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.

Number lines (1-100) (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.