<jupyter_start><jupyter_text>Day 1: Report RepairAfter saving Christmas five years in a row, you've decided to take a vacation at a nice resort on a tropical island. Surely, Christmas will go on without you.Before you leave, the Elves in accounting just need you to fix your expense report (your puzzle input); apparently, something isn't quite adding up.Specifically, they need you to find the two entries that sum to `2020` and then multiply those two numbers together.For example, suppose your expense report contained the following:```1721979366299675145```In this list, the two entries that sum to `2020` are `1721` and `299`. Multiplying them together produces `1721 * 299 = 514579`, so the correct answer is `514579`.Of course, your expense report is much larger. Find the two entries that sum to 2020; what do you get if you multiply them together?<jupyter_code>input_file = open("inputs/01.txt")
puzzle_input = input_file.read()
string_numbers = puzzle_input.split()
numbers = []
for line in string_numbers:
    numbers.append(int(line))
    
print("numbers[0] + numbers [1] = ", numbers[0] + numbers[1])
input_file.close()

def x():
    for number1 in numbers:
        for number2 in numbers:
            for number3 in numbers:
                if number1 + number2 + number3 == 2020:
                    print(number1, number2, number3, number1 * number2 * number3)
                    return
x()
input_file.closed
from itertools import combinations
from math import prod
with open("inputs/01.txt") as input_file:
    puzzle_input = input_file.read()
    
numbers = [int(entry) for entry in puzzle_input.splitlines()]
for combination in combinations(numbers, 2):
    if sum(combination) == 2020:
        print(combination, prod(combination))<jupyter_output>(1018, 1002) 1020036
{'fraction_non_alphanumeric': 0.08557046979865772, 'fraction_numerical': 0.05704697986577181, 'mean_word_length': 4.011204481792717, 'pattern_counts': {'":': 0, '<': 4, '<?xml version=': 0, '>': 4, 'https://': 0, 'lorem ipsum': 0, 'www.': 0, 'xml': 0}, 'pii_count': 0, 'substrings_counts': 0, 'word_list_counts': {'cursed_substrings.json': 0, 'profanity_word_list.json': 0, 'sexual_word_list.json': 0, 'zh_pornsignals.json': 0}}
{'chain_length': '1', 'dir': 'jupyter-structured-clean-dedup', 'id': '642501', 'license': 'MIT', 'n_tokens_mistral': 569, 'n_tokens_neox': 493, 'n_words': 209, 'path': '01.ipynb', 'repo_name': 'danialeshete/AdventOfCode2020', 'size': '1788'}