Dados los números del 1 al 3n, construya n ecuaciones de la forma a + b = c o axb = c de modo que cada número se use exactamente una vez. Por ejemplo:

 n=1 => 1+2=3 n=2 => 1+4=5, 2x3=6 n=3 => 4+5=9, 1+7=8, 2x3=6

La pregunta es, ¿existe una solución para cada n?

Intenté escribir un programa básico y se vuelve demasiado lento después de n = 14. Estas son las soluciones que tengo hasta ahora:

 1 ['1+2=3'] 2 ['2*3=6', '1+4=5'] 3 ['4+5=9', '1+7=8', '2*3=6'] 4 ['3+6=9', '1+10=11', '4+8=12', '2+5=7'] 5 ['2+8=10', '3+6=9', '1+13=14', '5+7=12', '11+4=15'] 6 ['3*5=15', '2+8=10', '4+14=18', '6+11=17', '7+9=16', '1+12=13'] 7 ['6+12=18', '3*5=15', '7+10=17', '1+20=21', '4+9=13', '2+14=16', '8+11=19'] 8 ['8+14=22', '6+12=18', '7+10=17', '2+19=21', '1+15=16', '11+13=24', '4+5=9', '3+20=23'] 9 ['6+19=25', '8+14=22', '4+13=17', '2+18=20', '1+26=27', '3+7=10', '9+15=24', '5+16=21', '11+12=23'] 10 ['6+19=25', '14+15=29', '11+17=28', '4+26=30', '2+18=20', '1+21=22', '3*9=27', '8+16=24', '5+7=12', '10+13=23'] 11 ['10+23=33', '6+19=25', '14+15=29', '11+17=28', '4+26=30', '2+18=20', '5+27=32', '1+12=13', '9+22=31', '3*7=21', '16+8=24'] 12 ['10+23=33', '3+29=32', '6+19=25', '15+21=36', '11+17=28', '8+14=22', '4+16=20', '7+27=34', '2*12=24', '1+30=31', '5+13=18', '9+26=35'] 13 ['10+23=33', '3+29=32', '7+30=37', '6+19=25', '5+34=39', '15+21=36', '11+17=28', '18+20=38', '4+31=35', '1+26=27', '9+13=22', '8+16=24', '2+12=14'] 14 ['10+23=33', '4+37=41', '3+29=32', '9+25=34', '15+21=36', '11+17=28', '8+14=22', '6+24=30', '13+27=40', '5*7=35', '2+18=20', '1+38=39', '12+19=31', '16+26=42']

Aquí está el código para el programa:

 import sys from itertools import combinations def main(n): r = set(range(1, n*3+1)) print(n, solve(n, r, [])) def solve(n, lst, solution): if not lst: if len(solution) != n: return False return solution for c in combinations(lst, 3): valid_solution = valid(c) if valid_solution: new_solution = solution + [valid_solution] result = solve(n, set(lst) - set(c), new_solution) if result: return result return False def valid(lst): a = lst[0] b = lst[1] c = lst[2] if a + b == c: return "%s+%s=%s" % (a, b, c) if a * b == c: return "%s*%s=%s" % (a, b, c) return False if __name__ == "__main__": n = int(sys.argv[1]) main(n)