For some reason Problem 61 of Project Euler is a problem that not so many people have solved compared to the problems in the sixties range. However, I think that it was a quite approachable problem which was fun to solve. The problem reads

**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*(3*n*-1)/2 |
**1, 5, 12, 22, 35, …** |

**Hexagonal** |
**P**_{6,n}=*n*(2*n-*1) |
**1, 6, 15, 28, 45, …** |

**Heptagonal** |
**P**_{7,n}=*n*(5*n-*3)/2 |
**1, 7, 18, 34, 55, …** |

**Octagonal** |
**P**_{8,n}=*n*(3*n-*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 (P**_{3,127}=8128), square (P_{4,91}=8281), and pentagonal (P_{5,44}=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.**

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