moly | Python Fiddle

#https://projecteuler.net/  --> source of this script

#Problem 1 : Multiples of 3 and 5
#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.
#Find the sum of all the multiples of 3 or 5 below(미만) 1000.

def mul_sum(n):
    result = 0
    for i in range(1,n):
        if i % 3 == 0 or i % 5 == 0:
            result = result + i
    return result

#print(mul_sum(1000))

#Problem 2 : Even Fibonacci numbers
#Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
#1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
#By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
def max_fib(n, m):
    a = [1, 2]
    while a[-1] + a[-2] < n + 1:
        a.append(a[-1] + a[-2])
    if m == 1:
        result = 0
        for i in range(0, len(a)):
            if a[i] % 2 == 0:
                result = result + a[i]
        return result
    if m == 2:
        result = 0
        for i in range(0, len(a)):
            if a[i] % 2 != 0:
                result = result + a[i]
        return result
    else:
        return "second value must be 1 or 2"
    
#print(max_fib(4000000, 1))

#Problem 3: Largest prime factor
#The prime factors of 13195 are 5, 7, 13 and 29.
#What is the largest prime factor of the number 600851475143 ?
def max_pf(n):
    a = list()
    b = set()
    for i in range(1, n + 1):
        if n % i == 0:
            a.append(i)
    print(a)
    for i in range(0, len(a)):
        chk = True
        for j in range(2, a[i]):
            if a[i] % j == 0:
                chk = False
                break
            if chk:
                b.add(a[i])
    return b

#print(max_pf(10))
def prime_factor(n):
    a = list()
    for i in range(2, n+1):
        chk = True
        for j in range(2, i):
            if i % j == 0:
                chk = False
                break
    	if chk:
            a.append(i)
    return a

print(prime_factor(10))

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