Don’t Require Students to Read Your Mind!

Find the next term in the sequence: 9, 4, 1, 8, 27,       

When I’ve proposed this problem before, the most common response I get is “64” (do you see how someone might make that guess?), but it’s wrong.  Would you have guessed that the next two numbers after the blank are 6 and 14?

The number that correctly fills in the blank is provided (and explained) in the following paragraph.  Trust me, you’re probably not going to guess the answer – nobody ever has (who hasn’t known – or figured out – the “trick”).  But if you insist on thinking about it, don’t go to the next paragraph until you give up.


The following excerpt from a team chronology that used to be available on the official Boston Red Sox website gives away the answer:

1931: The Red Sox player first wear numbers on their uniforms (in 1931). Since then, the Red Sox have retired five uniform numbers: Ted Williams’ No. 9 and Joe Cronin’s No. 4 officially retired May 29, 1984; Bobby Doerr’s No. 1 retired May 21, 1988; Carl Yastrzemski’s No. 8 retired August 6, 1989; and Carlton Fisk’s No. 27 retired September 4, 2000. Major League Baseball retired the No. 42 of Jackie Robinson for all teams.

Subsequently, the numbers of Johnny Pesky (6), Jim Rice (14), Pedro Martinez (45),Wade Boggs (26), and David Ortiz (34) have also been retired.


“But wait a second”, you say, “this problem isn’t fair. I have to know how you picked the numbers in order to correctly guess the next term in the sequence.”

Exactly!

In fact, if we’re honest, every problem of the type, “What’s the next term in the sequence?” is really a mind reading problem!.

 I could go even further and point out that, on an all-or-nothing high-stakes test, all “what’s the next term” problems are inherently unfair, since they require the student to read the mind of the test writer.

Let me be clear.  By no means do I think “guess the next term” problems are bad classroom problems, especially if students are asked to explain the pattern(s) they observe and how their pattern gives them their “answer”.  These types of problems provide excellent opportunities for students to flex their problem solving muscles as well as giving them a chance to explain their solution method. The unfairness arises when we insist that everyone notice the same pattern.

A “real-life example”, shown below, provides a cautionary tale from the 2002 grade 10 Massachusetts Comprehensive Assessment System (MCAS – the test that must be passed before a student is allowed to graduate from a public high school in Massachusetts). While the story had a happy ending, think about what would have happened had the young woman not questioned AUTHORITY and spoken out about what she thought was a correct solution?

From the Massachusetts Department of Education press release (December 4, 2002)

Points Awarded To Students Who Selected Alternate Answer On Grade 10 MCAS Exam

MALDEN – An additional 449 students in the classes of 2003 and 2004 have earned a competency determination, thanks to the ingenuity of a Whitman-Hanson Regional High School student who found a unique method of answering a math question on the 10th grade exam, education officials announced on Wednesday.

Because of this finding, a second answer on the question is now counted as correct. As a result, an additional 136 students in the class of 2003 and 421 students in the class of 2004 have now passed the math exam, and will not have to take the MCAS math retest being given next week.

“Although the answer we had marked as the right one is correct, it’s clear now that what this student found is also right, and I think that’s terrific,” said Education Commissioner David P. Driscoll. “This girl was able to take a typical math question and come up with a completely unique method of solving it that even our math experts, teachers in the field and our test reviewers never considered. This is a great example of just how creative our students can be, and I applaud her efforts.”

The question presented a real-world application of the binary, or base two, number system, such as the one used by computers. The numbers zero through 10 were shown as a sequence of on and off switches in four-switch panels. Students were asked to identify the panel that would represent number 11, the next in the sequence. The correct answer could be found either by knowing the binary system, or by recognizing changes in place value throughout the series.

However, when the student looked at the panels, she saw something different: a spatial sequence in the on and off switches, rather than the numeric pattern envisioned by the developers and reviewers of the item. After conferring with mathematicians, education officials this week determined that the student’s answer is also a viable second solution to the question.

 

See below for the problem in question

Example from 2002 MCAS

How are your visual estimation skills?

A typical 3-ball tennis ball canister is shown on the right.  If you happen to have such a canister, pick it up and perform the following investigation:

Without actually measuring anything, make your best guess as to which is larger: the height of the canister or the circumference of its lid.

Do you have an answer?  Using a piece of string or some similarly flexible, but not stretchable object, check your answer by marking the circumference of the lid on the string, then compare it to the height of the canister.

Most people are surprised by the fact that the circumference of the lid is larger than the height of the can, but in fact, the reason is as “easy as π.”  Here’s a way to think about it (think about it before scrolling down and reading the solution):  The height of the canister is equivalent to 3 ball diameters.  How is the lid’s circumference related to the diameter of a ball?

 

 

 

 

 

As mentioned above, the height of the canister is 3 ball diameters, while the diameter of the lid (or canister) is the same as the diameter of a ball.  Therefore, if D is the diameter of a ball, then the height of the canister is 3D, while the circumference of the lid is πD.  Since π>3, the circumference of the canister is greater than its height.

Why do most people guess wrong?  I think the most feasible reason is that it’s really hard to estimate curved lengths, since it requires the ability to “straighten out” the curve to a segment of equal length.  Our brains just don’t work that way!  Of course, many people just have a difficult time estimating even straight line distances, but that’s a topic for another day (or post)!

One of my favorite problems (and solutions!)

51 numbers are chosen at random from the counting numbers 1, 2, …, 100.  Will those 51 numbers always contain a “consecutive pair”?  That is, will there always be an N so that N and N+1 are in among your 51 numbers?

There are a number of reasons that I like this problem, but the main one is the surprising result that’s proven in the process of solving the problem.

We’ll use the Pigeonhole Principle, which is a common sense result that can be summarized as follows:  If you have more objects than places to put them, when all of the objects are “put away”,  at least one of the places must contain at least two objects.

To model the problem, imagine that you have 100 ping pong balls 50 small boxes (large enough to hold 2 ping pong balls).  The ping pong balls are individually labeled with the natural numbers from 1 to 100, then placed in a large container and the boxes are labeled in pairs: the first box is labeled [1 2], the second box is labeled [3 4], and so on (officially, box K is labeled [2K-1  2K]).  Notice that each ball will have exactly one box whose label includes the number on the ball.

Now, pick 51 balls from the container (therefore choosing 51 numbers from 1 to 100).  Taking each of the 51 balls one at a time, place them in the box whose label includes the number of the ball.  Since there are 51 balls and only 50 boxes, at least one of the boxes must contain more than one ball, which mean at least one of the boxes has to contain both balls whose numbers match the the label on the box.  That is, at at least one pair of the balls are labeled with consecutive numbers, which solves the problem. Namely,

If 51 numbers are chosen at random from the natural numbers 1, 2, … , 100, then there will always be a consecutive pair of numbers among the 51 chosen numbers.

Interesting result, right?  Well, wait, because there’s a lot more to this than might first appear.  In fact, what we proved was that there will always be a consecutive pair chosen for which the smaller of the two numbers is ODD.

Now, that on its own might not seem so out of the ordinary, since you might expect that, since there are the same number of odds and evens from 1 to 100, there must also be a consecutive pair (among the 51 chosen numbers) in which the smaller of the two numbers is EVEN.  Shouldn’t there be some sort of symmetry to this problem so that anything true of odd numbers must also be true for evens? However, this is NOT guaranteed – in fact there are many examples of 51 numbers (from 1 to 100) among which there are no consecutive pairs that “start” with an even number.  For example, the 51 chosen numbers COULD BE 1, all the even numbers, for which the only consecutive pair is 1 and 2.

So, there will ALWAYS be a pair of consecutive numbers in which the smaller is odd, but there MIGHT NOT be a pair of consecutive numbers in which the smaller of the numbers is even.  Why is our intuition that evens and odds should somehow be “the same” so wrong in this case?

Well, in a nutshell, it’s fairly easy to see what the big difference is: In the numbers from 1 to 100, there are 50 consecutive pairs in which the smaller of the pair is odd (1 and 2, 3 and 4, 5 and 6, … , 99 and 100), while there are only 49 consecutive pairs in which the smaller of the two numbers is even (2 and 3, 4 and 5, … , 98 and 99), so it’s more likely to pick an “odd starting consecutive pair” than to pick an “even starting consecutive pair”.

In terms of the method of our proof, if we instead forced our pairs to start with an even number, our boxes would be labeled [2 3], [4 5], [6 7], … , [98  99], but this collection of boxes has a significant problem: two of the numbers (1 and 100) are missing, so we need another box, which we’ll have to label [1 100].  Now, when our 51 numbers are chosen, at least one of them must have 2 ping pong balls in it. However, this box MIGHT BE the one labeled [1   100], so there might not be a consecutive pair “starting” with an even number.

 

A Mathematical Hike in the Woods

The following is an excerpt from an article I wrote a few years ago: Problem Solving By Analogy/ Problem Solving As Analogy,” The Mathematics Educator (2007), Vol. 17, No. 2, 2–6

Click here to download a PDF of the article

There are at least three different ways to go hiking in the woods.

  • One is to be led down a previously created path, often by an expert who’s taken the path before.
  • Another is to follow a path with which you are already familiar, perhaps after being led along the path several times.
  • The third is to be willing to leave a familiar path to try a completely new trail when the need arises (or just for the heck of it).

Being led on the hike is efficient, if your goal is to get to the end of the trail. You can see some sights but only those you are led to. It is probably the most comfortable method for the novice hikers, because it isn’t necessary for them to keep track of where they are. However, this also doesn’t help them learn to get to places that are not on the trail.

Traveling alone on a path with which you are familiar is less efficient, since now you don’t have an expert to keep you on the trail. However, it’s definitely more interesting, since you can choose when and where to stop and how quickly to walk. Of course, in order get to a location you haven’t already been to, you must be willing to stray off the path and possibly blaze a new trail every now and then.

Some of these new locations might be just off the path, while others may be far away from your comfort zone, but these less-traveled sights are often the most interesting (and educational). And each new trail you forge provides you with new locations you know how to get to (and return to later).

However, many hikers aren’t natural explorers. It isn’t likely that someone who has always been led through the woods will stray far from the known path. It takes a rare person to feel confident enough to take over leading the group to the end of a trail (or even back to where they started) or to choose to lead the group entirely off the path just to explore, unless they had been given the chance to explore in the past. Unless it is your responsibility to get everyone back to the trailhead, your mind is usually focused on following the leader.

The second method (taking a path with which you are familiar but without a leader) is more work, since you need to pay closer attention to where you are and which branches of the path you could choose. You might feel limited to taking the particular path you are on, but the real fun comes when you veer from the known path and explore, knowing that if you choose the left fork and arrive at a dead end, you can always find your way back and choose the other fork.

When you’re the leader of the hike (or at least an active participant in the decision making), whenever you break off a familiar path and look for new sights to see, you must be aware of where the familiar path is (in relation to your current location) so you don’t get lost.

Learning to be comfortable with straying off the path (or becoming a trailblazer) comes from experience, but that experience need not be a solo effort. A good hike leader will point out trail markers and share his or her decisions with fellow hikers, letting them in on the thought processes being used as options are chosen and decisions are made. And novices could be encouraged to take charge under the watchful eye of the hike leader, who allows them to make the decisions. Even if the novices get lost, the leader can keep track of their location and bring them back to familiar territory (or help them solve the problem of finding their way back themselves).

As novices become more comfortable (and experienced) with making decisions and realizing that every decision is reversible, they are more willing (and able) to explore on their own. If novice hikers ever find themselves in unfamiliar territory, they will have a very difficult time finding their way out if their only hiking experiences involved having been led or traveling on familiar paths.

Therein lies the key connection with problem solving. If we are to help students learn to solve new problems that they haven’t seen before, they need experiences—guided and otherwise—that allow them to try and fail, try something else, and eventually arrive at a solution to the problem.

They aren’t alone, though, because the teacher/hike leader is there as a safety net—not to solve the problem for them, but to serve as a mirror, reflecting their strategies and progress, asking probing questions that encourage the novices to think through their options.

Experienced hikers, like expert problem solvers, are able to keep track of where they are, where they need to be, and the options available to them at any given moment. By explicitly discussing these options and decisions with novices (hikers and problem solvers), the novices gain an understanding of the process and are more likely to be able to navigate the paths themselves.

George Pòlya, in Mathematics and Plausible Reasoning, spoke of the importance of intellectual courage—being ready to revise ones beliefs. I like to expand this notion a little further to include a willingness to persevere even when you don’t know whether your strategy will lead to a solution or to a dead end. Isn’t that the goal we have for all our students?