The other day, a friend of mine who does not consider himself to be a “math person” and who is somewhat amused that I edit books about math, sent me this question, which supposedly came from a 5th-grade classroom:
This is not a trick question.
There are 7 DNR wildlife managers in the woods in
Each manager has set up 7 elk traps.
In each trap, there are 7 elk.
For every elk, there are 7 deer.
How many legs are there?
Ten years ago, being someone who would have told you I was not a “math person,” I would have looked at this problem, started thinking about all those legs in traps, and lain down my pencil in defeat. Today, I look at a problem like this and can’t wait to get to work (although I would beg to differ that it isn’t a trick question. It doesn’t tell you how many whose legs you’re counting. Does it expect you to include the wildlife managers’ legs? Is it total legs or legs in each trap? It could have been worded better, but we’ll assume it means total human and nonhuman legs). Anyway, my guess is that the majority of you are having the same reaction I would have had ten years ago (if you're even still with me at this point), but I don’t believe you need be defeated. I am absolutely confident you can do this problem (unless you suffer from dyscalculia, which is highly unlikely if you can do things like tell time and read a train schedule).
What happened to you, then, to convince you that you can’t do it and that you are not a "math person?" Most likely, you were presented with horribly-worded questions such as this one when you were in 5th grade. It probably showed up on a test or work sheet, and you were told to solve it in isolation. If you bothered to ask for help (which, if you were me, you most certainly didn’t), your teacher made you feel like an idiot when you didn’t even know where to begin to answer this question. Most likely, you couldn’t care less about wildlife managers and elk and deer (or, again, if you were me, you were distracted by the fact that the poor things were all caught in traps, and you hoped they weren’t going to be killed). If you managed to get away from those traumatic numbers (7 is a bad number, all around, as those who study numbers and cognition can tell you) long enough (me again. Anything to distract me from the actual equation I was magically meant to create), you were wondering what kinds of traps these were. How could you possibly fit so many large animals into one trap?
Then, having automatically marked this question wrong on your paper without even waiting for the answer, you heard the teacher ask if anyone had the answer. Some Smarty Pants you always despised raised a hand and before being called on, blurted out, “10,990 legs.” “Correct,” your teacher said. What? You were flabbergasted. Where was the potion that had been dumped on that huge number to bring it into sight? If you were lucky, maybe your teacher (or Smarty Pants) explained a little more:
“You’ve got 14 human legs
1,372 elk legs
9,604 deer legs
14 + 1,372 + 9,604 = 10,990.”
You were none the wiser and hadn’t learned a thing or understood the problem any better than when it had been handed to you. Nobody discussed the problem. Nobody was encouraged to show how it could be done another way. You just moved right onto more problems with invisible answers for which you were given no revealing potion.
Today, I’m going to put what I learn from the books I edit into practice. I’m going to see what I can do to help you understand this problem better. First of all, we’re going to take into consideration situated cognition, which is learning and problem solving in particular contexts. Even those who flunked math in school can develop powerful problem-solving skills when they are doing so in their own contexts, like when they're camping, if that's what they like to do, or cooking, or building a doll house. In fact, even when doing things we might not like to do, but need to do, such as attending staff meetings, we can be pretty good at it. If not, most of us wouldn't last too long in our jobs.
I’m pretty sure that you, my blogging audience, is likely to have the same reaction as my fifth-grade self would have to the elk and deer problem. Interest is half the battle, and most of you don’t find yourselves in situations in which you have to (or care to) count large animal legs. So, what interests you? Let’s take a wild guess here: books? And if you are interested in books, you’re familiar with walking into bookstores and skimming pages of books you might want to buy. Forget elk and deer then. Here’s my problem for you:
(First of all, before we get started, get out your calculators, because, even though I am going to make these numbers much easier to work with than Smarty Pants did, there is absolutely no reason for those of you who had that sort of teacher for fifth-grade math to try to add and multiply numbers greater than 100 without the use of a calculator.)
7 book addicts belong to a book swap group. Every so often, they get together, each bringing 2 books they’ve read to swap with others in the group (isn’t that a neat idea? I just made it up, but I’d like to belong to a group like that). Everyone gets to choose 2 books from the pile of fourteen. Today they’ve met for lunch at Good Enough to Eat in NYC. They all trust each other by now, confident no one is bringing a real dud to this meeting, so each one skims one page of each of the books he or she has chosen, and then they get down to the business of eating and catching up with each other. Near the end of the meal, one of them says, “Hey, why don’t we all go to The Strand today?” They all agree this is a great idea, and they make their way to the bookstore with “18 miles of books.” Once at The Strand, they go their separate ways, and:
Each addict visits 7 sections.
In each section, each addict looks at 7 hardcover books.
For each hardcover book, each addict looks at 7 paperback books.
Each addict skims four pages of each book to decide which ones to buy.
How many total pages do all of the addicts skim that day?
Now, let’s pretend I’m one of the addicts. I head first to the mystery section, because I’ve had so much fun reading mysteries for my mystery book discussion group. I check to see if there are any Ian Rankins. I’m in luck. I find one hardcover Rankin and seven paperbacks I haven’t read (not hard to do, since I’ve only read one Rankin so far, and he’s an extremely prolific writer). That’s 1 HC + 7 PB = 8 Ian Rankins. I skim 4 pages in each for a total of 32 pages from Rankin. Next, I cruise the shelves for Dorothy Sayers and again find one HC and seven PB I haven’t read. That makes 64 pages I’ve now skimmed. I do this with 5 more mystery writers for a total of 32 x 5, which is 160. Add the 64 Rankin and Sayers pages to that, and I’ve skimmed 224 pages before I leave the mystery section. In other words 7 (authors) * 8 (books) * 4 (pages) = 224 pages in one section.
Next, it was onto literature. Again, I skimmed 4 pages from 8 books from 7 different authors for a total of 224 pages. Then I visited the cookbook section where I found one hardcover book of vegetarian recipes and seven vegetarian paperbacks to skim, one hardcover and seven paperbacks on chocolate, one hardcover and seven paperbacks on soups, one hardcover and seven paperbacks on Chinese food, one hardcover and seven paperbacks on Indian food, one hardcover and seven paperbacks on bread, and one hardback and seven paperbacks on chili peppers. (I also found a whole shelf of Rachael Ray books, but I purposely ignored those.) By the time I left the cookbook section, I’d skimmed another 224 pages. I then moved onto the psychology, science, history, and religion sections, where I again skimmed 224 pages in each.
Uh-oh, after all that, I realized I’d better get to the checkout, because it was time to meet all the other addicts (after all, we’d been here for nearly five hours). Lo and behold, when we all got back together, we discovered each and every one of us had skimmed the exact same number of pages from the exact same number of HC and PB books. I skimmed 224 pages in seven different sections, and so did each of the others. That makes 7 (addicts) * 7 (sections) * 224 (pages) = 10,796 pages. But don’t forget, before we even got to The Strand, we’d each skimmed one page from each of our two swapped books. In other words, we’d each already skimmed two pages before we'd even walked through the doors of The Strand. That’s 7 * 2 = 14. We needed to add that 14 to the 10,796 to get 10,990.
The formula for this is (7*7)(7 * 8 * 4) + (2 * 7) = (49)(224) + (2 * 7) = 10,976 + 14 = 10,990. If you were talking about elk and deer in that original problem, that 224 would be the number of elk and deer legs in each trap. The 49 would be 7 (traps) * 7 (wildlife managers). The 14 would be the number of wildlife managers' legs. As you can see by Smarty Pants’s answer, neither one of us did this problem the same way (and there are other ways to do it, too). However, aren’t my numbers a little more friendly? Wouldn’t you rather tap in 49 * 224 on your calculator than 1372 + 9604? And, even if you hate math, isn’t it easier to get that 49 and that 224 than to get that 1372 and that 9604? And that, my friends, is what a good teacher would have helped you see.
Oh, and for those of you who wonder how many books I actually bought, well, you know the answer to that: I bought all 56 books from each section and found myself bringing home 392 (Strand purchases) + 2 (swapped) books! And yes, of course, I know real life is rarely this un-messy. I’m sure all the Ian Rankins would have been gone, and I wouldn’t have been able to find a single hardcover Dorothy Sayers, and one of my fellow addicts would have found nine hardcover M.F.K. Fisher’s, while I found none, not to mention the fact that I can’t possibly carry 394 books around NYC by myself, but that’s where fantasy intersects with math.