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id	challengeType	title	forumTopicId
5900f3a91000cf542c50febc	5	Problem 61: Cyclical figurate numbers	302173

Description

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.

Instructions

Tests

tests:
  - text: <code>euler61()</code> should return 28684.
    testString: assert.strictEqual(euler61(), 28684);

Challenge Seed

function euler61() {
  // Good luck!
  return true;
}

euler61();

Solution

// solution required

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