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We had so many well-explained solutions to this problem - thank you all very much.
Adeline told us:
According to the first and second statements the only possibilities are 3, 9, 15, 21, 27 and 33.Kaitlyn from Mason Middle School made a table to help her:
The answer is 27. I know this because I used the process of elimination with the given hints. The way I did this was:Hint | Possible numbers |
The number is odd | 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35 |
The number is a multiple of three | 3,9,15,21,27,33 |
It is smaller than 7x4 (=28) | 3,9,15,21,27 |
The tens digit is even | 21,27 |
It is the greater of the two possibilities | 27 because that is greater than 21 |
Dylan from Oakland Township Elementary used a similar method, but he explains how he used the number grid itself to write on:
Thank you, Dylan. Crossing out or circling the numbers as you go along can help you keep track of what you're doing. Daniel from Engelfield Green Infants School did this too. Here is what Daniel wrote:
Here is a picture of Daniel's work:
Thank you again for all your contributions to this problem. I'm afraid there are too many to mention you all!
Arrange the numbers 1 to 6 in each set of circles below. The sum of each side of the triangle should equal the number in its centre.
Pat counts her sweets in different groups and both times she has some left over. How many sweets could she have had?