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.
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.