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Red Even
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?
Prime Magic
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?
Stage: 2 Challenge Level: 
Somin, Kim and Georgia from the American International School Chennai in India sent in the following:-
To do this problem, we started with the lowest circle on the V, because this would affect both arms. Then, we took the highest of the remaining numbers, and the lowest of the remaining numbers and added them together. We then compared the sum of those to the sum of the two middle/in-between remaining numbers. If the sums were the same, we entered the addins used to make those two sums into the 2 spots on each arm of the V.
We know that we found every solution because when trying to mix up the pairs of numbers to make other sums, none of the sums were equal. The only way you could make another V is if you mirrored one we already made, or switched the places of the top and middle circles of each arm. A pattern we noticed during this project is that every even number in the sequence (the 2nd and 4th numbers, not 2 and 4, for example) did not work as a center number. The other numbers could not be paired to create equal sums. We are including a picture of a small picture we made representing our work.
From Sudbury Primary School in Suffolk we had solutions sent in from Maisy, Jessica, Luke, Amelia, Molli & Amelia, Elena, Lili, Ben, Tashi, Megan C, Charlotte and Belle.
Here is Maisy and Jessica's submission;
We first found our first three Vs then realised we could find 8 solutions, by just changing one V. This also happened with the other two Vs. We multiplied 8 by 3 because our three vs had 8 solutions. Our answer was 24 solutions.
E.g. To start with, we had 23541 and we swapped the sides to get 41532.
Then we changed round the sides to get 14523. Afterwards, we just swapped one side then another till we couldn't swap it anymore without getting a solution we already had.
Here is Mollie and Amelia's results
To solve all of the solutions,first we decided to put 1 as the bottom number and worked systematically to get all 8 solutions. Then, we picked then number 3 for the bottom number as we knew it couldn't be even. After finding 8 answers, we got 5 for the number at the bottom. We found all the answers for that number. We stopped here since we couldn't go above 5.There are 24 solutions.
Matty from The British School Manila in the Philippines sent in the following exhaustive solution
If we have the numbers 1, 2, 3, 4 and 5, and our goal is to arrange all 5 numbers in a way such that the 2 'arms' have the same total. If we get the total of all the give numbers, we get 15. In the middle, we must have an odd number, because if we have an even, say 2, 15-2=13 and 13 is not divisible among the other 2 circles of arms. If we pick 1 as the middle,
then each side is required to have a total of 7 on the other 2 circles. For 3, we need a total of 6 on the 2 circles, and for 5, we need a total of 5 on the other 2 circles. Thanks for the question, Nrich. ( which was published at http://bit.ly/2pCieMH)
5,2,1,3,4 5,2,1,4,3 2,5,1,3,4 2,5,1,4,3 3,4,1,5,2 3,4,1,2,5
4,3,1,5,2 4,3,1,2,5 1,5,3,2,4 1,5,3,4,2 5,1,3,2,4 5,1,3,4,2
2,4,3,1,5 2,4,3,5,1 4,2,3,1,5 4,2,3,5,1 1,4,5,2,3 1,4,5,3,2
4,1,5,2,3 4,1,5,3,2 2,3,5,1,4 2,3,5,4,1 3,2,5,1,4 3,2,5,4,1
We have 24 different permutations. Here is the formula:
Slot 1: 4 possibilities
Slot 2: 1 possibility
Slot 3: 2 possibilities
Slot 4: 1 possibility
Middle number: 3 possibilities (3 odd numbers between 1 to 5, inclusive)
So 3*(4*1*2*1) = 24.
We also had some very interesting solutions from North London Collegiate School that are certainly worth viewing here.
magic Vs Y4.pdf magic Vs 3S group.pdf magic Vs Ronis group.pdf
Thank you all so much for these solutions you certainly went about it in good ways to find how many there were.
To do this problem, we started with the lowest circle on the V, because this would affect both arms. Then, we took the highest of the remaining numbers, and the lowest of the remaining numbers and added them together. We then compared the sum of those to the sum of the two middle/in-between remaining numbers. If the sums were the same, we entered the addins used to make those two sums into the 2 spots on each arm of the V.
We know that we found every solution because when trying to mix up the pairs of numbers to make other sums, none of the sums were equal. The only way you could make another V is if you mirrored one we already made, or switched the places of the top and middle circles of each arm. A pattern we noticed during this project is that every even number in the sequence (the 2nd and 4th numbers, not 2 and 4, for example) did not work as a center number. The other numbers could not be paired to create equal sums. We are including a picture of a small picture we made representing our work.
From Sudbury Primary School in Suffolk we had solutions sent in from Maisy, Jessica, Luke, Amelia, Molli & Amelia, Elena, Lili, Ben, Tashi, Megan C, Charlotte and Belle.
Here is Maisy and Jessica's submission;
We first found our first three Vs then realised we could find 8 solutions, by just changing one V. This also happened with the other two Vs. We multiplied 8 by 3 because our three vs had 8 solutions. Our answer was 24 solutions.
E.g. To start with, we had 23541 and we swapped the sides to get 41532.
Then we changed round the sides to get 14523. Afterwards, we just swapped one side then another till we couldn't swap it anymore without getting a solution we already had.
Here is Mollie and Amelia's results
To solve all of the solutions,first we decided to put 1 as the bottom number and worked systematically to get all 8 solutions. Then, we picked then number 3 for the bottom number as we knew it couldn't be even. After finding 8 answers, we got 5 for the number at the bottom. We found all the answers for that number. We stopped here since we couldn't go above 5.There are 24 solutions.
Matty from The British School Manila in the Philippines sent in the following exhaustive solution
If we have the numbers 1, 2, 3, 4 and 5, and our goal is to arrange all 5 numbers in a way such that the 2 'arms' have the same total. If we get the total of all the give numbers, we get 15. In the middle, we must have an odd number, because if we have an even, say 2, 15-2=13 and 13 is not divisible among the other 2 circles of arms. If we pick 1 as the middle,
then each side is required to have a total of 7 on the other 2 circles. For 3, we need a total of 6 on the 2 circles, and for 5, we need a total of 5 on the other 2 circles. Thanks for the question, Nrich. ( which was published at http://bit.ly/2pCieMH)
5,2,1,3,4 5,2,1,4,3 2,5,1,3,4 2,5,1,4,3 3,4,1,5,2 3,4,1,2,5
4,3,1,5,2 4,3,1,2,5 1,5,3,2,4 1,5,3,4,2 5,1,3,2,4 5,1,3,4,2
2,4,3,1,5 2,4,3,5,1 4,2,3,1,5 4,2,3,5,1 1,4,5,2,3 1,4,5,3,2
4,1,5,2,3 4,1,5,3,2 2,3,5,1,4 2,3,5,4,1 3,2,5,1,4 3,2,5,4,1
We have 24 different permutations. Here is the formula:
Slot 1: 4 possibilities
Slot 2: 1 possibility
Slot 3: 2 possibilities
Slot 4: 1 possibility
Middle number: 3 possibilities (3 odd numbers between 1 to 5, inclusive)
So 3*(4*1*2*1) = 24.
We also had some very interesting solutions from North London Collegiate School that are certainly worth viewing here.
magic Vs Y4.pdf magic Vs 3S group.pdf magic Vs Ronis group.pdf
Thank you all so much for these solutions you certainly went about it in good ways to find how many there were.
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NRICH is part of the family of activities in the Millennium Mathematics Project.
NRICH is part of the family of activities in the Millennium Mathematics Project.