>>12120676This might be overly complicated, but here's how I did it the first time around.
Three pairs of numbers on opposite sides need to add up to the same number: top left and bottom right, top and bottom, top right and bottom left. Those six numbers need to be symmetrical around some axis, so their consecutive differences need to be a palindrome. Consecutive differences of the seven given numbers are 2, 4, 2, 4, 2, 4. For the ends of the consecutive differences of the six numbers to be the same (necessary to be a palindrome), one of the ends (5 or 23) has to be removed from the list and put in the center of the diagram (to get rid of the 2 at the start or the 4 at the end).
Case 1: 5 in the center. Axis of symmetry of the remaining numbers is 15, so the magic total is 2*15+5=35=5 mod 6. Numbers on the ends mod 6 are 1, 5, 1, 5, 1, 5, so we need one 1 and two 5s to reach the desired total on both top and bottom rows. But there are three 1s and three 5s in the six numbers, not the two 1s and four 5s we need.
Case 2: 23 in the center. Axis of symmetry of the remaining numbers is 12, so the magic total is 2*12+23=47=5 mod 6 as well. Numbers on the ends mod 6 are 5, 1, 5, 1, 5, 1. Same argument as Case 1 for the rest.
Anyone have a shorter proof of impossibility?
