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Find the number which has 8 divisors, such that the product of the divisors is 331776. Skeleton

Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.

One O Five

Age 11 to 14 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 ÷ 3 = 6 Remainder = 0 0 * 70 = 0 18 ÷ 5 = 3 Remainder = 3 3 * 21 = 63 18 ÷ 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 ÷ 3 = 6 Remainder = 2 2 * 70 = 140 20 ÷ 5 = 4 Remainder = 0 0 * 21 = 21 20 ÷ 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 ÷ 3 = 20 Remainder = 0 0 * 70 = 0 60 ÷ 5 = 12 Remainder = 0 0 * 21 = 0 60 ÷ 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 ÷ 3 = 4 Remainder = 1 1 * 70 = 70 13 ÷ 5 = 2 Remainder = 3 3 * 21 = 63 13 ÷ 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