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:
Type of Number | Formula | Sequence |
---|---|---|
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.
Test
{{test}}Console output