Two ladybirds, Freda and Fred, are playing a game on a numberline.
Fred can jump three numbers at a time and Freda can only jump two.
Fred starts at 1 and Freda starts at 30.
If they both jump together, who gets to 100 first and how long do they have to wait for the other one?
This problem involves students in finding number patterns and solving algebraic problems. Have the students first make an estimate and prediction. Students may approach the problem in a range of ways including drawing jumps along a number line, making a table, seeing a relationship and using guess and check, using division.
Note that In Extension 1, Fred and Freda don’t land exactly on the number 100.
Copymaster of the problem (English)
Number lines (1100) (or use metre rulers)
Problem
Two ladybirds, Freda and Fred, are playing a game on a numberline. Fred can jump three numbers at a time and Freda can only jump two. Fred starts at 1 and Freda starts at 30. If they both jump together, who gets to 100 first and how long do they have to wait for the other one?
Teaching sequence
 Introduce the 2 characters.
 Pose the problem and have students demonstrate and explain the movement of each character.
 Discuss possible strategies and ways in which the students will record their solutions.
 As the students work ask questions that focus on the thinking that they are using.
What are you doing? Why are you solving it this way?
Who do you think will get there first? Why do you think that?
What can you tell me about the numbers in Freda's pattern?
What can you tell me about the numbers in Fred's pattern?  Share solutions
 If the students have all acted or drawn the problem ask them to look back and think about other ways that they could have used to solve the problem eg, use division.
Extension to the problem

Let Freda start on 51 and jump two numbers at a time. Let Fred start on 1 and jump four numbers at a time. Who is first to 100?

In Extension 1, on what number does the overtaking take place?
Solution
This can be done by using equipment, by drawing, by algebra (see Toothpick Squares problem), or by using a table such as this.

0

1

2

3

4

5

6

7

8

9

10

Freda

30

32

34

36

38

40

42

44

46

48

50

Fred

1

4

7

10

13

16

19

22

25

28

31














11

12

13

14

15

16

17

18

19

20

21

Freda

52

54

56

58

60

62

64

66

68

70

72

Fred

34

37

40

43

46

49

52

55

58

61

64














22

23

24

25

26

27

28

29

30

31

32

Freda

74

76

78

80

82

84

86

88

90

92

94

Fred

67

70

73

76

79

82

85

88

91

94

97














33

34

35









Freda

96

98

100









Fred

100










The table shows that Fred gets to the 100^{th} square and has to wait two jumps for Freda to catch up.
The table is a valid (if tedious) way to solve the problem. Freda is jumping on the squares numbered 2# + 30, then she gets to the 100th square when 2# + 30 = 100. This is when # = 35 (check this with the table).
On the other hand, Fred is using the pattern 3# + 1. So he gets to 100 when 3# + 1 = 100. In other words when 3# = 99 or when # = 33. The table shows that Fred gets to the 100th square in 33 jumps, two ahead of Freda.
Solution to the extensions:
Using the equation 2# + 51 = 100 for Freda, guess and check can be used to see that # must be bigger than 24 (2 x 24 + 51 = 99) and less than 25 (2 x 25 + 51 = 101). So Freda will need 25 steps to get to the 100th square.
Fred's equation is 4# + 1= 100. By using guess and check, a table, or some other means, it can be seen that # must be more than 24 (4 x 24 + 1 = 97) and less than 25 (4 x 25 + 1 = 101).
They both landed on the 101st square on their 25th jump.
If 2# + 51 = 4# + 1, then 2# = 50, so # = 25. They land together at the end of the 25th jump but that is the first time that they are together. Freda is ahead up to that point.