Ten Tiles II
Identify numbers divisible by 5;
Count large but relatively straightforward sets.
Devise and use problem solving strategies to explore situations mathematically (be systematic, think, draw a diagram).
This problem is the second of a series of three Ten Tiles I, Ten Tiles II and Ten Tiles III. If you haven’t done any systematic counting for a while it might be a good idea to start with At The Movies, (Level 3) though.
This problem is the second of a series of problems involving numbers with a given factor. Students first of all have to be able to identify the property that a number must have, to have this number as a factor. In this first problem, this is relatively easy. A number is divisible by 5 if its last digit is 0 or 5.
The next step is to be able to count all the numbers with this property in an efficient manner. As the answer to this question is nearly 6,000 the counting has to be done by some efficient method. It may help to look back at At The Movies or Ten Tiles I to see how this might be done. On the other hand, using a simpler case might be a good strategy to use here.
Counting is an important part of mathematics and many methods of counting sets of objects have been devised. They are usually to be found in a course on Discrete Mathematics or Combinatorics. Such courses are usually only found at university level.
This particular problem is about systematic counting. What has to be done is to find ways to count all possible numbers of a given type without writing them all down. This is necessary because writing them all down is a tedious proposition. What’s more, there are often so many of them that in writing them all down it is highly likely that some will be missed.
The Problem
Jim has ten tiles with a different digit on each of them. He plays around and discovers all sorts of things with them. Josey joins in.
When Frances sees the other two playing with the tiles she gets interested. She can make a lot of five-digit numbers that are divisible by five. How many can she make altogether?
Teaching sequence
- Use a classroom discussion to revise the idea of factors and how you can identify numbers that have given factors.
What are the factors of 24? 36?
What does it mean to say that 35 is divisible by 7?
How do you know if a number is divisible by 2? 10? 5? 3? - Pose the problem.
Can you give me a number that is divisible by 5?
Can you give me a five-digit number that is divisible by 5?
How about a five-digit number that contains each of 0, 1, 2, 3, 4, 5, 6, 7, 8, 9? - Let the students tackle the problem in their groups. For groups that are having trouble getting started ask
How many numbers can you make using just one 0, 1 and 2, that are divisible by 5?
How about if we use 0, 1, 2, 3 just once each? How many then?
Can you see a pattern here? - After a suitable time, bring the students back together to discuss how they solved the problem. The quicker students might also have had time to tackle the Extension problem.
- Get the students to write up their answers.
Extension to the problem
Using the Jim’s tiles, how many four-digit numbers are there that are divisible by four?
Solution
For a number to be divisible by five, it has to end with a 5 or a 0. So we have two choices for the last digit. If 0 is the last digit, then there are 9 choices for the first digit of the five-digit number, 8 for the second digit, 7 for the third and 6 for the fourth. So altogether we have 9 x 8 x 7 x 6 = 3024 five-digit numbers that are divisible by five with a 0 in the units position. On the other hand, if 5 is the last digit, there are 8 possibilities for the first digit (all ten digits less the 5 and the 0), then 8 for the second (0 is allowed here), 7 for the third and 6 for the fourth to give a total of 8 x 8 x 7 x 6 = 2688.
Altogether there are 3024 + 2688 = 5712 five-digit numbers divisible by five.
(Again this can be approached by a subtraction method – see Ten Tiles I.)
Solution to the extension
Four-digit numbers divisible by four: Here the rule is that the number is divisible by 4 if the last two digits are divisible by four. In total there are 25 potential last two digits that are divisible by four but three of these (00, 44 and 88) repeat a digit and so cannot be used here. There are 6 endings that have a 0 (04, 08, 20, 40, 60, 80). These give rise to 8 ´ 7 numbers each. There are 16 endings that do not have a 0 (12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 86) and these give rise to 7 x 7 numbers. Altogether then there are 6 x 8 x 7 + 16 x 7 x 7 = 1120 four-digit numbers divisible by four.
| Attachment | Size |
|---|---|
| TenTiles2.pdf | 61.21 KB |
| TenTiles2Maori.pdf | 77.45 KB |
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