Project Euler getting smallest multiple in python

Question

I am doing problem five in Project Euler: "2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?"

I have constructed the following code which finds the correct value 2520 when using 1 - 10 as divisors but code seems to be going on forever when using 1 - 20. Again I don't want the code just a pointer or two on where I am going wrong. Thanks

def smallestDiv(n):
    end=False
    while end == False:
        divisors = [x for x in range(1,21)]    # get divisors
        allDivisions = zip(n % i for i in divisors)    # get values for  n % all integers in divisors
        check = all(item[0]  == 0 for item in allDivisions )   # check if all values of n % i are equal to zero
        if check:         # if all values are equal to zero return n
            end = True
            return n
        else:             # else increase n by 1
            n +=1

EDIT:

I used some code I found relating to LCM and used reduce to solve the problem:

def lcm(*values):
    values = [value for value in values]
    if values:
        n  = max(values)
        m = n
        values.remove(n)
        while any( n % value for value in values ):
            n +=m
        return n
    return 0

print reduce(lcm, range(1,21))

Answer 1

If a problem is hard, trying solving a simpler version. Here, how to calculate the lowest common multiple of two numbers. If you've read any number theory book (or think about prime factors), you can do that using the greatest common divisor function (as implemented by the Euclidean algorithm).

from fractions import gcd
def lcm(a,b):
    "Calculate the lowest common multiple of two integers a and b"
    return a*b//gcd(a,b)

Observing lcm(a,b,c) ≡ lcm(lcm(a,b),c) it's simple to solve your problem with Python's reduce function

>>> from functools import reduce
>>> reduce(lcm, range(1,10+1))
2520
>>> reduce(lcm, range(1,20+1))
232792560

Answer 2

You are doing a brute force search, so it can get arbitrary long. You should read about LCM (least common multiple) in order to code an efficient solution.(which I believe is 232792560)