## Cyclical figurate numbers

### Problem 61

Published on Friday, 16th January 2004, 06:00 pm; Solved by 14880; Difficulty rating: 20%Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:

Triangle | P_{3,n}=n(n+1)/2 |
1, 3, 6, 10, 15, ... | ||

Square | P_{4,n}=n^{2} |
1, 4, 9, 16, 25, ... | ||

Pentagonal | P_{5,n}=n(3n−1)/2 |
1, 5, 12, 22, 35, ... | ||

Hexagonal | P_{6,n}=n(2n−1) |
1, 6, 15, 28, 45, ... | ||

Heptagonal | P_{7,n}=n(5n−3)/2 |
1, 7, 18, 34, 55, ... | ||

Octagonal | P_{8,n}=n(3n−2) |
1, 8, 21, 40, 65, ... |

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

- 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).
- 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. - 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.