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 | P3,n=n(n+1)/2 | 1, 3, 6, 10, 15, … |
| Square | P4,n=n2 | 1, 4, 9, 16, 25, … |
| Pentagonal | P5,n=n(3n−1)/2 | 1, 5, 12, 22, 35, … |
| Hexagonal | P6,n=n(2n−1) | 1, 6, 15, 28, 45, … |
| Heptagonal | P7,n=n(5n−3)/2 | 1, 7, 18, 34, 55, … |
| Octagonal | P8,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 (P3,127=8128), square (P4,91=8281), and pentagonal (P5,44=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统称为多边形数。它们分别由如下的公式给出：
| 三角形数 | P3,n=n(n+1)/2 | 1, 3, 6, 10, 15, … |
| 正方形数 | P4,n=n2 | 1, 4, 9, 16, 25, … |
| 五边形数 | P5,n=n(3n−1)/2 | 1, 5, 12, 22, 35, … |
| 六边形数 | P6,n=n(2n−1) | 1, 6, 15, 28, 45, … |
| 七边形数 | P7,n=n(5n−3)/2 | 1, 7, 18, 34, 55, … |
| 八边形数 | P8,n=n(3n−2) | 1, 8, 21, 40, 65, … |
由三个4位数8128、2882、8281构成的有序集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）。
- 每种多边形数——三角形数（P3,127=8128）、正方形数（P4,91=8281）和五边形数（P5,44=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位数的有序循环集，每种多边形数——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。