90 lines
1.7 KiB
Plaintext
90 lines
1.7 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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"# Euler Problem 1\n",
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"\n",
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"If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.\n",
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"\n",
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"Find the sum of all the multiples of 3 or 5 below 1000."
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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": 1,
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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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"def get_sum_of_natural_dividable_by_3_and_5_below(m):\n",
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" return sum([x for x in range(m) if x % 3 == 0 or x % 5 == 0])"
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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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"Test example provided in problem statement:"
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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": 2,
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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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"assert(get_sum_of_natural_dividable_by_3_and_5_below(10) == 23)"
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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": 3,
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"metadata": {
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"collapsed": false
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},
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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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"233168\n"
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]
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}
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],
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"source": [
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"print(get_sum_of_natural_dividable_by_3_and_5_below(1000))"
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]
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}
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],
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"metadata": {
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"completion_date": "Tue, 19 Aug 2014, 20:11",
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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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": "3.5.4"
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},
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"tags": [
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"brute force"
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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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