Challenge Level

Class 2YP at Madras College carried out their own investigation based on this problem and came up with some intriguing and surprising results. One problem they set themselves was to find the number of ways of writing 1999 as the sum of three odd numbers. We leave this for you as a challenge now. The account of 2YP's work will be published on 1 September.

Solutions accompanied by some explanation were sent in by Kang Hong Joo, age 14 of the Chinese High School, Singapore, by Daniel King and Chao Yu, year 7, Comberton Village College Cambridge and by class 8x from the Key Stage 3 Maths Club at Strabane Grammar School. Well done everyone who found all these solutions and also proved that it is impossible to find 6 odd numbers adding up to 15.

Class 8X said:

"We knew every solution had to contain a 1 since the next odd number is 3 and 4*3 = 12 and 8*3 = 24.

It is impossible to get a total of 15 with 6 odd numbers since adding 2 odd numbers gives an even number so if we add them in pairs we get 3 even numbers and adding even numbers gives an even number."

This is Kang Hong Joo's solution:

3 ways to add 4 **odd** numbers to get 10

1. | 1 | + | 1 | + | 1 | + | 7 | = | 10 |

2. | 1 | + | 1 | + | 3 | + | 5 | = | 10 |

3. | 1 | + | 3 | + | 3 | + | 3 | = | 10 |

11 ways to add 8 **odd** numbers to get 20

1. | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 13 | = | 20 |

2. | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 3 | + | 11 | = | 20 |

3. | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 5 | + | 9 | = | 20 |

4. | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 7 | + | 7 | = | 20 |

5. | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 3 | + | 3 | + | 9 | = | 20 |

6. | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 3 | + | 5 | + | 7 | = | 20 |

7. | 1 | + | 1 | + | 1 | + | 1 | + | 1 | + | 5 | + | 5 | + | 5 | = | 20 |

8. | 1 | + | 1 | + | 1 | + | 1 | + | 3 | + | 3 | + | 3 | + | 7 | = | 20 |

9. | 1 | + | 1 | + | 1 | + | 1 | + | 3 | + | 3 | + | 5 | + | 5 | = | 20 |

10. | 1 | + | 1 | + | 1 | + | 3 | + | 3 | + | 3 | + | 3 | + | 5 | = | 20 |

11. | 1 | + | 1 | + | 3 | + | 3 | + | 3 | + | 3 | + | 3 | + | 3 | = | 20 |

It is impossible to add 6 odd numbers to get 15 because the sum of an even number of odd numbers will always be an even number. The diagram shows how pairs of odd numbers give even numbers so the total must be even.

odd + odd | + | odd + odd | + | odd + odd |

V | V | V | ||

even | + | even | + | even |

W | ||||

even |