Follow the clues to find the mystery number.
You have 4 red and 5 blue counters. How many ways can they be
placed on a 3 by 3 grid so that all the rows columns and diagonals
have an even number of red counters?
Place the numbers 1, 2, 3,..., 9 one on each square of a 3 by 3 grid so that all the rows and columns add up to a prime number. How many different solutions can you find?
I am extremely impressed by the solutions sent in to the Magic Vs problem, so well done to you all. Tom from Crudgington Primary, Krishan from Buckingham College Prep School, Harriet and Laura from Shincliffe C of E Primary, Nicholas and Jared from Clifton Hill Primary, Andy from Garden International School, Jonathan from DGS, Lucas from Grinling Gibbons Primary and
Scott from Echline Primary explained that there were three different Magic Vs for the numbers 1-5.
Emma and Charlotte from Greenacre School for Girls said:
This is very clearly explained, thank you. Yujin from Dubai International Academy explained this in a slightly different way:
... if there are two even numbers and three odd numbers; you put the odd number at the bottom: even + even = odd + odd.
You can't put the even at the bottom because even + odd does not equal to odd + odd.
That's right. Gemma, Dylan, Faazil and Gabriel from Dubai English Speaking School had a good way of working which helped them explain why odd numbers have to go at the bottom:
Bazahir from Ray Lodge had a good way of approaching the problem:
That's a good idea, Bazahir.
So, the three solutions will be one with 1 at the bottom of the V, one with 3 at the bottom and one with 5 at the bottom:
However, some of you decided that you can rearrange the numbers on the arms of the V, keeping the same number at the bottom, to make different solutions. Alex, Zara, Frankie, Akeel and Molly from Talbot Primary School, Yujin and Gijs from Dubai International Academy and James from Crudgington Primary argued that there are 24 Magic Vs in total for the numbers 1-5. Gijs
took the example with the 3 at the bottom of the V. He wrote:
Here are the eight different Magic Vs with 3 in the bottom:
Can you see how all these have been generated? Perhaps you can do the same with 1 at the bottom of the V to convince yourself. So, if we assume these are all different Magic Vs, there would indeed be 24 in total.
Many of you then looked into Magic Vs which used different numbers and Magic Vs which were different sizes.Gemma, Dylan, Faazil and Gabriel from the Dubai English Speaking School did a very thorough investigation:
Very interesting! Did you try seven digits which started with an even number, I wonder? What would happen with 2, 3, 4, 5, 6, 7, 8, for example? They continued:
Many of you I've already mentioned, and Milne from East Hoathly, agreed with what they have said about the numbers 2-6.
You can see a summary of George's results here . Will it always be possible to create Magic Vs in this way, do you think? How do we know that we're going to be able to get each arm to add to the same total?
Ross and Brandon from Ashford Hill represented the five-digit Magic Vs alebraically which you may find helpful. They said:
What a wealth of responses - thank you to everyone who sent something in and I'm sorry that we can't mention everyone. If you would like to contribute anything more to the solution, do email us here firstname.lastname@example.org