174 lines
4.3 KiB
Plaintext
174 lines
4.3 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Sub-string divisibility (Euler Problem 43)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {
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"collapsed": true
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},
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"source": [
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"[https://projecteuler.net/problem=43](https://projecteuler.net/problem=43)\n",
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"\n",
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"The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.\n",
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"\n",
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"Let $d_1$ be the 1st digit, $d_2$ be the 2nd digit, and so on. In this way, we note the following:\n",
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"\n",
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"$d_2d_3d_4$=406 is divisible by 2\n",
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"\n",
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"$d_3d_4d_5$=063 is divisible by 3\n",
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"\n",
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"$d_4d_5d_6$=635 is divisible by 5\n",
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"\n",
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"$d_5d_6d_7=357$ is divisible by 7\n",
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"\n",
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"$d_6d_7d_8=572$ is divisible by 11\n",
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"\n",
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"$d_7d_8d_9=728$ is divisible by 13\n",
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"\n",
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"$d_8d_9d_{10} =289$ is divisible by 17\n",
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"\n",
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"Find the sum of all 0 to 9 pandigital numbers with this property."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"We write a function that tests for the sub-string divisibility property. Then we simply brute force."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 14,
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"metadata": {},
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"outputs": [],
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"source": [
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"def is_sub_string_divisible(s):\n",
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" s = \"\".join(s)\n",
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" if int(s[1:4]) % 2 != 0:\n",
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" return False\n",
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" if int(s[2:5]) % 3 != 0:\n",
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" return False\n",
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" if int(s[3:6]) % 5 != 0:\n",
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" return False\n",
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" if int(s[4:7]) % 7 != 0:\n",
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" return False\n",
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" if int(s[5:8]) % 11 != 0:\n",
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" return False\n",
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" if int(s[6:9]) % 13 != 0:\n",
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" return False\n",
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" if int(s[7:10]) % 17 != 0:\n",
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" return False\n",
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" return True\n",
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"\n",
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"assert(is_sub_string_divisible(\"1406357289\"))\n",
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"assert(is_sub_string_divisible(\"1406357298\") == False)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 15,
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"metadata": {},
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"outputs": [],
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"source": [
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"import itertools\n",
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"\n",
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"s = sum([int(\"\".join(p)) for p in itertools.permutations(\"0123456789\") if is_sub_string_divisible(p)])"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 17,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"16695334890\n"
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]
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}
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],
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"source": [
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"print(s)\n",
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"assert(s == 16695334890)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"I just want to write my own permutations functions to see that I am able to do it quickly. We pick each element once and combine it with the permutations for the remaining items:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 32,
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"metadata": {},
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"outputs": [],
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"source": [
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"def permutations(items):\n",
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" items = list(items)\n",
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" if not items:\n",
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" return [[]]\n",
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" perms = []\n",
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" for item in items:\n",
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" items_new = list(items)\n",
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" items_new.remove(item)\n",
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" permutations_new = permutations(items_new)\n",
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" for perm in permutations_new:\n",
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" perms.append([item] + perm)\n",
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" return perms\n",
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"\n",
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"\n",
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"assert(permutations([]) == [[]])\n",
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"assert(permutations([1]) == [[1]])\n",
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"assert(permutations([1,2]) == [[1,2], [2,1]])\n",
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"assert(permutations([1,2,3]) == [[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]])\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"collapsed": true
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},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"completion_date": "Fri, 21 Dec 2018, 16:28",
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python3.6",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.6.5"
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},
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"tags": [
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"pandigital",
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"divisibility",
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"permutations"
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]
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}
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