### Solving the plus sign problem

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?

This is a fifth grade problem taken from Word Problems for Kids by Canada's SchoolNet.

The first step, as always, is to understand the problem. The student needs to know what is an "addition sign" and a "digit". We're simply asked to put plus signs in between those numbers and add them up, and try to come up with 99.

Then, after we have a basic idea of what the problem is about, is the time to do something. You know, often the child may say, "I don't know how to start. I have no idea what to do!"

But in this case, as often happens, you'll get somewhere as soon as you'll do something. It's really simple: put some plus signs in there and just see what happens. Let's simply put the plus sign in between every digit:

Great, we got something. We got 45 which is too small. Now the student should start thinking HOW to make the sum bigger?

Obviously, the only way to do that is to use ONE OR SOME two-digit numbers. We need to omit at least one of those plus signs!

So try something. For example:

98 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 136.

9 + 8 + 7 + 6 + 5 + 4 + 3 + 21 = 63.

Encourage the student to try a few other possibilities. He/she should notice that if you make the two-digit number using the large digits, the sum will be MORE than if you use the smaller digits for that.

(That is, of course, really simple.)

At this point the student might simply use "brute force" and write out all the possibilities with one two-digit number. That's alright; that will get him an answer!

However, there is also a quicker way, if we use some thinking. The target answer 99 is about half-way in between the two sums I have above (136 and 63), so I would try next to use the "middle digits" such as 4, 5, and 6 as the two-digit number:

9 + 8 + 7 + 6 + 54 + 3 + 2 + 1 = 90, which is too little.

9 + 8 + 7 + 65 + 4 + 3 + 2 + 1 = 99, which is the right answer!

Any other possibility with one two-digit number will either be less than 90 or more than 99, so that is the only solution using one two-digit number and seven plus signs.

But, good problem solvers will also consider checking if this is the ONLY solution. There is the possibility of using TWO two-digit numbers. Indeed, I quickly stumbled upon another solution that way:

It's easy to note that this is the only possibility using six plus signs, since if you put the plus sign between some other digits, your sum will be MORE than this sum.

I hope this is helpful to some of you. I know the problem is quite easy. My intention is simply to point out how a typical problem solving process can go. Observing the "tricks of the trade" can help you to solve problems, and to teach others do the same.

## Comments

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.

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.

I've posted about this problem on my blog: rationalmathed.blogspot.com

With 9+8 converted to 98

we lose 17 and gain 98 for a net increase of 86. ( too darn big)

We are looking for a net increase of 54 to increase from the current sum of 45 to 99.

Choose from the following table of differences:

98 -17 = 81

87 -15 = 72

76 -13 = 63

65 -11 = 54

54 - 9 = 45

43 - 7 = 36

32 - 5 = 27

21 - 3 = 18

Pretty cool table eh?..

So 65 -11 = 54 is an instant winner in the 54 sweepstakes.

Thus:

9+8+7+65+4+3+2+1 = 45 +54 = 99

Since 36+18 = 54

43 and 21 are a winning duo

as:

9+8+7+6+5+43+21 = 45 + 36 + 18 = 99

None of the other differences can be combined to give 54 (which is 6 sets of 9)

If we view the above differences as sets of 9 we have:

81 -- 9 sets

72 -- 8 sets

63 -- 7 sets

54 -- 6 sets

45 -- 5 sets

36 -- 4 sets

27 -- 3 sets

18 -- 2 sets

Only two ways to get 6 sets.