Challenge Level

Thanks to the group of pupils from River Valley High School, Singapore, (Chong Ching Tong, Chen Wei Jian and Teo Seow Tian). They demonstrated that the rule worked for a range of numbers and identified some patterns; namely that 70, 21 and 15 are multiples of combinations of two of the divisors (3, 5 and 7) and that 105 is the lowest common multiple of 3, 5 and 7. They looked at combinations of multiples of 3, 5 and 7, some of which are given below. However, they were not quite able to generalise what they had discovered so I think there is a little more work to do on this. I have given you a hint for the next step at the end.

After testing out a few times, we managed to explain how it works for multiples of 3:

18 Ã· 3 = 6 Remainder = 0 | 0 * 70 = 0 |

18 Ã· 5 = 3 Remainder = 3 | 3 * 21 = 63 |

18 Ã· 7 = 2 Remainder = 4 | 4 * 15 = 60 |

- The reason 15 and 21 was used because they are the multiples of three
- The reason why 70 was used because 70 cannot be divided by 3
- Therefore the sum of 60 and 63 is actually a multiple of 3
- Besides, 105 is also the lowest common multiple of 3, 5 and 7
- That was why it worked as the sum which is a multiple of 3 minus 105, we will get back the original answer

After testing out a few times, we managed to explain how it works for multiples of 5:

20 Ã· 3 = 6 Remainder = 2 | 2 * 70 = 140 |

20 Ã· 5 = 4 Remainder = 0 | 0 * 21 = 21 |

20 Ã· 7 = 2 Remainder = 6 | 6 * 15 = 90 |

- The reason 15 and 70 was used because they are the multiples of five
- The reason why 21 was used because 21 cannot be divided by 5
- Therefore the sum of 90 and 140 is actually a multiple of 5
- Besides, 105 is also the lowest common multiple of 3, 5 and 7
- That was why it worked as the sum which is a multiple of 5 minus 105, we will get back the original answer
- Besides, 105 is also the lowest common multiple of 3, 5 and 7
- That was why it worked as the sum which is a multiple of 7 minus 105, we will get back the original answer

After testing out a few times, we managed to explain how it works for multiples of 3 and 5:

60 Ã· 3 = 20 Remainder = 0 | 0 * 70 = 0 |

60 Ã· 5 = 12 Remainder = 0 | 0 * 21 = 0 |

60 Ã· 7 = 8 Remainder = 4 | 4 * 15 = 60 |

- The reason 15 was used because 15 is the lowest common multiple of three and five
- The reason why 21 and 70 was used because 21 and 70 cannot be divided by both 3 and 5
- Therefore without subtracting from 105, we can get back the original number

After testing out a few times, we managed to explain how it works for prime numbers:

13 Ã· 3 = 4 Remainder = 1 | 1 * 70 = 70 |

13 Ã· 5 = 2 Remainder = 3 | 3 * 21 = 63 |

13 Ã· 7 = 1 Remainder = 6 | 6 * 15 = 90 |

- In this case, we have to use 210 to be subtracted from the sum of 70, 63 and 90, which is 223
- Besides, 210 is also a common multiple of 3, 5 and 7
- Then we can get back the original number

**Hint**

A possible route to the solution might be to use the idea that when one number (n say) is divided by another number (d) the answer can be written as a whole number (q) with a remainder (r).

That is n/d = q Remainder r

This can also be thought of as: n = qd + r

So, for example, 32/5 = 6 Remainder 2

This can be thought of as : 32 = 6 x 5 + 2