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	$P_{3, n} = n (n + 1) / 2$	$1, 3, 6, 10, 15, \dots$
Square	$P_{4, n} = n^{2}$	$1, 4, 9, 16, 25, \dots$
Pentagonal	$P_{5, n} = n (3 n - 1) / 2$	$1, 5, 12, 22, 35, \dots$
Hexagonal	$P_{6, n} = n (2 n - 1)$	$1, 6, 15, 28, 45, \dots$
Heptagonal	$P_{7, n} = n (5 n - 3) / 2$	$1, 7, 18, 34, 55, \dots$
Octagonal	$P_{8, n} = n (3 n - 2)$	$1, 8, 21, 40, 65, \dots$

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	$P_{3, n} = n (n + 1) / 2$	$1, 3, 6, 10, 15, \dots$
正方形数	$P_{4, n} = n^{2}$	$1, 4, 9, 16, 25, \dots$
五边形数	$P_{5, n} = n (3 n - 1) / 2$	$1, 5, 12, 22, 35, \dots$
六边形数	$P_{6, n} = n (2 n - 1)$	$1, 6, 15, 28, 45, \dots$
七边形数	$P_{7, n} = n (5 n - 3) / 2$	$1, 7, 18, 34, 55, \dots$
八边形数	$P_{8, n} = n (3 n - 2)$	$1, 8, 21, 40, 65, \dots$

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。