Problem 61 - Find the sum of the only set of six 4-digit figurate numbers with a cyclic property.

링크

Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:

Triangle		P3,n=n(n+1)/2		1, 3, 6, 10, 15, ...
Square		P4,n=n2		1, 4, 9, 16, 25, ...
Pentagonal		P5,n=n(3n1)/2		1, 5, 12, 22, 35, ...
Hexagonal		P6,n=n(2n1)		1, 6, 15, 28, 45, ...
Heptagonal		P7,n=n(5n3)/2		1, 7, 18, 34, 55, ...
Octagonal		P8,n=n(3n2)		1, 8, 21, 40, 65, ...

The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.

1. The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
2. Each polygonal type: triangle (P_3,127=8128), square (P_4,91=8281), and pentagonal (P_5,44=2882), is represented by a different number in the set.
3. This is the only set of 4-digit numbers with this property.

Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.

Python

Tri = [x*(x+1)/2 for x in range(45, 141)] Squ = [x*x for x in range(32, 100)] Pen = [x*(3*x-1)/2 for x in range(26, 82)] Hex = [x*(2*x-1) for x in range(23, 71)] Hep = [x*(5*x-3)/2 for x in range(21, 64)] Oct = [x*(3*x-2) for x in range(19, 59)] List = [Tri, Squ, Pen, Hex, Hep, Oct] Use = [0] * 6 def go(x, l, last, sum): if l == 6: if str(Use[0])[:2] == str(Use[last])[2:]: print Use, sum return for b in range(6): if Use[b] > 0: continue; for y in List[b]: if str(x)[2:] == str(y)[:2]: Use[b] = y go(y, l+1, b, sum + y) Use[b] = 0 for x in List[0]: Use[0] = x go(x, 1, 0, x)