Question

Dear Mitch,

I oversee the math curriculum at a k-8 school in Minnesota and over the last few years I have been using the internet to find most of the interesting problems that engage students in the way the math teachers in my department find most conducive to teaching for true understanding.

Somehow, the textbook companies seem to be lagging behind and present concepts in the same dry manner that hasn't worked too well for years.

This is about the point in your replies where you sometimes write "enough preamble", which always makes me smile. (Do you realize that most people never use that word and have to look it up when studying the United States Constitution?)

So, enough preamble. My question is simple. I've noticed that you try to cover as many of the holidays as you can without ignoring the particular theme you've been covering (SAT questions, for a while, preceded by PSAT questions, and now, I've been seeing you cover that very important area of "Problem-solving" for students working towards their college application, which I have collected and will go over them with my daughter when her time comes.) So, now, would you be kind enough to start early (for you, anyway!) and provide some Valentine's Day problems? Just a thought, but I can tell you that I've been searching the internet, and I just haven't found the kinds of problems that feel new and different, the way ALL your problems always manage to be!

Thank You for contributing what you contribute to our math community; I for one look forward to reading your answers every week.

Marilyn Cst.

Answer

Dear Ms. Cst.,

It would be my pleasure to begin St. Valentine's Day Math on the 'early side'. That's the 'good news'. The not-so-good news, at least from the way I'm reading your letter, is that while it is true that I almost always enjoy coming up with my own questions that no one has heard before (because I do my best when I come up with them for the first time as direct responses to people' questions), this time I am going to do the opposite.

WHAT??

Here's the situation. A few evenings ago, I was at a dinner party when someone asked me what I thought was the most interesting math problem/brain twister I had ever heard. And, guess what? For the first time I can recall, I was 'speechless'. Why? Because I happen to have many favorites. In fact, when I got home, I found myself scribbling down a key word or two that would be enough to remind me of each problem I'd put on my list of favorites. And I've been adding to the list as I recall more and more. I ended up crossing off a few that didn't meet my usual criteria for that category, or just seemed a little less exciting. Right now, I have the list narrowed down to 387 favorites. And while some of these are ones I have come up with over the years, MANY are such classics that everyone ought to have a chance to hear them at least once. Like a comedian I used to know often said, "No matter how 'old' a joke is, if a person hasn't heard it before, then it's not old to him" (or her).

So, before I start writing out all the cool new problems I have scribbled on the napkin next to my keypad in front of me, ones that I'm excited to share with you in the next couple of weeks, I have been bursting to give my Valentine's day twist to a problem that is so famous (and probably so old), that at least three or four times a year – every year – someone comes up to me and goes through it as though he had just heard it for the first time and was STUMPED. And I mean *stumped. F*rom young children to high school teachers, anyone who hasn't heard this classic problem will find it to be as interesting as its 20 or 30 year popularity would indicate. And lastly, it has the quality that many of my 387 favorites share: You do not need to know much math at all, but you do have to try to think mathematically (or logically, or REASONABLY). It's simple, like most of the great mathematical questions are; even though it is fair to say that many of life's simple questions have complicated answers. But not this one!

O.K., enough preamble...

To get a good view of the Saint Valentine's Day parade, three friends decide to avoid the crowd and treat themselves to a hotel room right on the town's main avenue so they could stay warm and dry (if it rained) and still enjoy the view from a near-perfect location.

The hotel room is $30 (a Valentine's Day Special), and so the three men agree to chip in $10.00 each.

The bellhop, who is also the hotel's manager, shows them to their room and collects their money. He wishes them a pleasant afternoon and goes back into the elevator. But as the elevator begins to descend, the man realizes that the room he gave the men was really only a $25 room. He feels he should do the right thing and return the five dollars to them even though they do not know it is on sale. So he rides the elevator back up to give them their refund, and as the doors reopen in front of their room, he realizes he does not know how to split $5 evenly between three people. (He tells himself he was absent that day in math, because he was pulled out for an orthodontist's appointment). He begins to get nervous, does not want to embarrass himself, and thinks quickly. Then he decides what to do.

He has several one-dollar bills in his wallet, so he'll tell the men that the room is reduced from the thirty, but he'll only tell them it's reduced by three dollars. The room is twenty-seven dollars, he'll tell them, instead of thirty, and give each man one dollar back. He will keep the two dollars for himself as a little tip (that's stealing, but on Valentine's Day, we won't focus on that part. After all, he figures, the men will be happy they're each getting a dollar back, making each of their share 9 dollars each, which is a pretty good savings.)

He'll keep the two remaining dollars for himself, he decides, which he'll use to buy a math book and brush up on basic facts. (Either that, or candy, and he can decide later).

Indeed, he gives each of the three men one dollar back, and they thank him. The room only cost each one of them nine dollars instead of 10. After all, it makes sense:

9 x 3 = 27

And, of course, that $27 + the $2 in the bellhop's pocket makes $29.

The only mystery is this: They started with $30, and now altogether have $29.

*Where is the missing dollar?*

Tune in next time, (a day or two) same math channel, same math concept, and we will answer this perplexing mystery...

Until then, I hope this helps,

Mitch