Always 3 solutions
If the 5 numbers contain more evens than odds, the number at the botttom will be even. Otherwise odd.
If the average of the five numbers is n, the totals of each arm will be 3n-1, 3n and 3n+1
Alison's method to find a magic V with a particular arm total:
Say we wanted a magic V with an arm total of 157. There are three possible 1-5 magic Vs with totals of 8, 9 and 10. 157 is one more than a multiple of 3, so I need to start from the magic V with a total of 10. To get from 10 to 157, I need to add 147, which is 3 times 49. So if I add 49 to each of the numbers in the magic V with a total of 10, I'll get a total of 157.
| 1 | 3 | 50 | 52 | |||||||
| 4 | 2 | 53 | 51 | |||||||
| 5 | 54 |
Sometimes 3 solutions (e.g. when using the numbers 2 to 7)
Sometimes 2 solutions (e.g. when using the numbers 4 to 9)
Sometimes 1 solutions (e.g. when using the numbers 6 to 11)
Sometimes no solutions (e.g. when using the numbers 11 to 16)
The number at the vertex will always be odd.
Always 7 solutions.
If the 7 consecutive numbers contain 4 evens and 3 odds, the number at the centre of the N will be odd. Otherwise even.