The 400 Problem Part I – A Staff Seminar

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Introduction

The two seminars, The 400 Problem Part I and The 400 Problem Part II, of which this is obviously the first, are based on the Problem Solving lesson ‘The 400 Problem’ Level 5. The aim in these two seminars is to both solve the problem and to show how mathematics develops.

Here we start by looking at the highlights and key points of the investigation. This is how you might lead your colleagues through a seminar. While working through the steps below, it is important to focus the discussion on what your students would do, and how you could best help them to understand, and solve the problem and its extensions.

In Part II we go on to talk about the creative way new mathematics is produced and use the 400 Problem to illustrate the various steps.

The problem

Can you find the numbers a and b so that the two subtraction problems below have exactly the same answer?


 

It is worth pointing out at the start that the solution of this problem is only the first step of a much bigger investigation. Once this problem is solved the mathematicians’ stock in trade of extension and generalisation kick in.

 

The key points

 

Work through these points with your colleagues.

 

1. Play. Just get a feel for the problem and how you might solve it. Don’t worry about solving it. Just think how you might solve it. It can be solved by guess and check; by clever use of the normal algorithm for subtracting two 3-digit numbers; by realising the limits on a and b; by algebra; and possibly by other methods. No one of these is any better than any other.  The following describes how an application of the written algorithm for subtraction produces the result.

 

Start with the left hand sum; you can’t take 4 from 0 so you have to make it 10 and change the 40 to 39.

So 4 from 10 is 6.

We know that the two sums are the same. In the right sum the digits number is clearly ‘b’. So as the two sums give the same result, b must be 6.

So if b is now 6 and 6 from 9 is 3.

In the right sum that means that a = 3.

So now we know that a = 3 and b = 6.

 

2. Extend. Once you have solved this problem you might realise that you can solve a lot of similar problems: What happens if you replace all of the 4s by 1s, 2s, 3s, 5s, …?

 

Draw up a table something like this. Don’t worry about getting a complete table at this point. Note that we’ve called the number that started off as 4 and then got changed to 5 and so on, n.

 

n

1

2

3

4

5

6

7

8

9

a

 

 

2

3

4

5

 

 

 

b

 

 

7

6

5

4

 

 

 

 

Actually you might do better to have the n, a and b as the headers of columns. It’s a bit easier to see the possible patterns if the table is this way up.

 

3. Patterns. Once you have collected all these answers together, you should start to see patterns.

 

Two patterns are commonly found. One is that the a numbers start at 0 and go up by 1 each time, or a = n – 1; while the b numbers start at 9 and go down by 1 each time, or b = 10 – n. Another pattern is that this table is the 9 times table, so 9n = ab, or if you prefer, 9n = 10a + b.

 

Right so we have two guesses. But mathematicians don’t call guesses guesses, they call them Conjectures.  So one Conjecture is that in this problem, for n = 1, 2, up to 9, 9n = 10a + b.

 

4. Justification. Can you justify the guesses? Have you justified them already?

 

This particular Conjecture can be proved by exhaustion – complete the table and you will show that 9 times every n is 10a + b.

 

5. Extend again. There are lots of ways to get a new problem from this old one. You might ask if there are any other 3-digit subtractions like this that give some nice patterns. What can you change to get similar results?

 

You might also think of setting up some addition sums whose answers are the same. Can you make any progress here?

 

Another successful line is to make the problem a 4-digit one. Well maybe it’s not so successful. There are no answers for a and b in the corresponding 4-digit problem (the 4000 problem). When you think about it it’s fairly uncommon to find questions that don’t have answers. But mathematicians come across problems that they can’t answer all the time.

 

So try the 40000 problem. You’ll make progress on that.

 

It may be easiest to simply tell your staff that there are no solutions for the variations with an odd number of 0s, and to work on the even variations, though with a class of students this is certainly something you would want them to discover for themselves.

 

6. Super-Generalisation.  But can you generalise the 400 problem to any number of digits? What would you guess would be the answers to the 4000…00 problem? Does that guess fit in with all that you have done so far? Is it possible to justify your guess?

 

If you have the time and patience to try a range of numbers of 0s added to the problem you may be able to produce the Conjectures below and even prove them:

 

Conjecture 1: If we look at the general problem with an even number of zeros (and an even number of unknowns), then 10a + b = 10c + d = … = 9n.

 

Conjecture 2: If we look at the general problem with an odd number of zeros, there is no answer for the unknowns a, b, c, d, …

 

+       +       +       +       +       +       +       +       +       +

 

Now you’ve got the idea of how this small problem can develop into a largish investigation, you might have some idea of how mathematics develops from single problems to whole areas of the subject. To see further down this track, go to The 400 Problem Part II.

 

If you decide to stop there, your colleagues will have had a chance to tackle a non-trivial maths problem that doesn’t require more maths than they know. The next step that we offer in Part II looks at what we have done meta-mathematically and draws some general conclusions about the way that research mathematics is done. Surprisingly, in one sense, there is not very much difference between what we did above and what research mathematicians do. Teachers ought to know this. And it’s not too difficult to see. Have a look at Part II and see if you agree.

 

Oh, by the way, if you have a really bright ‘student’ you might find that they prove some of these Conjectures using algebra. This method can be used to solve Conjectures 1 and 2 but it’s not the nicest way. We show some methods in Part II.