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Megan from the Thomas Deacon Academy used a spreadsheet to find all 28 pairs of numbers that add up to a multiple of 11. This is how she did it:
To work out the first box I started with 9, adding the rest of the numbers, and then moved on to 46, but didn't do 46 add 9 since it had already been done. Ithen carried this out through out the table but making sure I had not done any in front of the number I was working on as it would have already been done.
Then I highlighted all the answers that are in the 11 times table.
In the attached document are my notes.
Alex from St. Anne's School noticed something special about the numbers in some of the pairs:
There were over 25 different pairs of numbers wich totalled a multiple of 11.
We noticed that the numbers we added to 9, 20 and 31 were all the same:
9+46=55
9+79=88
9+13=22
9+90=99
9+2=11
20+46=66
20+79=99
20+13=33
20+90=110
20+2=22
31+46=77
31+79=110
31+13=44
31+90=121
31+2=33
The difference between 9 and 20 is 11 and the difference between 20 and 31 is 11.
When we added 11 to 31 and made 42. We added 46, 79, 13, 90 and 2 to this number and found that each result was a multiple of 11.
Jack and Zaim from London sent us this very clear explanation of why this happens.
Curtis from Shatin College used a similar strategy :
I divided all of the numbers by 11 and wrote down their remainders. Then I wrote a chart of them, in the same spot. After that, I checked in the remainder box for any pairs that added up to 11. Finally I transfered the numbers back on to the provided grid, and came up with 28 solutions for both the 11's and the 13's.
Adil from Valentines High School discovered the same property of the numbers that could be paired:
Well done to you all.
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 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.