Jim has ten tiles with a different digit on each of them.

He discovers that he can use the tiles to make a lot of nine-digit numbers that are divisible by nine.

How many can he make?

This problem investigates numbers with a given factor. Students must first be able to identify the property of a number to determine its factors. For example: A number is divisible by 9 if the sum of its digits is divisible by 9. The same test holds for divisibility by 3. Students must then apply a systematic approach to count in an efficient manner all the numbers with this property, ensuring that none has been missed.

Related problems include: Ten Tiles I and Ten Tiles III.

Note: At The Movies, (Level 3) may be a useful starting place for this problem.

### The Problem

Jim has ten tiles with a different digit on each of them.

He discovers that he can use the tiles to make a lot of nine-digit numbers that are divisible by nine.

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?*

How do you know if a number is divisible by 2? 10? 5? 3? - Pose the problem and ask:
*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? - As students work on the problem, if any has 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? - As solutions emerge, share and discuss as a class how they solved the problem.

As appropriate, encourage students to try the Extension problem. - Have the students to write up their answers.

#### Extension

Josey uses Jim’s tiles to 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: With divisibility by three, 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 | 42 | 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 | 48 | 0,6,9 0,7,8 | 8 |

18 | 1,8,9 2,7,9 3,6,9 3,7,8 4,5,9 4,6,8 5,6,7 | 42 | ||

21 | 4,8,9 5,7,9 6,7,8 | 18 | ||

24 | 7,8,9 | 6 | ||

Total of numbers divisible by three | 180 | 48 |