Choose any three by three square of dates on a calendar page. Circle any number on the top row, put a line through the other numbers that are in the same row and column as your circled number. Repeat this for a number of your choice from the second row. You should now have just one number left on the bottom row, circle it. Find the total for the three numbers circled. Compare this total with the number in the centre of the square. What do you find? Can you explain why this happens?
Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!
What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?
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:
After testing out a few times, we managed to explain how it works for multiples of 5:
After testing out a few times, we managed to explain how it works for multiples of 3 and 5:
After testing out a few times, we managed to explain how it works for prime numbers:
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