diff --git a/ipython/EulerProblem028.ipynb b/ipython/EulerProblem028.ipynb
new file mode 100755
index 0000000..e71f4cf
--- /dev/null
+++ b/ipython/EulerProblem028.ipynb
@@ -0,0 +1,94 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 28\n",
+    "\n",
+    "Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:\n",
+    "\n",
+    "~~~\n",
+    "21 22 23 24 25\n",
+    "20  7  8  9 10\n",
+    "19  6  1  2 11\n",
+    "18  5  4  3 12\n",
+    "17 16 15 14 13\n",
+    "~~~\n",
+    "\n",
+    "$1 + 3 + 5 + 7 + 9 + 13 + 17 + 21 + 25 = 101$\n",
+    "\n",
+    "It can be verified that the sum of the numbers on the diagonals is 101.\n",
+    "\n",
+    "What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "I would try to create a function $f(n)$ which yields the sum of the outmost ring of a n by n spiral.\n",
+    "\n",
+    "For example:\n",
+    "\n",
+    "$f(1) = 1$\n",
+    "\n",
+    "$f(3) = 3 + 5 + 7 + 9 = 24$\n",
+    "\n",
+    "$f(5) = 13 + 17 + 21 + 25 = 76$\n",
+    "\n",
+    "When we have this function we calculate the solution simply by\n",
+    "\n",
+    "~~~\n",
+    "s = sum([f(n) for n in range(1, 1002, 2)])\n",
+    "~~~\n",
+    "\n",
+    "For each outer ring there is an initial corner value c ($c_3 = 3, c_5 = 76$). Once we have this value we can caluclate f like $f(n) = c_{n} + (c_n + n - 1) + (c_n + 2(n-1)) + (c_n + 3(n-1)) = 4c_n + 6 (n-1)$"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": [
+    "def f(n):\n",
+    "    if n == 1:\n",
+    "        return 1\n",
+    "    return 0\n",
+    "\n",
+    "s = sum([f(n) for n in range(1, 1002, 2)])\n",
+    "assert(s == 669171001)\n",
+    "s"
+   ]
+  }
+ ],
+ "metadata": {
+  "completion_date": "Wed, 23 Aug 2017, 15:54",
+  "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": [
+   "spiral",
+   "diagonals"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}
diff --git a/ipython/EulerProblem029.ipynb b/ipython/EulerProblem029.ipynb
new file mode 100755
index 0000000..0062700
--- /dev/null
+++ b/ipython/EulerProblem029.ipynb
@@ -0,0 +1,59 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 29\n",
+    "\n",
+    "Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:\n",
+    "\n",
+    "$2^2=4, 2^3=8, 2^4=16, 2^5=32$\n",
+    "\n",
+    "32=9, 33=27, 34=81, 35=243\n",
+    "\n",
+    "42=16, 43=64, 44=256, 45=1024\n",
+    "\n",
+    "52=25, 53=125, 54=625, 55=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",
+    "\n",
+    "How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "comopletion_date": "",
+  "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": []
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}