# Hexagonal numbers

## Project Euler 61: Find the sum of the only set of six 4-digit figurate numbers with a cyclic property

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 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.

Posted by Kristian in Project Euler, 4 comments

## Project Euler 45: After 40755, what is the next triangle number that is also pentagonal and hexagonal?

Problem 45 of Project Euler reads

Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

 Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, … Pentagonal Pn=n(3n-1)/2 1, 5, 12, 22, 35, … Hexagonal Hn=n(2n-1) 1, 6, 15, 28, 45, …

It can be verified that T285 = P165 = H143 = 40755.

Find the next triangle number that is also pentagonal and hexagonal.

Posted by Kristian in Project Euler, 13 comments