I received an email a few weeks ago from a math teacher:

Hello, Dr. Wiggins —

My students and I sent a question to Prentice Hall in hopes that they would forward it on to the book’s editors. As several weeks have passed, we were not sure that the message was sent or received. So out of the textbook editors list, we chose you as a possible contact about our question … partly because we read that you are in a rock band and figured you’d be fun to talk to.

Tom Rice & the Algebra I Class, East Tipp Middle School

Here is the start of their letter:

Dear Editors –

My name is Tom Rice; I teach first-year algebra to seventh- and eighth-graders at East Tipp Middle School in Lafayette, lndiana, using Prentice Hall’s Algebra. Conner, Molly, Madi, Shelby, Kensington, Kelsey, Aidan, Kirby, Jon, Kyle, Zach, Jackson, Maya, Tyler, Meredith, Elliot, Justin, Leonard, Grace, Drew, Josie, and I enjoy your textbook very much, especially when we get a chance to go off on tangents related to your word problems.

We are in the middle of a study of Chapter 7: Exponents and Exponential Functions. When reviewing an assignment made for the night before, Justin asked about question 42 on page 460:

*Suppose you start a lawn-mowing business and make a profit of $400 in the first year. Each year, your profit increases 5%.*

*a. Write a function that models your annual profit.*

Just like your text, Justin and I came up with p = 400(1.05)n. And then came the moment that changed us forever. Satisfied with our work with part a, we moved on to part b. (Even when typing it above, Microsoft Word automatically started my next line with a part b.) We expected a part b; we anticipated a part b; we yearned for a part b.

There was no part b.

Some kids were excited; one was frightened. “Hold on a minute, kids,” I said, “Let me check my book. Sometimes they get edited separately … maybe I have 42b. I didn’t have 42b. Most of the kids were a bit disappointed, but not surprised. On further inspection, however, I noticed that l *did* have an answer for 42b: $5031.16.

Kyle announced, “A mystery is afoot, friendsl” (you’d have to know Kyle.)

To read the full letter and the solution of the mystery (as far as Mr. Rice’s class is concerned) download this pdf file: In Search of 42b.

Well, I was so tickled by their letter after sending it along to the real textbook editors I sent the class back a problem to solve. As people who know me and read this blog know, I am a hopeless sucker for Car Talk in general and the Puzzler in particular. (It’s an acquired taste for some, I know, including my wife and kids).

Here’s my email back to Tom and Co.:

Thanks so much for taking the time to write and for turning your kids into great math detectives. Have you and they considered a career in CSI – Math?

I am afraid that I am as much at a loss as you are as to what has happened to the mysterious 42b. My role is more on the educational advice side than true editing.

Frankly, I think whatever the truth is, it CAN’T be as interesting and fun as what you guys did with it as a class. In fact, in future textbooks I will strongly recommend to the authors and editors that they *routinely* leave out sentences, phrases, and entire questions since it seems clear that the investigations posed by you and your students are far more interesting than most of the ones in the textbook! (Did you know that dictionary and encyclopedia makers always put in at least one totally bogus entry to catch unscrupulous plagiarizers of their work?)

Since your students are clearly excellent at making conjectures and investigating problems, I have a challenging problem for them:

*Farmer John has long weighed his hay bales by using a balance beam scale and a rock. The rock is exactly 40 pounds, which perfectly balances the right amount of hay for bales placed on the other pan of the beam.*

*Farmer Joe comes by one day and is very impressed with the rock and asks Farmer John if he can borrow it. Farmer John says, Sure… *

*But you know what happened next, I’ll bet: Farmer Joe borrowed the rock and broke it by accident, into 3 pieces.*

*Farmer Joe profusely apologizes to Framer John for breaking the rock, but as it turns out it’s all for the good: A day later Farmer John reports to Farmer Joe that he can now weigh ANY item accurately that weighs between 1 ands 40 pounds using only the 3 rock pieces (and the balance beam scale)!*

*What are the weights of the 3 pieces of broken rock?*

And off went the email.

YIKES! Check your sources…. I had sent the email from memory of the problem, without going back to the Car Talk Puzzler site to check the terms. At 3 AM – I’m serious – I woke up, bolted upright: It’s 4 pieces, not 3, dummy!! I hastily wrote back the next morning:

Tom and class:

OOPS.

*Never* give a problem from memory.

I awoke at 3 AM and thought – it’s 4 pieces of rock, not 3. Let your kids know!

Ya gotta love Tom’s answer:

Hey, Dr. Wiggins –

Thanks for the update … but if it’s all the same to you, we’re going to analyze the problem as originally written as well. We know it’s impossible, but so was 42b! (Danger is our middle name.) We’ll let you know when we have something.

And ‘have something’ they did. Here is the full report back, including their clever stab at the problem stated incorrectly by me. (Tom clearly could quit his day job to become a story-teller). Rocks for Dr Wiggins

So, off went another email:

Tom & Co.:

You guys are too much! Fantastic work, especially taking my bonehead memory lapse and having some fun with it in Part 2.

Here is one more challenge for your whiz kids, one of my all-time favorites.

*We all know that the Pythagorean Theorem states that A ^{2} + B^{2} = C^{2} in right-angled triangles in Euclidean geometry.*

*But you may have forgotten that the “squared” is meant literally here: the square on side A (i.e. the area of that square) + the square on side B ((i.e. the area of that square) = the square on side C, i.e. the area of that square.*

*QUESTION: can the theorem be generalized to shapes other than squares?*

*For example, what about RHOMBUS on side A + RHOMBUS on side B? Do the areas of the 2 rhombuses (rhombi?) = the area of the rhombus on side C? Find out. Does anything follow?*

*Obviously a RECTANGLE as a shape is problematic because the possibilities are unlimited. A key one is the special case of the square, of course, but messing around with other rectangles with proportional relationships of width vs. length is interesting.*

*Investigate other areas of other shapes and their areas, regular and irregular.*

*So, how general can we make the theorem?*

Stay tuned. Tom says a report will be coming soon. I can’t wait.

PS: Here is the official answer to the mystery of 42b, courtesy of textbook editor Dennis Slattery. The conjecture the class made was sound:

Dear Mr. Rice and East Tipp students,

Grant Wiggins forwarded your questions about Exercise 42b in Lesson 7-7 of our Algebra 1 text. I apologize for the fact that the exercise was missing. It got covered up by the bottom frame in the digital file of the page. That error has been corrected in subsequent editions.

The text of 42b is “If you continue your business for 10 yr, what will your total profit be?” And so you correctly figured out the mystery. I understand your reasoning about the rounding of the answer, and I think you make a good case. However, I believe an equally good case can be made for $5031.16, which you cover in your analysis. I’m pretty sure that if we changed the answer to $5031.15 we’d get letters from someone who felt cheated out of the penny.

Thanks very much for your query, and I hope you get as much from the rest of the text as you did from exercise 42.

Hmmm: mystery solved??? What do we make of that last penny, readers?

David

said:Being a software engineer and not a mathematician, I couldn’t remember the best way (any way) to solve this without brute forcing it (this being the 40 pound rock problem). However, I always enjoy writing a program that can figure it out for me and that’s sometimes as much a puzzle in itself as the actual problem. Luckily, I realized quickly that the rocks could be placed on either side of the scale to counter-balance themselves to get the intermediate weights of hay.

So I wrote myself a little program that increments 4 integers (much like a combination lock, but in base 40 instead of base 10) with a method that returns true when the sum of any subset of {±a, ±b, ±c, ±d} equals result where result is everything from 1 to 40.

With that being said I was able to find the answer, but I would love to know a better way than brute forcing through it.

Claude

said:what a cliffhanger …looking forward to the reply

J Martens

said:Thank you — this post made my day!

David Wees

said:Hi Grant,

I’ve published a link to this post, as well as a quote from the letter sent to you by the students on a site where I am compiling mathematical investigations and examples of students “thinking like mathematicians.” See http://maththinking.org/. I hope you don’t mind. This is a fabulous example of both.

David

grantwiggins

said:Of course – no problem.

Jeanette

said:I absolutely loved reading this! What an amazing boost of motivation! I will be using some of your questions in my class. And I will definitely be looking for more ways bring some of Tom’s energy to my teaching! Thanks so much for sharing!