Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems.
In this second post, I'm going to solve problems six to ten. Again, I will code the remaining solutions as time permits.
The sum of the squares of the first ten natural numbers is, 1^2 + 2^2 + ... + 10^2 = 385 The square of the sum of the first ten natural numbers is, (1 + 2 + ... + 10)^2 = 55^2 = 3025 Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640. Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.
def sum_square_difference(n): return (sum(range(1, n+1))**2) - sum(i**2 for i in range(1,n+1)) print(sum_square_difference(100))
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10 001st prime number?
def nth_prime(n): def is_prime(num): if num < 2: return False for i in range(2, int(num ** 0.5) + 1): if num % i == 0: return False return True primes =  i = 2 while len(primes) < n: if is_prime(i): primes.append(i) i += 1 return primes[-1] print(nth_prime(10001))
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which, a^2 + b^2 = c^2 For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2. There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.
def pythagorean_triplet(): for a in range(1, 1000): for b in range(a, 1000): c = 1000 - a - b if a**2 + b**2 == c**2: return a*b*c print(pythagorean_triplet())
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million.
def is_prime(num): if num < 2: return False for i in range(2, int(num ** 0.5) + 1): if num % i == 0: return False return True def prime_sum(n): return sum(i for i in range(2, n) if is_prime(i)) print(prime_sum(2000000))