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 5 if its last digit is 0 or 5. They must then apply a systematic approach to count in an efficient manner all the numbers with this property, ensuring that none has been missed.
Jim has ten tiles with a different digit on each of them. He makes a lot of five-digit numbers that are divisible by five. How many can he make altogether?
- 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 a number that is divisible by 5?
Can you give a five-digit number that is divisible by 5?
- 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 5?
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 to the problem
Using the Jim’s tiles, how many four-digit numbers are there that are divisible by four?
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.