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# 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.

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 (P3,127=8128), square (P4,91=8281), and pentagonal (P5,44=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.

1. 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）。
2. 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。
3. 这是唯一一个满足上述性质的4位数有序集。