euler/ipython/EulerProblem043.ipynb

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Sub-string divisibility (Euler Problem 43)"
]
},
{
"cell_type": "markdown",
"metadata": {
"collapsed": true
},
"source": [
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"[https://projecteuler.net/problem=43](https://projecteuler.net/problem=43)\n",
"\n",
"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",
"\n",
"Let $d_1$ be the 1st digit, $d_2$ be the 2nd digit, and so on. In this way, we note the following:\n",
"\n",
"$d_2d_3d_4$=406 is divisible by 2\n",
"\n",
"$d_3d_4d_5$=063 is divisible by 3\n",
"\n",
"$d_4d_5d_6$=635 is divisible by 5\n",
"\n",
"$d_5d_6d_7=357$ is divisible by 7\n",
"\n",
"$d_6d_7d_8=572$ is divisible by 11\n",
"\n",
"$d_7d_8d_9=728$ is divisible by 13\n",
"\n",
"$d_8d_9d_{10} =289$ is divisible by 17\n",
"\n",
"Find the sum of all 0 to 9 pandigital numbers with this property."
]
},
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{
"cell_type": "markdown",
"metadata": {},
"source": [
"We write a function that tests for the sub-string divisibility property. Then we simply brute force."
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {},
"outputs": [],
"source": [
"def is_sub_string_divisible(s):\n",
" s = \"\".join(s)\n",
" if int(s[1:4]) % 2 != 0:\n",
" return False\n",
" if int(s[2:5]) % 3 != 0:\n",
" return False\n",
" if int(s[3:6]) % 5 != 0:\n",
" return False\n",
" if int(s[4:7]) % 7 != 0:\n",
" return False\n",
" if int(s[5:8]) % 11 != 0:\n",
" return False\n",
" if int(s[6:9]) % 13 != 0:\n",
" return False\n",
" if int(s[7:10]) % 17 != 0:\n",
" return False\n",
" return True\n",
"\n",
"assert(is_sub_string_divisible(\"1406357289\"))\n",
"assert(is_sub_string_divisible(\"1406357298\") == False)"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {},
"outputs": [],
"source": [
"import itertools\n",
"\n",
"s = sum([int(\"\".join(p)) for p in itertools.permutations(\"0123456789\") if is_sub_string_divisible(p)])"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"16695334890\n"
]
}
],
"source": [
"print(s)\n",
"assert(s == 16695334890)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"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:"
]
},
{
"cell_type": "code",
"execution_count": 32,
"metadata": {},
"outputs": [],
"source": [
"def permutations(items):\n",
" items = list(items)\n",
" if not items:\n",
" return [[]]\n",
" perms = []\n",
" for item in items:\n",
" items_new = list(items)\n",
" items_new.remove(item)\n",
" permutations_new = permutations(items_new)\n",
" for perm in permutations_new:\n",
" perms.append([item] + perm)\n",
" return perms\n",
"\n",
"\n",
"assert(permutations([]) == [[]])\n",
"assert(permutations([1]) == [[1]])\n",
"assert(permutations([1,2]) == [[1,2], [2,1]])\n",
"assert(permutations([1,2,3]) == [[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]])\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": []
}
],
"metadata": {
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"completion_date": "Fri, 21 Dec 2018, 16:28",
"kernelspec": {
"display_name": "Python 3",
"language": "python3.6",
"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": "3.6.5"
},
"tags": [
"pandigital",
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"divisibility",
"permutations"
]
},
"nbformat": 4,
"nbformat_minor": 2
}