Ten Tiles III
Identify numbers divisible by 9
Count large but relatively straightforward sets.
Devise and use problem solving strategies to explore situations mathematically (be systematic, think).
This problem is the last of a series of three Ten Tiles I, Ten Tiles II and Ten Tiles III. (It might be a good idea to start with At The Movies, (Level 3) though, especially if you haven’t done much on counting so far.)
This problem is the third 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 not quite so easy. A number is divisible by 9 if its last digit is if the sum of its digits is divisible by 9. (Check this out.) The same test holds for divisibility by 3.
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 700,000 the counting has to be done by some efficient method. It may help to look back at Ten Tiles I or Ten Tiles II to see how this might be done. On the other hand, using a simpler case might be a good strategy to use here.
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.
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.
The Problem
Jim has ten tiles with a different digit on each of them. He plays around and discovers that he can make quite a lot of nine-digit numbers that are divisible by nine by using the tiles. In fact how many can he make?
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? 9? 3? - Pose the problem.
Can you give me a number that is divisble by 9?
Can you give me a nine-digit number that is divisible by 9?
How about a nine-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 9?
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
Josey looks at Jim’s tiles and sees that she can make a lot of three-digit numbers that are divisible by three. How many can she make?
Solution
Nine-digit numbers divisible by nine: The rule is that the sum of the digits has to be divisible by 9. If 0 is not used then we use the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 and these add to 45. Any number using these nine digits is divisible by 9. There are 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362880.
If 0 is used then the other digits than may be used are 1, 2, 3, 4, 5, 6, 7 and 8. Since 0 can’t be used in the left hand most position, there are 8 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 322560 such numbers. Hence we have 685440 altogether.
Solution to the extension
Three-digit numbers divisible by three: The trick with divisibility by three is that the sum of the digits is divisible by three. So we need to find all numbers abc where a + b + c is a number divisible by 3. Since we are limited to the numbers 3, 6, 9, 12, 15, 18, 21 and 24. Now if none of a, b or c is 0, then we get 6 numbers for each sum (abc, acb, bac, bca, cab, cba). If one of a, b or c is 0, then we only get four numbers (ab0, a0b, ba0, b0a).
The sums have to be listed systematically. We show these in the table below along with the number of numbers that we get for each possibility. There are 158 three-digit numbers that are divisible by three altogether.
|
a + b + c |
No zero |
Number of numbers |
One 0 |
Number of numbers |
|
3 |
0,1,2 |
4 |
||
|
6 |
1,2,3 |
6 |
0,1,5 0,2,4 |
8 |
|
9 |
1,2,6 1,3,5 2,3,4 |
18 |
0,1,8 0,2,7 0,3,6 0,4,5 |
16 |
|
12 |
1,2,9 1,3,8 1,4,7 1,5,6 2,3,7 2,4,6 3,4,5 |
21 |
0,3,9 0,4,8 0,5,7 |
12 |
|
15 |
1,5,9 1,6,8 2,4,9, 2,5,8 2,6,7 3,4,8, 3,5,7 4,5,6 |
24 |
0,6,9 |
4 |
|
18 |
1,8,9 2,7,9 3,6,9 3,7,8 4,5,9 4,6,8 5,6,7 |
21 |
||
|
21 |
4,8,9 5,7,9 6,7,8 |
18 |
||
|
24 |
7,8,9 |
6 |
||
|
Total of numbers divisible by three |
114 |
44 |
| Attachment | Size |
|---|---|
| Ten Tiles III.pdf | 32.78 KB |
| Ten Tiles IIIMaori.pdf | 50.86 KB |
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