Problem 61
Cyclical figurate numbers
Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:
| Triangle | $P_{3,n}=n(n+1)/2$ | $1, 3, 6, 10, 15, \ldots$ |
| Square | $P_{4,n}=n^2$ | $1, 4, 9, 16, 25, \ldots$ |
| Pentagonal | $P_{5,n}=n(3n−1)/2$ | $1, 5, 12, 22, 35, \ldots$ |
| Hexagonal | $P_{6,n}=n(2n−1)$ | $1, 6, 15, 28, 45, \ldots$ |
| Heptagonal | $P_{7,n}=n(5n−3)/2$ | $1, 7, 18, 34, 55, \ldots$ |
| Octagonal | $P_{8,n}=n(3n−2)$ | $1, 8, 21, 40, 65, \ldots$ |
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.
循环的多边形数
三角形数、正方形数、五边形数、六边形数、七边形数和八边形数等等统称为多边形数。它们分别由如下的公式给出：
| 三角形数 | $P_{3,n}=n(n+1)/2$ | $1, 3, 6, 10, 15, \ldots$ |
| 正方形数 | $P_{4,n}=n^2$ | $1, 4, 9, 16, 25, \ldots$ |
| 五边形数 | $P_{5,n}=n(3n−1)/2$ | $1, 5, 12, 22, 35, \ldots$ |
| 六边形数 | $P_{6,n}=n(2n−1)$ | $1, 6, 15, 28, 45, \ldots$ |
| 七边形数 | $P_{7,n}=n(5n−3)/2$ | $1, 7, 18, 34, 55, \ldots$ |
| 八边形数 | $P_{8,n}=n(3n−2)$ | $1, 8, 21, 40, 65, \ldots$ |
由三个$4$位数$8128$、$2882$、$8281$构成的有序集合有如下三个有趣的性质。
- 这个集合是循环的，每个数的后两位是后一个数的前两位（最后一个数的后两位也是第一个数的前两位）。
- 前三种多边形数——三角形数（$P_{3,127}=8128$）、正方形数（$P_{4,91}=8281$）和五边形数（$P_{5,44}=2882$）——在其中各有一个代表。
- 这是唯一一个满足上述性质的$4$位数有序集。
存在唯一一个包含六个$4$位数的循环有序集合，满足前六种多边形数——三角形数、正方形数、五边形数、六边形数、七边形数和八边形数——在其中各有一个代表。求这个集合的元素和。