add problem 43

notebooks/problem0043.ipynb

@@ -0,0 +1,127 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "aac67a5d",
+   "metadata": {},
+   "source": [
+    "# [Sub-string Divisibility](https://projecteuler.net/problem=43)\n",
+    "\n",
+    "If you *really* wanted to, you could work this out with pen and paper using [divisibility rules](https://en.wikipedia.org/wiki/Divisibility_rule). Alternatively, it's easy to brute force. First we write a function to test for the given property."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "71136ed1",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def is_substring_divisible(digits):\n",
+    "    for idx, divisor in enumerate((2, 3, 5, 7, 11, 13, 17), start=1):\n",
+    "        x, y, z = digits[idx:idx+3]\n",
+    "        n = 100*x + 10*y + z\n",
+    "        if n % divisor != 0:\n",
+    "            return False\n",
+    "    \n",
+    "    return True"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "3560bcb4",
+   "metadata": {},
+   "source": [
+    "Then we test every permutation of 0 through 9. There's $10! = 3628800$ such permutations, which may seem like a lot, and you could apply some logic to reduce the search space some (e.g. $d_4$ must be 0, 2, 4, 6, or 8), but the test runs relatively quickly for each permutation as-is."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "dc866d6b",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "{1406357289, 1430952867, 1460357289, 4106357289, 4130952867, 4160357289}"
+      ]
+     },
+     "execution_count": 2,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "from itertools import permutations\n",
+    "\n",
+    "numbers = set()\n",
+    "for digits in permutations((0,1,2,3,4,5,6,7,8,9)):\n",
+    "    if is_substring_divisible(digits):\n",
+    "        n = sum(10^k * digit for (k, digit) in enumerate(reversed(digits)))\n",
+    "        numbers.add(n)\n",
+    "        \n",
+    "numbers"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "fca588e4",
+   "metadata": {},
+   "source": [
+    "Turns out there's only six of these numbers."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "b6275692",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "16695334890"
+      ]
+     },
+     "execution_count": 3,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "sum(numbers)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "bfaa46b7",
+   "metadata": {},
+   "source": [
+    "## Relevant sequences\n",
+    "* Pandigital numbers: [A050278](https://oeis.org/A050278)"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "SageMath 9.5",
+   "language": "sage",
+   "name": "sagemath"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.11.2"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}