lessons
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Counting Cubes                                    Number, Level 2

Kelsey loves to count. One day Ms Martin put a pile of cubes on Kelsey’s desk. Kelsey began to count the cubes. She told Ms Martin the following facts about the cubes:
When I count the cubes by two I have one left over.
When I count the cubes by three I have one left over.
When I count the cubes by four I have one left over.
When I count by seven I have none left over.

From this information can Ms Martin work out how many cubes Kelsey has?

What is this problem about?

This problem involves skip counting and implies division with remainders. Using the concept of the least common multiple can solve problems like this.

This kind of problem is very useful for finding number patterns. As the children go through the problem they should begin to see some patterns that will tell them more about the solution to Kelsey’s problem.

The other interesting thing about this problem is that there is redundant information. Not all of the information is of help because it is already contained as part of other information. If children can see this it will cut down the number of calculations that they will have to make.

Achievement Objectives
Number, level 2
- demonstrate the ability to use the multiplication facts

Mathematical Process
- devise and use problem solving strategies (make an organised list, think)

Resources
Hundreds board
Cubes to introduce the problem
Blackline master of the problem (English)

Specific Learning Outcomes
The children will be able to:
- skip count in 2s, 3s, 4s and 7s

Lesson Sequence

  1. Start counting backwards in 2s from 30. Ask the children to join in when they have worked out the pattern.
    30 ..28..26..24..22..
    Try other skip counting patterns for the children to follow.
    1..3..5..7..9
    Let the children invent their own counting patterns.
  2. Introduce the problem using a pile of cubes. Check that the children understand the problem before getting them to solve it with a partner.
  3. As the children solve the problem encourage them to use words or diagrams to explain their thinking.
  4. Ask questions that allow you to determine if they can understand the ideas of counting by 2s and having a remainder. Do they realise that if one cube was left after counting by 4s then the number of cubes must be in the sequence 1..5..9..13..?
  5. Make sure that the children are recording their solution so that they can convince themselves and others that they have attained a correct solution.
  6. Share records.

Other Contexts
Tapes
Shells

Extension
Is there more than solution? How many solutions are there?

Solution
The lowest number of cubes that Kelsey could have is 49.
The children could show this on a number line or in an organised list. I have used a list to show that number 49 is the first number to appear in each of the number sequences described by Kelsey.


2s
1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
31
33
35
37
39
41
43
45
47
49
3s
1
4
7
10
13
16
19
22
25
28
31
34
37
40
43
46
49
4s
1
5
9
13
17
21
25
29
33
37
41
45
49
7s
7
14
21
28
35
42
49

And now for some thinking! If you look at the first and third columns you will see that the numbers in the third column are already in the first column. This suggests that we don’t need the third column. However, the redundant column is actually the first column. This is because only numbers in both the columns fit the information. But the numbers in the first column are not in both columns whereas the numbers in the third column are. Hence we can forget about the numbers in the first column as far as the calculations go. As there are a lot more numbers to calculate in the first column, this observation will save a lot of work, especially in the attempt to do the Extension.

Extension: By continuing the organised list, we see that the next answer comes at 49 + 84 = 133 (or 21 + 28 if the ‘count by three and have one left over’ option is omitted). What is the next answer? You need to go a further 84 to get 133 + 84 = 217. Answers keep coming every 84 from there. What is important about 84? Well it is 3 ´ 4 ´ 7. You see, any further answer has to be 49 plus a multiple of 3 to be included in the 3s column; and it has to be 49 plus a multiple of 4 to be included in the 4s column; and it has to be 49 plus a multiple of 7 to be included in the 7s column. These numbers only coincide every 84 = 3 ´ 4 ´ 7 numbers.

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