Added some further ipython solutions.

ipython/EulerProblem029.ipynb

@@ -10,31 +10,41 @@
     "\n",
     "$2^2=4, 2^3=8, 2^4=16, 2^5=32$\n",
     "\n",
-    "32=9, 33=27, 34=81, 35=243\n",
+    "$3^2=9, 3^3=27, 3^4=81, 3^5=243$\n",
     "\n",
-    "42=16, 43=64, 44=256, 45=1024\n",
+    "$4^2=16, 4^3=64, 4^4=256, 4^5=1024$\n",
     "\n",
-    "52=25, 53=125, 54=625, 55=3125\n",
+    "$5^2=25, 5^3=125, 5^4=625, 5^5=3125$\n",
     "\n",
     "If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:\n",
     "\n",
-    "4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125\n",
+    "$4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125$\n",
     "\n",
-    "How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?"
+    "How many distinct terms are in the sequence generated by $a^b$ for $2 ≤ a ≤ 100$ and $2 ≤ b ≤ 100$?"
    ]
   },
   {
    "cell_type": "code",
-   "execution_count": null,
-   "metadata": {
-    "collapsed": true
-   },
-   "outputs": [],
-   "source": []
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "9183\n"
+     ]
+    }
+   ],
+   "source": [
+    "s = len(set([a**b for a in range(2, 101) for b in range(2, 101)]))\n",
+    "assert(s == 9183)\n",
+    "print(s)"
+   ]
   }
  ],
  "metadata": {
-  "comopletion_date": "",
+  "comopletion_date": "Fri, 25 Aug 2017, 10:03",
   "kernelspec": {
    "display_name": "Python 3",
    "language": "python",
@@ -52,7 +62,10 @@
    "pygments_lexer": "ipython3",
    "version": "3.6.3"
   },
-  "tags": []
+  "tags": [
+   "distinct",
+   "powers"
+  ]
  },
  "nbformat": 4,
  "nbformat_minor": 2

ipython/EulerProblem030.ipynb

@@ -0,0 +1,83 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 30\n",
+    "\n",
+    "Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:\n",
+    "\n",
+    "$1634 = 1^4 + 6^4 + 3^4 + 4^4$\n",
+    "\n",
+    "$8208 = 8^4 + 2^4 + 0^4 + 8^4$\n",
+    "\n",
+    "$9474 = 9^4 + 4^4 + 7^4 + 4^4$\n",
+    "\n",
+    "As $1 = 1^4$ is not a sum it is not included.\n",
+    "\n",
+    "The sum of these numbers is $1634 + 8208 + 9474 = 19316$.\n",
+    "\n",
+    "Find the sum of all the numbers that can be written as the sum of fifth powers of their digits."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "This is an eays brute force. We look up the fifth powers."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 7,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "443839\n"
+     ]
+    }
+   ],
+   "source": [
+    "fifth_power_lookup = {str(i): i**5 for i in range(0,10)}\n",
+    "\n",
+    "def is_number_sum_of_fiths_powers_of_digits(n):\n",
+    "    return n == sum([fifth_power_lookup[d] for d in str(n)])\n",
+    "\n",
+    "s = sum([i for i in range(1, 1000000) if is_number_sum_of_fiths_powers_of_digits(i) and not i is 1])\n",
+    "print(s)\n",
+    "assert(s == 443839)"
+   ]
+  }
+ ],
+ "metadata": {
+  "completion_date": "Fri, 25 Aug 2017, 12:13",
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.6.3"
+  },
+  "tags": [
+   "digit",
+   "powers",
+   "brute force"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}

ipython/EulerProblem031.ipynb

@@ -0,0 +1,65 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 31\n",
+    "\n",
+    "In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:\n",
+    "\n",
+    "1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).\n",
+    "\n",
+    "It is possible to make £2 in the following way:\n",
+    "\n",
+    "1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p\n",
+    "\n",
+    "How many different ways can £2 be made using any number of coins?"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "I really have mental issues to this in a non recursive way."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "\n",
+    "        "
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.6.3"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}