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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 ÷ 3 = 6 Remainder = 0 | 0 * 70 = 0 |
18 ÷ 5 = 3 Remainder = 3 | 3 * 21 = 63 |
18 ÷ 7 = 2 Remainder = 4 | 4 * 15 = 60 |
After testing out a few times, we managed to explain how it works for multiples of 5:
20 ÷ 3 = 6 Remainder = 2 | 2 * 70 = 140 |
20 ÷ 5 = 4 Remainder = 0 | 0 * 21 = 21 |
20 ÷ 7 = 2 Remainder = 6 | 6 * 15 = 90 |
After testing out a few times, we managed to explain how it works for multiples of 3 and 5:
60 ÷ 3 = 20 Remainder = 0 | 0 * 70 = 0 |
60 ÷ 5 = 12 Remainder = 0 | 0 * 21 = 0 |
60 ÷ 7 = 8 Remainder = 4 | 4 * 15 = 60 |
After testing out a few times, we managed to explain how it works for prime numbers:
13 ÷ 3 = 4 Remainder = 1 | 1 * 70 = 70 |
13 ÷ 5 = 2 Remainder = 3 | 3 * 21 = 63 |
13 ÷ 7 = 1 Remainder = 6 | 6 * 15 = 90 |
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
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!
Choose two digits and arrange them to make two double-digit numbers. Now add your double-digit numbers. Now add your single digit numbers. Divide your double-digit answer by your single-digit answer. Try lots of examples. What happens? Can you explain it?
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.