Maria Miller, seen above, has an interesting and useful blog on mathematics for home schoolers, as well as other web resources, not the least of which appears to be a recent web page called "Math word problems — the do's and dont's of teaching problem solving in math" that I've just started looking at today.

The most recent entries on her blog address this vital issue of teaching problem solving, via a couple of problems for 5th graders taken from a Canadian on-line resource.

The first post Maria made on these problems reads as follows:

I'd like for you to try solve these two 5th grade problems. They are from a very nice collection of word problems for kids by Canada's SchoolNet.I set about attacking these problems as soon as I saw them, both because they appealed to me and because I wanted to be ready for the ensuing discussion. You may want to take some time to think about the first one, for which solutions and comments appeared on her blog today.

I'd like to use them to illustrate problem solving. But before I do that and give you the solutions, please try to solve them yourself. And don't post solutions here for now; I will have to reject such comments.

Solutions and discussion will follow in a few days.

THE FIRST PROBLEM IS:

* How many addition signs should be put between digits of the number 987654321 and where should we put them to get a total of 99?

THE SECOND IS:

* Divide the face of the clock into three parts with two lines so that the sum of the numbers in the three parts are equal.

Maria's comments can be read by clicking on that last link ("today") or going to her blog and finding the entry for 16 January 2008. You will also find a response from "mathmom" and one from me. By the time you read this, there will likely be others.

SPOILER WARNING: if you read past here, assuming you've not yet clicked on the last three links, you're going to see solutions to the plus sign problem.

Here is what "mathmom" had to say:

I approached this one a little more methodically. I first summed the 9 digits to get 45. I found that the required sum was 54 greater than this number.

I then thought about what would happen if I made a two-digit number. The digit in the ones place would be added into the sum just the same as before. But the new tens digit would count for 10 times as much as it did before. So if I "promoted" the digit D to a tens place, I would add 10D and subtract D (since it would no longer be counted as a ones digit anymore) for a net increase of 9D. So if I need to increase the sum by 54, I need 9D = 54 and D=6. Thus if I "promote" the 6 by making 6+5 into 65, I arrive at the required sum.

I didn't think about the fact that I might have promoted more than one digit and used more than one two-digit number, but my son came up with the second solution, where the two-digit numbers used are 42 and 21. Here he promoted two digits, a 4 and a 2, with the same net effect as promoting just the 6. He had computed that he needed to increase the sum by 54, and then found this solution by a guess-and-check method.

To show that these are the only possible solutions, consider the possible combinations of digits that sum to 6:

6

5 1

4 2

3 3

3 2 1

We can't promote the 1, since there's nothing after it in the row to serve as its ones digit. (If you add a zero to the end of the line, this admits the 5 1 solution, but not the 3 2 1 solution, because you can't promote both 3 and 2 at once.) And we can't promote 3 twice, so the only combinations that work are 6 alone, or 2 and 4.

Here is my comment:

One correction to mathmom's comment: in the second solution, the first two-digit number is 43, not 42.

I had done something similar to arrive at her first solution (and later formalized it algebraically, though in this case I'm not convinced I gained a lot by doing so), and I also missed the second solution by not thinking further about the problem, to my embarrassment. But in reading the third paragraph of her post, I noticed before reading further that since D = 6, using her notation, and the other solution involves using D1 = 4 and D2 = 2, with D1 + D2 = 6, I then wondered about using D1 = 5 and D2 = 1, where again D1 + D2 = 6. Before I could think through why that didn't work (duh!), I saw what mathmom had added and realized the issue, which she neatly pointed out could be solved by including 0 as a tenth digit.

So a great continuation/extension question would be (for students who have been exposed to negative integers or who wouldn't be troubled by thinking about them), can we find more solutions if we append more consecutive integers to the list on EITHER end (or both ends)? Why or why not?

Since I've just thought of this, I don't have an answer. But I raise the issue because I'm always looking for extensions, not just as challenges, but because extending and generalizing problems is one of the major activities of professional mathematicians. Doing so is one way to help develop new theorems, methods, and, sometimes, entirely new areas of mathematics. I look back on my own K-12 mathematics education with a great deal of regret, not the least of which grounded in the fact that I was never led to realize any of that aspect of mathematics. Like most Americans, I was taught in a way that made mathematics appear to be a dead, completely closed subject in which everything was already known (by the professionals), and where my job was to accept received knowledge and be able to show that I "got it" by solving textbook problems that were, for the most part, strictly computational in nature. By 9th grade, the approach simply didn't grab me any longer, and not because I didn't generally do extremely well in mathematics up to that point: in fact, I was an A student through the end of 8th grade. While there were other factors having little or nothing to do with the instruction or content that led to my decreasing interest and effort in math starting in 9th grade, I wish I'd been shown something along the lines of what I mention above: the idea of doing what might be something new and original, or finding creative solutions to problems that already had known solutions, may well have kept me "in the game" at a time when I was growing restless, impatient and bored when it came to math and science.

Now, since posting the above comment this afternoon, I've had a chance to play a bit with my extension questions, as you may already have done or are about to do. I will not post my additional findings right now, but will wait to let folks have fun with the extension question on their own.

It would also be great if anyone has other extension ideas for this. Don't feel limited to the alleged "5th grade" level of the original problem, but it would bear thinking about the target for any such extensions. The one I'd suggest right away is: What happens if we change 99 to other values?