Reorganization
This commit is contained in:
David Doblas Jiménez
parent
7c8165d9c2
commit
2a1db7eda3
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#!/usr/bin/python3
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"""
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Created on 14 Mar 2017
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 1 of Project Euler
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https://projecteuler.net/problem=1
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"""
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from utils import timeit
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@timeit("Problem 1")
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def compute():
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"""
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If we list all the natural numbers below 10 that are multiples of 3 or 5,
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we get 3, 5, 6 and 9. The sum of these multiples is 23.
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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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ans = sum(x for x in range(1000) if (x % 3 == 0 or x % 5 == 0))
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return ans
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if __name__ == "__main__":
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print(f"Result for problem 1: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 14 Mar 2017
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 2 of Project Euler
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https://projecteuler.net/problem=2
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"""
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from utils import timeit
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@timeit("Problem 2")
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def compute():
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"""
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Each new term in the Fibonacci sequence is generated by adding the
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previous two terms. By starting with 1 and 2, the first 10 terms will be:
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1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
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Find the sum of all the even-valued terms in the sequence which do not
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exceed four million.
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"""
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ans = 0
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limit = 4_000_000
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x, y = 1, 1
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z = x + y # Because every third Fibonacci number is even
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while z <= limit:
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ans += z
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x = y + z
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y = z + x
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z = x + y
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return ans
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if __name__ == "__main__":
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print(f"Result for problem 2: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 18 Mar 2017
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 3 of Project Euler
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https://projecteuler.net/problem=3
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"""
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from utils import timeit
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@timeit("Problem 3")
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def compute():
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"""
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The prime factors of 13195 are 5, 7, 13 and 29.
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What is the largest prime factor of the number 600851475143 ?
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"""
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ans = 600851475143
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factor = 2
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while factor * factor < ans:
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while ans % factor == 0:
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ans = ans // factor
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factor += 1
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return ans
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if __name__ == "__main__":
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print(f"Result for problem 3: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 18 Mar 2017
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 4 of Project Euler
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https://projecteuler.net/problem=4
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"""
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from utils import timeit
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@timeit("Problem 4")
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def compute():
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"""
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A palindromic number reads the same both ways. The largest palindrome made
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from the product of two 2-digit numbers is 9009 = 91 x 99.
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Find the largest palindrome made from the product of two 3-digit numbers.
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"""
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ans = 0
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for i in range(100, 1000):
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for j in range(100, 1000):
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palindrome = i * j
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s = str(palindrome)
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if s == s[::-1] and palindrome > ans:
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ans = palindrome
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return ans
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if __name__ == "__main__":
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print(f"Result for problem 4: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 23 Apr 2017
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 5 of Project Euler
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https://projecteuler.net/problem=5
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"""
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import math
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from utils import timeit
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# The LCM of two natural numbers x and y is given by:
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# def lcm(x, y):
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# return x * y // math.gcd(x, y)
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# It is possible to compute the LCM of more than two numbers by iteratively
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# computing the LCM of two numbers, i.e. LCM(a, b, c) = LCM(a, LCM(b, c))
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@timeit("Problem 5")
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def compute():
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"""
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2520 is the smallest number that can be divided by each of the numbers
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from 1 to 10 without any remainder.
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What is the smallest positive number that is evenly divisible by all of
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the numbers from 1 to 20?
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"""
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ans = 1
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for i in range(1, 21):
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ans *= i // math.gcd(i, ans)
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return ans
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if __name__ == "__main__":
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print(f"Result for problem 5: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 17 Jun 2017
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 6 of Project Euler
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https://projecteuler.net/problem=6
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"""
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from utils import timeit
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@timeit("Problem 6")
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def compute():
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"""
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The sum of the squares of the first ten natural numbers is,
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1^2 + 2^2 + ... + 10^2 = 385
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The square of the sum of the first ten natural numbers is,
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(1 + 2 + ... + 10)^2 = 55^2 = 3025
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Hence the difference between the sum of the squares of the first ten
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natural numbers and the square of the sum is 3025 − 385 = 2640.
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Find the difference between the sum of the squares of the first one
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hundred natural numbers and the square of the sum. Statement
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"""
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n = 100
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square_of_sum = sum(i for i in range(1, n+1))**2
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sum_squares = sum(i**2 for i in range(1, n+1))
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diff = square_of_sum - sum_squares
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return diff
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if __name__ == "__main__":
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print(f"Result for Problem 6: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 28 Jun 2017
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 7 of Project Euler
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https://projecteuler.net/problem=7
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"""
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from utils import timeit, is_prime
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@timeit("Problem 7")
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def compute():
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"""
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# Statement
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"""
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number = 2
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primeList = []
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while len(primeList) < 10001:
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if is_prime(number):
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primeList.append(number)
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number += 1
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ans = primeList[len(primeList)-1]
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return ans
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if __name__ == "__main__":
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print(f"Result for Problem 7: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 28 Abr 2017
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 8 of Project Euler
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https://projecteuler.net/problem=8
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"""
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from utils import timeit
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@timeit("Problem 8")
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def compute():
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"""
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The four adjacent digits in the 1000-digit number that have the
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greatest product are 9 × 9 × 8 × 9 = 5832.
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731671...963450
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Find the thirteen adjacent digits in the 1000-digit number that have
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the greatest product. What is the value of this product?
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"""
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NUM = """
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73167176531330624919225119674426574742355349194934
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96983520312774506326239578318016984801869478851843
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85861560789112949495459501737958331952853208805511
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12540698747158523863050715693290963295227443043557
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66896648950445244523161731856403098711121722383113
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62229893423380308135336276614282806444486645238749
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30358907296290491560440772390713810515859307960866
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70172427121883998797908792274921901699720888093776
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65727333001053367881220235421809751254540594752243
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52584907711670556013604839586446706324415722155397
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53697817977846174064955149290862569321978468622482
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83972241375657056057490261407972968652414535100474
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82166370484403199890008895243450658541227588666881
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16427171479924442928230863465674813919123162824586
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17866458359124566529476545682848912883142607690042
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24219022671055626321111109370544217506941658960408
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07198403850962455444362981230987879927244284909188
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84580156166097919133875499200524063689912560717606
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05886116467109405077541002256983155200055935729725
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71636269561882670428252483600823257530420752963450
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"""
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num = NUM.replace('\n', '').replace(' ', '')
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adjacent_digits = 13
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target = len(num) - adjacent_digits #- 1
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ans, i = 0, 0
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while i < target:
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a, j = 1, i
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while j < (i + adjacent_digits):
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a = a * int(num[j])
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j += 1
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if a > ans:
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ans = a
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i += 1
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return ans
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if __name__ == "__main__":
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print(f"Result for Problem 8: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 26 Aug 2017
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 9 of Project Euler
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https://projecteuler.net/problem=9
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"""
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from utils import timeit
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@timeit("Problem 9")
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def compute():
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"""
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A Pythagorean triplet is a set of three natural numbers, a < b < c,
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for which a^2 + b^2 = c^2
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For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
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There exists exactly one Pythagorean triplet for which a + b + c = 1000.
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Find the product abc.
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"""
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upper_limit = 1000
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for a in range(1, upper_limit + 1):
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for b in range(a + 1, upper_limit + 1):
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c = upper_limit - a - b
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if a * a + b * b == c * c:
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# It is now implied that b < c, because we have a > 0
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return a * b * c
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if __name__ == "__main__":
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print(f"Result for Problem 9: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 26 Aug 2017
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 10 of Project Euler
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https://projecteuler.net/problem=10
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"""
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import math
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from utils import timeit, list_primes
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@timeit("Problem 10")
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def compute():
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"""
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The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
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Find the sum of all the primes below two million.
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"""
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ans = sum(list_primes(1_999_999))
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return ans
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if __name__ == "__main__":
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print(f"Result for Problem 10: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 07 Jul 2021
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 11 of Project Euler
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https://projecteuler.net/problem=11
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"""
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from utils import timeit
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GRID = [
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[ 8, 2,22,97,38,15, 0,40, 0,75, 4, 5, 7,78,52,12,50,77,91, 8],
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[49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48, 4,56,62, 0],
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[81,49,31,73,55,79,14,29,93,71,40,67,53,88,30, 3,49,13,36,65],
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[52,70,95,23, 4,60,11,42,69,24,68,56, 1,32,56,71,37, 2,36,91],
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[22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],
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[24,47,32,60,99, 3,45, 2,44,75,33,53,78,36,84,20,35,17,12,50],
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[32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],
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[67,26,20,68, 2,62,12,20,95,63,94,39,63, 8,40,91,66,49,94,21],
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[24,55,58, 5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],
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[21,36,23, 9,75, 0,76,44,20,45,35,14, 0,61,33,97,34,31,33,95],
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[78,17,53,28,22,75,31,67,15,94, 3,80, 4,62,16,14, 9,53,56,92],
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[16,39, 5,42,96,35,31,47,55,58,88,24, 0,17,54,24,36,29,85,57],
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[86,56, 0,48,35,71,89, 7, 5,44,44,37,44,60,21,58,51,54,17,58],
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[19,80,81,68, 5,94,47,69,28,73,92,13,86,52,17,77, 4,89,55,40],
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[ 4,52, 8,83,97,35,99,16, 7,97,57,32,16,26,26,79,33,27,98,66],
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[88,36,68,87,57,62,20,72, 3,46,33,67,46,55,12,32,63,93,53,69],
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[ 4,42,16,73,38,25,39,11,24,94,72,18, 8,46,29,32,40,62,76,36],
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[20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74, 4,36,16],
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[20,73,35,29,78,31,90, 1,74,31,49,71,48,86,81,16,23,57, 5,54],
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[ 1,70,54,71,83,51,54,69,16,92,33,48,61,43,52, 1,89,19,67,48]
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]
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def grid_product(x, y, dx, dy, n):
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result = 1
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for i in range(n):
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result *= GRID[y + i * dy][x + i * dx]
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return result
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@timeit("Problem 11")
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def compute():
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"""
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In the 20x20 grid above, four numbers along a diagonal line have been
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marked in red.
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The product of these numbers is 26 x 63 x 78 x 14 = 1788696.
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What is the greatest product of four adjacent numbers in any direction
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(up, down, left, right, or diagonally) in the 20x20 grid?
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"""
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ans = 0
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width = height = len(GRID)
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adjacent_nums = 4
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for y in range(height):
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for x in range(width):
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if x + adjacent_nums <= width:
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ans = max(grid_product(x, y, 1, 0, adjacent_nums), ans)
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if y + adjacent_nums <= height:
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ans = max(grid_product(x, y, 0, 1, adjacent_nums), ans)
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if x + adjacent_nums <= width and y + adjacent_nums <= height:
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ans = max(grid_product(x, y, 1, 1, adjacent_nums), ans)
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if x - adjacent_nums >= -1 and y + adjacent_nums <= height:
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ans = max(grid_product(x, y, -1, 1, adjacent_nums), ans)
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return ans
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if __name__ == "__main__":
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print(f"Result for Problem 11: {compute()}")
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#!/usr/bin/env python3
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"""
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Created on 01 Jan 2018
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 12 of Project Euler
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https://projecteuler.net/problem=12
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"""
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import itertools
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from utils import timeit
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from math import sqrt, floor
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||||
# Returns the number of integers in the range [1, n] that divide n.
|
||||
def num_divisors(n):
|
||||
end = floor(sqrt(n))
|
||||
divs = []
|
||||
for i in range(1, end + 1):
|
||||
if n % i == 0:
|
||||
divs.append(i)
|
||||
if end**2 == n:
|
||||
divs.pop()
|
||||
return 2*len(divs)
|
||||
|
||||
|
||||
|
||||
@timeit("Problem 12")
|
||||
def compute():
|
||||
"""
|
||||
The sequence of triangle numbers is generated by adding the natural
|
||||
numbers. So the 7th triangle number would be:
|
||||
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
|
||||
|
||||
The first ten terms would be:
|
||||
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
|
||||
|
||||
Let us list the factors of the first seven triangle numbers:
|
||||
|
||||
1: 1
|
||||
3: 1,3
|
||||
6: 1,2,3,6
|
||||
10: 1,2,5,10
|
||||
15: 1,3,5,15
|
||||
21: 1,3,7,21
|
||||
28: 1,2,4,7,14,28
|
||||
|
||||
We can see that 28 is the first triangle number to have over five divisors.
|
||||
|
||||
What is the value of the first triangle number to have over five hundred
|
||||
divisors?
|
||||
"""
|
||||
|
||||
triangle = 0
|
||||
for i in itertools.count(1):
|
||||
# This is the ith triangle number, i.e. num = 1 + 2 + ... + i =
|
||||
# = i*(i+1)/2
|
||||
triangle += i
|
||||
if num_divisors(triangle) > 500:
|
||||
return str(triangle)
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 12: {compute()}")
|
||||
@ -1,32 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 1 Jan 2018
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 13 of Project Euler
|
||||
https://projecteuler.net/problem=13
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 13")
|
||||
def compute():
|
||||
"""
|
||||
Work out the first ten digits of the sum of the following one-hundred
|
||||
50-digit numbers.
|
||||
"""
|
||||
|
||||
with open("../files/Problem13.txt", "r") as f:
|
||||
num = f.readlines()
|
||||
result = 0
|
||||
for line in num:
|
||||
result += int(line)
|
||||
return str(result)[:10]
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 13: {compute()}")
|
||||
@ -1,63 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 7 Jan 2018
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 14 of Project Euler
|
||||
https://projecteuler.net/problem=14
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
def chain_length(n, terms):
|
||||
length = 0
|
||||
while n != 1:
|
||||
if n in terms:
|
||||
length += terms[n]
|
||||
break
|
||||
if n % 2 == 0:
|
||||
n = n / 2
|
||||
else:
|
||||
n = 3 * n + 1
|
||||
length += 1
|
||||
return length
|
||||
|
||||
@timeit("Problem 14")
|
||||
def compute():
|
||||
"""
|
||||
The following iterative sequence is defined for the set of positive
|
||||
integers:
|
||||
|
||||
n → n/2 (n is even)
|
||||
n → 3n + 1 (n is odd)
|
||||
|
||||
Using the rule above and starting with 13, we generate the following
|
||||
sequence:
|
||||
|
||||
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
|
||||
|
||||
It can be seen that this sequence (starting at 13 and finishing at 1)
|
||||
contains 10 terms. Although it has not been proved yet (Collatz Problem),
|
||||
it is thought that all starting numbers finish at 1.
|
||||
|
||||
Which starting number, under one million, produces the longest chain?
|
||||
|
||||
NOTE: Once the chain starts the terms are allowed to go above one million.
|
||||
"""
|
||||
ans = 0
|
||||
limit = 1_000_000
|
||||
score = 0
|
||||
terms = dict()
|
||||
for i in range(1, limit):
|
||||
terms[i] = chain_length(i, terms)
|
||||
if terms[i] > score:
|
||||
score = terms[i]
|
||||
ans = i
|
||||
return ans
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 14: {compute()}")
|
||||
@ -1,31 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 7 Jan 2018
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 15 of Project Euler
|
||||
https://projecteuler.net/problem=15
|
||||
"""
|
||||
|
||||
from math import factorial
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 15")
|
||||
def compute():
|
||||
"""
|
||||
Starting in the top left corner of a 2×2 grid, and only being able to
|
||||
move to the right and down, there are exactly 6 routes to the bottom
|
||||
right corner.
|
||||
|
||||
How many such routes are there through a 20×20 grid?
|
||||
"""
|
||||
n = 20
|
||||
return int(factorial(2*n) / (factorial(n) * factorial(2*n - n)))
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 15: {compute()}")
|
||||
@ -1,28 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 13 Jan 2018
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 16 of Project Euler
|
||||
https://projecteuler.net/problem=16
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 16")
|
||||
def compute():
|
||||
"""
|
||||
2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.
|
||||
|
||||
What is the sum of the digits of the number 2^1000?
|
||||
"""
|
||||
n = 1000
|
||||
return sum(int(digit) for digit in str(2**n))
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 16: {compute()}")
|
||||
@ -1,62 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 13 Jan 2018
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 17 of Project Euler
|
||||
https://projecteuler.net/problem=17
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
def num2letters(num):
|
||||
nums = {0: '', 1: 'one', 2: 'two', 3: 'three', 4: 'four', 5: 'five',
|
||||
6: 'six', 7: 'seven', 8: 'eight', 9: 'nine', 10: 'ten',
|
||||
11: 'eleven', 12: 'twelve', 13: 'thirteen', 14: 'fourteen',
|
||||
15: 'fifteen', 16: 'sixteen', 17: 'seventeen', 18: 'eighteen',
|
||||
19: 'nineteen', 20: 'twenty', 30: 'thirty', 40: 'forty',
|
||||
50: 'fifty', 60: 'sixty', 70: 'seventy', 80: 'eighty',
|
||||
90: 'ninety', 100: 'hundred', 1000: 'thousand'}
|
||||
|
||||
if num <= 20:
|
||||
return len(nums[num])
|
||||
elif num < 100:
|
||||
tens, units = divmod(num, 10)
|
||||
return len(nums[tens * 10]) + num2letters(units)
|
||||
elif num < 1000:
|
||||
hundreds, rest = divmod(num, 100)
|
||||
if rest:
|
||||
return num2letters(hundreds) + len(nums[100]) + len('and') +\
|
||||
num2letters(rest)
|
||||
else:
|
||||
return num2letters(hundreds) + len(nums[100])
|
||||
else:
|
||||
thousands, rest = divmod(num, 1000)
|
||||
return num2letters(thousands) + len(nums[1000])
|
||||
|
||||
@timeit("Problem 17")
|
||||
def compute():
|
||||
"""
|
||||
If the numbers 1 to 5 are written out in words: one, two, three, four,
|
||||
five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
|
||||
|
||||
If all the numbers from 1 to 1000 (one thousand) inclusive were written
|
||||
out in words, how many letters would be used?
|
||||
|
||||
NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and
|
||||
forty-two) contains 23 letters and 115 (one hundred and fifteen) contains
|
||||
20 letters. The use of "and" when writing out numbers is in compliance
|
||||
with British usage.
|
||||
"""
|
||||
n = 1000
|
||||
letters = 0
|
||||
for num in range(1, n + 1):
|
||||
letters += num2letters(num)
|
||||
return letters
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 17: {compute()}")
|
||||
@ -1,56 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 15 Sep 2018
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 18 of Project Euler
|
||||
https://projecteuler.net/problem=18
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
triangle = [ # Mutable
|
||||
[75],
|
||||
[95,64],
|
||||
[17,47,82],
|
||||
[18,35,87,10],
|
||||
[20, 4,82,47,65],
|
||||
[19, 1,23,75, 3,34],
|
||||
[88, 2,77,73, 7,63,67],
|
||||
[99,65, 4,28, 6,16,70,92],
|
||||
[41,41,26,56,83,40,80,70,33],
|
||||
[41,48,72,33,47,32,37,16,94,29],
|
||||
[53,71,44,65,25,43,91,52,97,51,14],
|
||||
[70,11,33,28,77,73,17,78,39,68,17,57],
|
||||
[91,71,52,38,17,14,91,43,58,50,27,29,48],
|
||||
[63,66, 4,68,89,53,67,30,73,16,69,87,40,31],
|
||||
[ 4,62,98,27,23, 9,70,98,73,93,38,53,60, 4,23],
|
||||
]
|
||||
|
||||
@timeit("Problem 18")
|
||||
def compute():
|
||||
"""
|
||||
By starting at the top of the triangle below and moving to adjacent
|
||||
numbers on the row below, the maximum total from top to bottom is 23.
|
||||
|
||||
3
|
||||
7 4
|
||||
2 4 6
|
||||
8 5 9 3
|
||||
|
||||
That is, 3 + 7 + 4 + 9 = 23.
|
||||
|
||||
Find the maximum total from top to bottom of the triangle above
|
||||
"""
|
||||
for i in reversed(range(len(triangle) - 1)):
|
||||
for j in range(len(triangle[i])):
|
||||
triangle[i][j] += max(triangle[i + 1][j], triangle[i + 1][j + 1])
|
||||
|
||||
return triangle[0][0]
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 18: {compute()}")
|
||||
@ -1,44 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 15 Sep 2018
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 19 of Project Euler
|
||||
https://projecteuler.net/problem=19
|
||||
"""
|
||||
|
||||
from datetime import date
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 19")
|
||||
def compute():
|
||||
"""
|
||||
You are given the following information, but you may prefer to do some
|
||||
research for yourself.
|
||||
|
||||
1 Jan 1900 was a Monday.
|
||||
Thirty days has September, April, June and November.
|
||||
All the rest have thirty-one,
|
||||
Saving February alone,
|
||||
Which has twenty-eight, rain or shine.
|
||||
And on leap years, twenty-nine.
|
||||
A leap year occurs on any year evenly divisible by 4, but not on a century
|
||||
unless it is divisible by 400.
|
||||
|
||||
How many Sundays fell on the first of the month during the twentieth
|
||||
century (1 Jan 1901 to 31 Dec 2000)?
|
||||
"""
|
||||
sundays = 0
|
||||
for y in range(1901, 2001):
|
||||
for m in range (1, 13):
|
||||
if date(y, m, 1).weekday() == 6:
|
||||
sundays += 1
|
||||
return sundays
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 19: {compute()}")
|
||||
@ -1,35 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 15 Sep 2018
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 20 of Project Euler
|
||||
https://projecteuler.net/problem=20
|
||||
"""
|
||||
|
||||
from math import factorial
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 20")
|
||||
def compute():
|
||||
"""
|
||||
n! means n × (n − 1) × ... × 3 × 2 × 1
|
||||
|
||||
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
|
||||
and the sum of the digits in the number 10! is:
|
||||
3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
|
||||
|
||||
Find the sum of the digits in the number 100!
|
||||
"""
|
||||
fact = factorial(100)
|
||||
sum_digits = sum(int(digit) for digit in str(fact))
|
||||
|
||||
return sum_digits
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 20: {compute()}")
|
||||
@ -1,42 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 15 Sep 2018
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 21 of Project Euler
|
||||
https://projecteuler.net/problem=21
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
def sum_divisors(n):
|
||||
return sum(i for i in range(1, n//2+1) if n%i==0)
|
||||
|
||||
@timeit("Problem 21")
|
||||
def compute():
|
||||
"""
|
||||
Let d(n) be defined as the sum of proper divisors of n (numbers
|
||||
less than n which divide evenly into n).
|
||||
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable
|
||||
pair and each of a and b are called amicable numbers.
|
||||
|
||||
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22,
|
||||
44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are
|
||||
1, 2, 4, 71 and 142; so d(284) = 220.
|
||||
|
||||
Evaluate the sum of all the amicable numbers under 10000.
|
||||
"""
|
||||
n = 10000
|
||||
sum_amicable = 0
|
||||
for i in range(1, n):
|
||||
value = sum_divisors(i)
|
||||
if i != value and sum_divisors(value) == i:
|
||||
sum_amicable += i
|
||||
return sum_amicable
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 21: {compute()}")
|
||||
@ -1,42 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 31 Dec 2018
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 22 of Project Euler
|
||||
https://projecteuler.net/problem=22
|
||||
"""
|
||||
|
||||
from pathlib import Path
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 22")
|
||||
def compute():
|
||||
"""
|
||||
Using names.txt, a 46K text file containing over five-thousand first names,
|
||||
begin by sorting it into alphabetical order. Then working out the
|
||||
alphabetical value for each name, multiply this value by its alphabetical
|
||||
position in the list to obtain a name score.
|
||||
|
||||
For example, when the list is sorted into alphabetical order, COLIN, which
|
||||
is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So,
|
||||
COLIN would obtain a score of 938 × 53 = 49714.
|
||||
|
||||
What is the total of all the name scores in the file?
|
||||
"""
|
||||
file = Path("/datos/Scripts/Gitea/Project_Euler/src/files/Problem22.txt")
|
||||
with open(file, 'r') as f:
|
||||
names = sorted(f.read().replace('"', '').split(','))
|
||||
|
||||
result = 0
|
||||
for idx, name in enumerate(names, 1):
|
||||
result += sum(ord(c) - 64 for c in name) * idx
|
||||
return result
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 22: {compute()}")
|
||||
@ -1,58 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 05 Jan 2019
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 23 of Project Euler
|
||||
https://projecteuler.net/problem=23
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 23")
|
||||
def compute():
|
||||
"""
|
||||
A perfect number is a number for which the sum of its proper divisors is
|
||||
exactly equal to the number. For example, the sum of the proper divisors
|
||||
of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect
|
||||
number.
|
||||
|
||||
A number n is called deficient if the sum of its proper divisors is less
|
||||
than n and it is called abundant if this sum exceeds n.
|
||||
|
||||
As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest
|
||||
number that can be written as the sum of two abundant numbers is 24. By
|
||||
mathematical analysis, it can be shown that all integers greater than 28123
|
||||
can be written as the sum of two abundant numbers. However, this upper
|
||||
limit cannot be reduced any further by analysis even though it is known
|
||||
that the greatest number that cannot be expressed as the sum of two
|
||||
abundant numbers is less than this limit.
|
||||
|
||||
Find the sum of all the positive integers which cannot be written as the
|
||||
sum of two abundant numbers.
|
||||
"""
|
||||
LIMIT = 28124
|
||||
divisorsum = [0] * LIMIT
|
||||
for i in range(1, LIMIT):
|
||||
for j in range(i * 2, LIMIT, i):
|
||||
divisorsum[j] += i
|
||||
abundantnums = [i for (i, x) in enumerate(divisorsum) if x > i]
|
||||
|
||||
expressible = [False] * LIMIT
|
||||
for i in abundantnums:
|
||||
for j in abundantnums:
|
||||
if i + j < LIMIT:
|
||||
expressible[i + j] = True
|
||||
else:
|
||||
break
|
||||
|
||||
ans = sum(i for (i, x) in enumerate(expressible) if not x)
|
||||
return ans
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 23: {compute()}")
|
||||
@ -1,37 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 11 Sep 2019
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 24 of Project Euler
|
||||
https://projecteuler.net/problem=24
|
||||
"""
|
||||
|
||||
from itertools import permutations
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 24")
|
||||
def compute():
|
||||
"""
|
||||
A permutation is an ordered arrangement of objects. For example, 3124 is
|
||||
one possible permutation of the digits 1, 2, 3 and 4. If all of the
|
||||
permutations are listed numerically or alphabetically, we call it
|
||||
lexicographic order. The lexicographic permutations of 0, 1 and 2 are:
|
||||
|
||||
012 021 102 120 201 210
|
||||
|
||||
What is the millionth lexicographic permutation of the digits
|
||||
0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
|
||||
"""
|
||||
digits = list(range(10))
|
||||
_permutations = list(permutations(digits))
|
||||
print(_permutations[0])
|
||||
return "".join(str(digit) for digit in _permutations[999999])
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 24: {compute()}")
|
||||
@ -1,54 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 11 Sep 2019
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 25 of Project Euler
|
||||
https://projecteuler.net/problem=25
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 25")
|
||||
def compute():
|
||||
"""
|
||||
The Fibonacci sequence is defined by the recurrence relation:
|
||||
|
||||
Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.
|
||||
|
||||
Hence the first 12 terms will be:
|
||||
|
||||
F1 = 1
|
||||
F2 = 1
|
||||
F3 = 2
|
||||
F4 = 3
|
||||
F5 = 5
|
||||
F6 = 8
|
||||
F7 = 13
|
||||
F8 = 21
|
||||
F9 = 34
|
||||
F10 = 55
|
||||
F11 = 89
|
||||
F12 = 144
|
||||
|
||||
The 12th term, F12, is the first term to contain three digits.
|
||||
|
||||
What is the index of the first term in the Fibonacci sequence to
|
||||
contain 1000 digits?
|
||||
"""
|
||||
a, b = 1, 1
|
||||
index = 2
|
||||
|
||||
while len(str(b)) < 1000:
|
||||
a, b = b, b+a
|
||||
index += 1
|
||||
|
||||
return index
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 25: {compute()}")
|
||||
@ -1,52 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 11 Sep 2019
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 26 of Project Euler
|
||||
https://projecteuler.net/problem=26
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 26")
|
||||
def compute():
|
||||
"""
|
||||
A unit fraction contains 1 in the numerator. The decimal representation
|
||||
of the unit fractions with denominators 2 to 10 are given:
|
||||
|
||||
1/2 = 0.5
|
||||
1/3 = 0.(3)
|
||||
1/4 = 0.25
|
||||
1/5 = 0.2
|
||||
1/6 = 0.1(6)
|
||||
1/7 = 0.(142857)
|
||||
1/8 = 0.125
|
||||
1/9 = 0.(1)
|
||||
1/10 = 0.1
|
||||
|
||||
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle.
|
||||
It can be seen that 1/7 has a 6-digit recurring cycle.
|
||||
|
||||
Find the value of d < 1000 for which 1/d contains the longest recurring
|
||||
cycle in its decimal fraction part.
|
||||
"""
|
||||
cycle_length = 0
|
||||
number_d = 0
|
||||
for number in range(3, 1000, 2):
|
||||
if number % 5 == 0:
|
||||
continue
|
||||
p = 1
|
||||
while 10**p % number != 1:
|
||||
p += 1
|
||||
if p > cycle_length:
|
||||
cycle_length, number_d = p, number
|
||||
return number_d
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 26: {compute()}")
|
||||
@ -1,58 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 15 Sep 2019
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 27 of Project Euler
|
||||
https://projecteuler.net/problem=27
|
||||
"""
|
||||
|
||||
from utils import timeit, is_prime
|
||||
|
||||
|
||||
@timeit("Problem 27")
|
||||
def compute():
|
||||
"""
|
||||
Euler discovered the remarkable quadratic formula:
|
||||
|
||||
n^2 + n + 41
|
||||
|
||||
It turns out that the formula will produce 40 primes for the consecutive
|
||||
integer values 0≤n≤39. However, when n=40, 40^2+40+41=40(40+1)+41 is
|
||||
divisible by 41, and certainly when n=41,41^2+41+41 is clearly divisible
|
||||
by 41.
|
||||
|
||||
The incredible formula n^2−79n+1601 was discovered, which produces 80
|
||||
primes for the consecutive values 0≤n≤79. The product of the coefficients,
|
||||
−79 and 1601, is −126479.
|
||||
|
||||
Considering quadratics of the form:
|
||||
|
||||
n^2 + an + b
|
||||
|
||||
where |a|<1000, |b|≤1000 and |n| is the modulus/absolute value of n
|
||||
e.g. |11|=11 and |−4|=4
|
||||
|
||||
Find the product of the coefficients, a and b, for the quadratic expression
|
||||
that produces the maximum number of primes for consecutive values of n,
|
||||
starting with n=0.
|
||||
"""
|
||||
LIMIT = 1000
|
||||
consecutive_values = 0
|
||||
|
||||
for a in range(-999, LIMIT):
|
||||
for b in range(LIMIT + 1):
|
||||
n = 0
|
||||
while is_prime(abs((n ** 2) + (a * n) + b)):
|
||||
n += 1
|
||||
if n > consecutive_values:
|
||||
consecutive_values = n
|
||||
c = a * b
|
||||
return c
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 27: {compute()}")
|
||||
@ -1,48 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 3 Jan 2020
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 28 of Project Euler
|
||||
https://projecteuler.net/problem=28
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 28")
|
||||
def compute():
|
||||
"""
|
||||
Starting with the number 1 and moving to the right in a clockwise
|
||||
direction a 5 by 5 spiral is formed as follows:
|
||||
|
||||
21 22 23 24 25
|
||||
20 7 8 9 10
|
||||
19 6 1 2 11
|
||||
18 5 4 3 12
|
||||
17 16 15 14 13
|
||||
|
||||
It can be verified that the sum of the numbers on the diagonals is 101.
|
||||
|
||||
What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral
|
||||
formed in the same way?
|
||||
"""
|
||||
size = 1001 # Must be odd
|
||||
ans = 1 # Special case for size 1
|
||||
step = 0
|
||||
i, cur = 1, 1
|
||||
while step < size-1:
|
||||
step = i * 2
|
||||
for j in range(1, 5):
|
||||
cur += step
|
||||
ans += cur
|
||||
i += 1
|
||||
|
||||
return ans
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 28: {compute()}")
|
||||
@ -1,38 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 3 Jan 2020
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 29 of Project Euler
|
||||
https://projecteuler.net/problem=29
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 29")
|
||||
def compute():
|
||||
"""
|
||||
Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:
|
||||
|
||||
2^2=4, 2^3=8, 2^4=16, 2^5=32
|
||||
3^2=9, 3^3=27, 3^4=81, 3^5=243
|
||||
4^2=16, 4^3=64, 4^4=256, 4^5=1024
|
||||
5^2=25, 5^3=125, 5^4=625, 5^5=3125
|
||||
|
||||
If they are then placed in numerical order, with any repeats removed, we
|
||||
get the following sequence of 15 distinct terms:
|
||||
|
||||
4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125
|
||||
|
||||
How many distinct terms are in the sequence generated by ab for 2≤a≤100
|
||||
and 2≤b≤100?
|
||||
"""
|
||||
terms = set(a**b for a in range(2, 101) for b in range(2, 101))
|
||||
return len(terms)
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 29: {compute()}")
|
||||
@ -1,42 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 3 Jan 2020
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 30 of Project Euler
|
||||
https://projecteuler.net/problem=30
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 30")
|
||||
def compute():
|
||||
"""
|
||||
Surprisingly there are only three numbers that can be written as the sum
|
||||
of fourth powers of their digits:
|
||||
|
||||
1634 = 1^4 + 6^4 + 3^4 + 4^4
|
||||
8208 = 8^4 + 2^4 + 0^4 + 8^4
|
||||
9474 = 9^4 + 4^4 + 7^4 + 4^4
|
||||
|
||||
As 1 = 14 is not a sum it is not included.
|
||||
|
||||
The sum of these numbers is 1634 + 8208 + 9474 = 19316.
|
||||
|
||||
Find the sum of all the numbers that can be written as the sum of fifth
|
||||
powers of their digits.
|
||||
"""
|
||||
|
||||
def power_digit_sum(pow, n):
|
||||
return sum(int(c)**pow for c in str(n))
|
||||
|
||||
ans = sum(i for i in range(2, 1000000) if i == power_digit_sum(5, i))
|
||||
|
||||
return ans
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 30: {compute()}")
|
||||
@ -1,43 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 24 Feb 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 31 of Project Euler
|
||||
https://projecteuler.net/problem=31
|
||||
"""
|
||||
|
||||
from itertools import product
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 31")
|
||||
def compute():
|
||||
"""
|
||||
In the United Kingdom the currency is made up of pound (£) and pence (p).
|
||||
There are eight coins in general circulation:
|
||||
|
||||
1p, 2p, 5p, 10p, 20p, 50p, £1 (100p), and £2 (200p).
|
||||
|
||||
It is possible to make £2 in the following way:
|
||||
|
||||
1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
|
||||
|
||||
How many different ways can £2 be made using any number of coins?
|
||||
"""
|
||||
no_ways = 0
|
||||
|
||||
coins = [2, 5, 10, 20, 50, 100]
|
||||
|
||||
bunch_of_coins = product(*[range(0, 201, i) for i in coins])
|
||||
for money in bunch_of_coins:
|
||||
if sum(money) <= 200:
|
||||
no_ways += 1
|
||||
# consider also the case for 200 coins of 1p
|
||||
return no_ways + 1
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 31: {compute()}")
|
||||
@ -1,45 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 26 Feb 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 32 of Project Euler
|
||||
https://projecteuler.net/problem=32
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 32")
|
||||
def compute():
|
||||
"""
|
||||
We shall say that an n-digit number is pandigital if it makes use of all
|
||||
the digits 1 to n exactly once; for example, the 5-digit number, 15234, is
|
||||
1 through 5 pandigital.
|
||||
|
||||
The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing
|
||||
multiplicand, multiplier, and product is 1 through 9 pandigital.
|
||||
|
||||
Find the sum of all products whose multiplicand/multiplier/product identity
|
||||
can be written as a 1 through 9 pandigital.
|
||||
HINT: Some products can be obtained in more than one way so be sure to only
|
||||
include it once in your sum.
|
||||
"""
|
||||
|
||||
ans = set()
|
||||
pandigital = ['1', '2', '3', '4', '5', '6', '7', '8', '9']
|
||||
|
||||
for x in range(1, 100):
|
||||
for y in range(100, 10000):
|
||||
# product = x * y
|
||||
if sorted(str(x) + str(y) + str(x * y)) == pandigital:
|
||||
ans.add(x * y)
|
||||
|
||||
return sum(ans)
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 32: {compute()}")
|
||||
@ -1,47 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 04 Mar 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 33 of Project Euler
|
||||
https://projecteuler.net/problem=33
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 33")
|
||||
|
||||
def compute():
|
||||
"""
|
||||
The fraction 49/98 is a curious fraction, as an inexperienced mathematician
|
||||
in attempting to simplify it may incorrectly believe that 49/98 = 4/8,
|
||||
which is correct, is obtained by cancelling the 9s.
|
||||
|
||||
We shall consider fractions like, 30/50 = 3/5, to be trivial examples.
|
||||
|
||||
There are exactly four non-trivial examples of this type of fraction, less
|
||||
than one in value, and containing two digits in the numerator and denominator.
|
||||
|
||||
If the product of these four fractions is given in its lowest common terms,
|
||||
find the value of the denominator.
|
||||
"""
|
||||
numerator = 1
|
||||
denominator = 1
|
||||
for x in range(10, 100):
|
||||
for y in range(10, 100):
|
||||
if x < y:
|
||||
if str(x)[1] == str(y)[0]:
|
||||
if int(str(y)[1]) != 0:
|
||||
if int(str(x)[0]) / int(str(y)[1]) == x / y:
|
||||
numerator *= x
|
||||
denominator *= y
|
||||
|
||||
return int(denominator/numerator)
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 33: {compute()}")
|
||||
@ -1,41 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 02 Apr 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 34 of Project Euler
|
||||
https://projecteuler.net/problem=34
|
||||
"""
|
||||
|
||||
import math
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 34")
|
||||
def compute():
|
||||
"""
|
||||
145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
|
||||
|
||||
Find the sum of all numbers which are equal to the sum of the factorial
|
||||
of their digits.
|
||||
|
||||
Note: As 1! = 1 and 2! = 2 are not sums they are not included.
|
||||
"""
|
||||
|
||||
ans = 0
|
||||
|
||||
for num in range(10,2540160):
|
||||
sum_of_factorial = 0
|
||||
for digit in str(num):
|
||||
sum_of_factorial += math.factorial(int(digit))
|
||||
if sum_of_factorial == num:
|
||||
ans += sum_of_factorial
|
||||
|
||||
return ans
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 34: {compute()}")
|
||||
@ -1,48 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 02 Apr 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 35 of Project Euler
|
||||
https://projecteuler.net/problem=35
|
||||
"""
|
||||
|
||||
from utils import timeit, is_prime
|
||||
|
||||
def circular_number(n):
|
||||
n = str(n)
|
||||
result = []
|
||||
for i in range(len(n)):
|
||||
result.append(int(n[i:] + n[:i]))
|
||||
return result
|
||||
|
||||
@timeit("Problem 35")
|
||||
def compute():
|
||||
"""
|
||||
The number, 197, is called a circular prime because all rotations of the
|
||||
digits: 197, 971, and 719, are themselves prime.
|
||||
|
||||
There are thirteen such primes below 100:
|
||||
2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
|
||||
|
||||
How many circular primes are there below one million?
|
||||
"""
|
||||
circular_primes = []
|
||||
for i in range(2, 1_000_000):
|
||||
if is_prime(i):
|
||||
all_primes = True
|
||||
for j in circular_number(i):
|
||||
if not is_prime(j):
|
||||
all_primes = False
|
||||
break
|
||||
if all_primes:
|
||||
circular_primes.append(i)
|
||||
|
||||
return len(circular_primes)
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 35: {compute()}")
|
||||
@ -1,35 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 08 Apr 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 36 of Project Euler
|
||||
https://projecteuler.net/problem=36
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
def is_palidrome(num):
|
||||
return str(num) == str(num)[::-1]
|
||||
|
||||
@timeit("Problem 36")
|
||||
def compute():
|
||||
"""
|
||||
The decimal number, 585 = 1001001001_2 (binary), is palindromic
|
||||
in both bases.
|
||||
|
||||
Find the sum of all numbers, less than one million, which are palindromic
|
||||
in base 10 and base 2.
|
||||
"""
|
||||
ans = 0
|
||||
for i in range(1, 1_000_001, 2):
|
||||
if is_palidrome(i) and is_palidrome(bin(i)[2:]):
|
||||
ans += i
|
||||
return ans
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 36: {compute()}")
|
||||
@ -1,46 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 09 Apr 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 37 of Project Euler
|
||||
https://projecteuler.net/problem=37
|
||||
"""
|
||||
|
||||
from utils import timeit, list_primes, is_prime
|
||||
|
||||
def is_truncatable_prime(number):
|
||||
num_str = str(number)
|
||||
for i in range(1, len(num_str)):
|
||||
if not is_prime(int(num_str[i:])) or not is_prime(int(num_str[:-i])):
|
||||
return False
|
||||
|
||||
return True
|
||||
|
||||
@timeit("Problem 37")
|
||||
def compute():
|
||||
"""
|
||||
The number 3797 has an interesting property. Being prime itself, it is
|
||||
possible to continuously remove digits from left to right, and remain
|
||||
prime at each stage: 3797, 797, 97, and 7.
|
||||
Similarly we can work from right to left: 3797, 379, 37, and 3.
|
||||
|
||||
Find the sum of the only eleven primes that are both truncatable from left
|
||||
to right and right to left.
|
||||
|
||||
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
|
||||
"""
|
||||
ans = 0
|
||||
primes = list_primes(1_000_000)
|
||||
# Statement of the problem says this
|
||||
for number in primes[4:]:
|
||||
if is_truncatable_prime(number):
|
||||
ans += number
|
||||
return ans
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 37: {compute()}")
|
||||
@ -1,56 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 03 Jun 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 38 of Project Euler
|
||||
https://projecteuler.net/problem=38
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 38")
|
||||
def compute():
|
||||
"""
|
||||
Take the number 192 and multiply it by each of 1, 2, and 3:
|
||||
|
||||
192 × 1 = 192
|
||||
192 × 2 = 384
|
||||
192 × 3 = 576
|
||||
|
||||
By concatenating each product we get the 1 to 9 pandigital,
|
||||
192384576. We will call 192384576 the concatenated product of
|
||||
192 and (1,2,3)
|
||||
|
||||
The same can be achieved by starting with 9 and multiplying by
|
||||
1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is
|
||||
the concatenated product of 9 and (1,2,3,4,5).
|
||||
|
||||
What is the largest 1 to 9 pandigital 9-digit number that can
|
||||
be formed as the concatenated product of an integer with
|
||||
(1,2, ... , n) where n > 1?
|
||||
|
||||
"""
|
||||
|
||||
results = []
|
||||
# Number must 4 digits (exactly) to be pandigital
|
||||
# if n > 1
|
||||
for i in range(1, 10_000):
|
||||
integer = 1
|
||||
number = ""
|
||||
while len(number) < 9:
|
||||
number += str(integer * i)
|
||||
if sorted(number) == list('123456789'):
|
||||
results.append(number)
|
||||
|
||||
integer += 1
|
||||
|
||||
return max(results)
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 38: {compute()}")
|
||||
@ -1,44 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 05 Jun 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 39 of Project Euler
|
||||
https://projecteuler.net/problem=39
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 39")
|
||||
def compute():
|
||||
"""
|
||||
If p is the perimeter of a right angle triangle with integral length sides,
|
||||
{a,b,c}, there are exactly three solutions for p = 120:
|
||||
|
||||
{20,48,52}, {24,45,51}, {30,40,50}
|
||||
|
||||
For which value of p ≤ 1000, is the number of solutions maximised?
|
||||
"""
|
||||
ans, val = 0, 0
|
||||
for p in range(2, 1001, 2):
|
||||
sol = 0
|
||||
for a in range(1, p):
|
||||
for b in range(a+1, p-2*a):
|
||||
c = p - (a + b)
|
||||
if a**2 + b**2 == c**2:
|
||||
sol += 1
|
||||
elif a**2 + b**2 > c**2:
|
||||
# As we continue our innermost loop, the left side
|
||||
# gets bigger, right gets smaller, so we're done here
|
||||
break
|
||||
if sol > ans:
|
||||
ans, val = sol, p
|
||||
|
||||
return val
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 39: {compute()}")
|
||||
@ -1,43 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 23 Jun 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 40 of Project Euler
|
||||
https://projecteuler.net/problem=40
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 40")
|
||||
def compute():
|
||||
"""
|
||||
An irrational decimal fraction is created by concatenating the positive integers:
|
||||
|
||||
0.123456789101112131415161718192021...
|
||||
|
||||
It can be seen that the 12th digit of the fractional part is 1.
|
||||
|
||||
If d_n represents the n^th digit of the fractional part, find the value of the following expression.
|
||||
|
||||
d_1 × d_{10} × d_{100} × d_{1_000} × d_{10_000} × d_{100_000} × d_{1_000_000}
|
||||
|
||||
"""
|
||||
|
||||
fraction = ''
|
||||
for i in range(1, 1_000_000):
|
||||
fraction += str(i)
|
||||
|
||||
ans = 1
|
||||
for i in range(7):
|
||||
ans*= int(fraction[10**i - 1])
|
||||
|
||||
return ans
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 40: {compute()}")
|
||||
@ -1,39 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 29 Jun 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 41 of Project Euler
|
||||
https://projecteuler.net/problem=41
|
||||
"""
|
||||
|
||||
from utils import timeit, is_prime
|
||||
|
||||
|
||||
@timeit("Problem 41")
|
||||
def compute():
|
||||
"""
|
||||
We shall say that an n-digit number is pandigital if it makes
|
||||
use of all the digits 1 to n exactly once. For example, 2143 is
|
||||
a 4-digit pandigital and is also prime.
|
||||
|
||||
What is the largest n-digit pandigital prime that exists?
|
||||
"""
|
||||
def is_pandigital(number):
|
||||
number = sorted(str(number))
|
||||
check = [str(i) for i in range(1, len(number)+1)]
|
||||
if number == check:
|
||||
return True
|
||||
return False
|
||||
|
||||
|
||||
for ans in range(7654321, 1, -1):
|
||||
if is_pandigital(ans):
|
||||
if is_prime(ans):
|
||||
return ans
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 41: {compute()}")
|
||||
@ -1,50 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 26 Jul 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 42 of Project Euler
|
||||
https://projecteuler.net/problem=42
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 42")
|
||||
def compute():
|
||||
"""
|
||||
The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1);
|
||||
so the first ten triangle numbers are:
|
||||
|
||||
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
|
||||
|
||||
By converting each letter in a word to a number corresponding to its alphabetical
|
||||
position and adding these values we form a word value. For example, the word value
|
||||
for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a triangle number then we
|
||||
shall call the word a triangle word.
|
||||
|
||||
Using words.txt, a 16K text file containing nearly two-thousand common English words,
|
||||
how many are triangle words?
|
||||
"""
|
||||
|
||||
def triangle_number(num):
|
||||
return int(0.5*num*(num+1))
|
||||
|
||||
def word_to_value(word):
|
||||
return sum(ord(letter)-64 for letter in word)
|
||||
|
||||
triangular_numbers = [triangle_number(n) for n in range(27)]
|
||||
ans = 0
|
||||
with open("files/Problem42.txt", "r") as f:
|
||||
words = f.readline().strip('"').split('","')
|
||||
for word in words:
|
||||
if word_to_value(word) in triangular_numbers:
|
||||
ans += 1
|
||||
|
||||
return ans
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 42: {compute()}")
|
||||
@ -1,55 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 03 Aug 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 43 of Project Euler
|
||||
https://projecteuler.net/problem=43
|
||||
"""
|
||||
|
||||
from itertools import permutations
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 43")
|
||||
def compute():
|
||||
"""
|
||||
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.
|
||||
|
||||
Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this
|
||||
way, we note the following:
|
||||
|
||||
d2d3d4=406 is divisible by 2
|
||||
d3d4d5=063 is divisible by 3
|
||||
d4d5d6=635 is divisible by 5
|
||||
d5d6d7=357 is divisible by 7
|
||||
d6d7d8=572 is divisible by 11
|
||||
d7d8d9=728 is divisible by 13
|
||||
d8d9d10=289 is divisible by 17
|
||||
|
||||
Find the sum of all 0 to 9 pandigital numbers with this property.
|
||||
"""
|
||||
ans = []
|
||||
pandigital = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
|
||||
|
||||
for n in permutations(pandigital):
|
||||
n_ = "".join(n)
|
||||
if n_[0] != "0" and sorted("".join(n_)) == pandigital:
|
||||
if int(n_[7:]) % 17 == 0:
|
||||
if int(n_[6:9]) % 13 == 0:
|
||||
if int(n_[5:8]) % 11 == 0:
|
||||
if int(n_[4:7]) % 7 == 0:
|
||||
if int(n_[3:6]) % 5 == 0:
|
||||
if int(n_[2:5]) % 3 == 0:
|
||||
if int(n_[1:4]) % 2 == 0:
|
||||
ans.append(int(n_))
|
||||
|
||||
return sum(ans)
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 43: {compute()}")
|
||||
@ -1,48 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 30 Aug 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 44 of Project Euler
|
||||
https://projecteuler.net/problem=44
|
||||
"""
|
||||
|
||||
from itertools import combinations
|
||||
from operator import add, sub
|
||||
from utils import timeit
|
||||
|
||||
def pentagonal(n):
|
||||
return int(n*(3*n-1)/2)
|
||||
|
||||
@timeit("Problem 44")
|
||||
def compute():
|
||||
"""
|
||||
Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2.
|
||||
The first ten pentagonal numbers are:
|
||||
|
||||
1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
|
||||
|
||||
It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their
|
||||
difference, 70 − 22 = 48, is not pentagonal.
|
||||
|
||||
Find the pair of pentagonal numbers, Pj and Pk, for which their
|
||||
sum and difference are pentagonal and D = |Pk − Pj| is minimised.
|
||||
|
||||
What is the value of D?
|
||||
"""
|
||||
dif = 0
|
||||
pentagonal_list = set(pentagonal(n) for n in range(1,2500))
|
||||
pairs = combinations(pentagonal_list, 2)
|
||||
for p in pairs:
|
||||
if add(*p) in pentagonal_list and abs(sub(*p)) in pentagonal_list:
|
||||
dif = (abs(sub(*p)))
|
||||
# the first one found would be the smallest
|
||||
break
|
||||
|
||||
return dif
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 44: {compute()}")
|
||||
@ -1,42 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 09 Sep 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 45 of Project Euler
|
||||
https://projecteuler.net/problem=45
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
def pentagonal(n):
|
||||
return int(n*(3*n-1)/2)
|
||||
|
||||
def hexagonal(n):
|
||||
return int(n*(2*n-1))
|
||||
|
||||
@timeit("Problem 45")
|
||||
def compute():
|
||||
"""
|
||||
Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
|
||||
Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ...
|
||||
Pentagonal Pn=n(3n−1)/2 1, 5, 12, 22, 35, ...
|
||||
Hexagonal Hn=n(2n−1) 1, 6, 15, 28, 45, ...
|
||||
|
||||
It can be verified that T285 = P165 = H143 = 40755.
|
||||
|
||||
Find the next triangle number that is also pentagonal and hexagonal.
|
||||
"""
|
||||
pentagonal_list = set(pentagonal(n) for n in range(2,100_000))
|
||||
# all hexagonal numbers are also triangle numbers!
|
||||
hexagonal_list = set(hexagonal(n) for n in range(2,100_000))
|
||||
|
||||
ans = sorted(hexagonal_list & pentagonal_list)
|
||||
# First one is already known
|
||||
return ans[1]
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 45: {compute()}")
|
||||
@ -1,48 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 12 Sep 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 46 of Project Euler
|
||||
https://projecteuler.net/problem=46
|
||||
"""
|
||||
|
||||
from utils import timeit, is_prime
|
||||
|
||||
def is_goldbach(number):
|
||||
for i in range(number - 1, 1, -1):
|
||||
if is_prime(i) and ((number - i) / 2)**0.5 % 1 == 0:
|
||||
return True
|
||||
return False
|
||||
|
||||
@timeit("Problem 46")
|
||||
def compute():
|
||||
"""
|
||||
It was proposed by Christian Goldbach that every odd composite number
|
||||
can be written as the sum of a prime and twice a square.
|
||||
|
||||
9 = 7 + 2×1^2
|
||||
15 = 7 + 2×2^2
|
||||
21 = 3 + 2×3^2
|
||||
25 = 7 + 2×3^2
|
||||
27 = 19 + 2×2^2
|
||||
33 = 31 + 2×1^2
|
||||
|
||||
It turns out that the conjecture was false.
|
||||
|
||||
What is the smallest odd composite that cannot be written as the sum
|
||||
of a prime and twice a square?
|
||||
"""
|
||||
ans = 9
|
||||
while True:
|
||||
ans += 2
|
||||
if not is_prime(ans) and not is_goldbach(ans):
|
||||
return ans
|
||||
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 46: {compute()}")
|
||||
@ -1,60 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 12 Sep 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 47 of Project Euler
|
||||
https://projecteuler.net/problem=47
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
def factor(n):
|
||||
ans = []
|
||||
d = 2
|
||||
while d * d <= n:
|
||||
if n % d == 0:
|
||||
ans.append(d)
|
||||
n //= d
|
||||
else:
|
||||
d += 1
|
||||
if n > 1:
|
||||
ans.append(n)
|
||||
return ans
|
||||
|
||||
@timeit("Problem 47")
|
||||
def compute():
|
||||
"""
|
||||
The first two consecutive numbers to have two distinct prime factors are:
|
||||
|
||||
14 = 2 × 7
|
||||
15 = 3 × 5
|
||||
|
||||
The first three consecutive numbers to have three distinct prime factors are:
|
||||
|
||||
644 = 2² × 7 × 23
|
||||
645 = 3 × 5 × 43
|
||||
646 = 2 × 17 × 19.
|
||||
|
||||
Find the first four consecutive integers to have four distinct prime factors each.
|
||||
What is the first of these numbers?
|
||||
"""
|
||||
ans = []
|
||||
|
||||
for number in range(1, 1_000_000):
|
||||
if len(ans) == 4:
|
||||
break
|
||||
elif len(set(factor(number))) == 4:
|
||||
ans.append(number)
|
||||
else:
|
||||
ans = []
|
||||
|
||||
return ans[0]
|
||||
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 47: {compute()}")
|
||||
@ -1,28 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 12 Sep 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 48 of Project Euler
|
||||
https://projecteuler.net/problem=48
|
||||
"""
|
||||
|
||||
from utils import timeit
|
||||
|
||||
|
||||
@timeit("Problem 48")
|
||||
def compute():
|
||||
"""
|
||||
The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
|
||||
|
||||
Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
|
||||
"""
|
||||
series = sum(i**i for i in range(1,1001))
|
||||
return str(series)[-10:]
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 48: {compute()}")
|
||||
@ -1,41 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 18 Sep 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 49 of Project Euler
|
||||
https://projecteuler.net/problem=49
|
||||
"""
|
||||
|
||||
from utils import timeit, list_primes
|
||||
|
||||
|
||||
@timeit("Problem 49")
|
||||
def compute():
|
||||
"""
|
||||
The arithmetic sequence, 1487, 4817, 8147, in which each of the terms
|
||||
increases by 3330, is unusual in two ways:
|
||||
(i) each of the three terms are prime, and,
|
||||
(ii) each of the 4-digit numbers are permutations of one another.
|
||||
|
||||
There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes,
|
||||
exhibiting this property, but there is one other 4-digit increasing sequence.
|
||||
|
||||
What 12-digit number do you form by concatenating the three terms in this sequence?
|
||||
"""
|
||||
ans = []
|
||||
primes_list = sorted(set(list_primes(10_000)) - set(list_primes(1_000)))
|
||||
|
||||
for number in primes_list:
|
||||
if set(list(str(number))) == set(list(str(number+3330))) == set(list(str(number+6660))):
|
||||
if number+3330 in primes_list and number+6660 in primes_list:
|
||||
ans.append(str(number)+str(number+3300)+str(number+6660))
|
||||
# return the second one
|
||||
return ans[1]
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 49: {compute()}")
|
||||
@ -1,51 +0,0 @@
|
||||
#!/usr/bin/env python3
|
||||
"""
|
||||
Created on 18 Sep 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for problem 50 of Project Euler
|
||||
https://projecteuler.net/problem=50
|
||||
"""
|
||||
|
||||
from utils import timeit, list_primes, is_prime
|
||||
|
||||
|
||||
@timeit("Problem 50")
|
||||
def compute():
|
||||
"""
|
||||
The prime 41, can be written as the sum of six consecutive primes:
|
||||
|
||||
41 = 2 + 3 + 5 + 7 + 11 + 13
|
||||
|
||||
This is the longest sum of consecutive primes that adds to a prime below one-hundred.
|
||||
|
||||
The longest sum of consecutive primes below one-thousand that adds to a prime,
|
||||
contains 21 terms, and is equal to 953.
|
||||
|
||||
Which prime, below one-million, can be written as the sum of the most consecutive primes?
|
||||
"""
|
||||
|
||||
ans = 0
|
||||
result = 0
|
||||
prime_list = list_primes(1_000_000)
|
||||
|
||||
for i in range(len(prime_list)):
|
||||
sum = 0
|
||||
count = 0
|
||||
for j in prime_list[i:]:
|
||||
sum += j
|
||||
count += 1
|
||||
if is_prime(sum) and count > result:
|
||||
result = count
|
||||
ans = sum
|
||||
# print(sum, result)
|
||||
if sum > 1_000_000:
|
||||
break
|
||||
return ans
|
||||
|
||||
|
||||
if __name__ == "__main__":
|
||||
|
||||
print(f"Result for Problem 50: {compute()}")
|
||||
Loading…
x
Reference in New Issue
Block a user