Solved problem 58.

ipython/EulerProblem061.ipynb

@@ -0,0 +1,73 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Cyclical figurate numbers (Euler Problem 61)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {
+    "collapsed": true
+   },
+   "source": [
+    "[https://projecteuler.net/problem=61](https://projecteuler.net/problem=61)\n",
+    "\n",
+    "Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:\n",
+    "\n",
+    "Triangle\t \tP3,n=n(n+1)/2\t \t1, 3, 6, 10, 15, ...\n",
+    "\n",
+    "Square\t \tP4,n=n2\t \t1, 4, 9, 16, 25, ...\n",
+    "\n",
+    "Pentagonal\t \tP5,n=n(3n−1)/2\t \t1, 5, 12, 22, 35, ...\n",
+    "\n",
+    "Hexagonal\t \tP6,n=n(2n−1)\t \t1, 6, 15, 28, 45, ...\n",
+    "\n",
+    "Heptagonal\t \tP7,n=n(5n−3)/2\t \t1, 7, 18, 34, 55, ...\n",
+    "\n",
+    "Octagonal\t \tP8,n=n(3n−2)\t \t1, 8, 21, 40, 65, ...\n",
+    "\n",
+    "The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.\n",
+    "\n",
+    "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).\n",
+    "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.\n",
+    "This is the only set of 4-digit numbers with this property.\n",
+    "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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+   "outputs": [],
+   "source": []
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