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Cyclical figurate numbers
Problem 61
Published on 16 January 2004 at 06:00 pm [Server Time]
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.
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