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Here are some splendid solutions, all from Moorfield Junior School. Well done Moorfield! Each month we get lots of superb work sent in by Moorfield; we hope they'll try some more challenges having made this impressive debut.
The first solution is from Adam and Anthony.
"We changed the rules by only being allowed to take 1, 2, 3 or 4 and we started with 25 and counted down to 0. We said the key numbers were in the 5 times tables. Your opponent has to go first then whatever your opponent takes you take the amount that adds up to 5. For example: if Mr C took 3 then I would take 2 so it equals one of the key numbers which are 20, 15, 10 and 5."
The second is much the same from Adam F and Niall.
"In the game we could only take 1, 2 ,3 or 4 cubes and we didn't go first. There were 25 cubes and every time our opponent took a number of cubes we took the number that made it up to 5. The key numbers we had to leave our opponent were: 20, 15, 10, 5."
Now a solution from Yuji, Matt and Jimbo all from Moorfield Junior School.
"We made our target number 98 and the maximum number we could pick was 7. The first pick was 2, then the computer picked 7 and we picked 1 which made our total 10. Then what ever the computer picked we made it to 8. We did that for 11 times because 8 times 11 equals 88, add to the 10 we made early and that makes 98, the total."
The last one is from Steven and Matthew.
"We made our target number 98. The highest number we could pick was 7 and the lowest number was 1. The key number was 90 because if we made the total to 90 then the lowest number he could pick was 1 so we would pick 7 to make it 98. We tried different numbers depending on what the computer chose. The numbers when the computer had to choose were: 10, 18, 26, 34, 42, 50, 58, 66, 74, 82 and 90. We tried going first and then going second but we found that going first was the better choice than going second. We made the first total to 10 and then we made it to 8 so the total would be 98."
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!
Is there an efficient way to work out how many factors a large number has?
15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?