, sadly, is 100% true.) The truth is that I learned this problem from other instructors when teaching at a gifted summer camp.

It may not be true. It is, however, one helluva great problem.

But it has always bothered me that the problem is so difficult. I’ve always wanted a simpler version, so that every student could have an entry point. Today, I spent some time creating a few.

Use the same set-up for each problem below… walk into a store… take some items to check-out counter… multiply instead of add… same total either way. The only difference is the number of items purchased and the total cost.

I’ve tried to rank the problems by level of difficulty. Below, I’ve given some additional explanation — but not the answers… you’ll have to figure them out on your own.

  • (trivial) Two items, $4.00.
  • (easy) Two items, $4.50.
  • (fun) Two items, $102.01.
  • (systematic) Two items, $8.41.
  • (perfect) Three items, $6.00.
  • (tough) Three items, $6.42.
  • (rough) Three items, $5.61.
  • (insane) Four items, $6.44.
  • (the one that started it all) Four items, $7.11.

Editor’s Notes

trivial — C’mon, now… even my seven-year-old sons figured this one out!

easy, fun, systematic — All of these are systems of two equations in two variables. Should be simple enough for anyone who’s studied basic algebra. All others can use guess-and-check.

perfect — Almost as easy as trivial, and the name is a hint.

tough — But not too tough. Finding one of the prices should be fairly easy. Once you have that, what’s left reduces to a system of equations in two variables.

rough — Much tougher than tough. None of the prices are easy to find in this one.

insane — Gridiculously hard, so how ’bout a hint? Okay. Each item has a unique price under $2.00. If you use brute force and try every possibility, that’s only about 1.5 billion combinations. Shouldn’t take too long to get through all of them…

the one that started it all — As tough as insane, and not for the faint of heart. But no hint this time. Good luck!

 

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2 Comments Add your own

  • 1. Patti Sewald  |  January 18, 2015 at 9:25 pm

    I’ve solved that 7-11 several ways now. The method that was most helpful was to prime factor 711,000,000.

    711\,000\,000 = 2^6 \times 5^6 \times 3^2 \times 79

    That gives a place to start because one of the prices has to be a multiple of 79 cents.

    Reply
    • 2. venneblock  |  January 20, 2015 at 9:18 pm

      Nice work, Patti! Even realizing that you need to factor, though, it’s still not easy. Good for you if you found the solution.

      Reply

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The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

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