Project Euler - Problem 5 - Smallest multiple

Today I came across this interesting problem. Project Euler's Problem 5 - Smallest Multiple.

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?

Definitely 20! (factorial of 20) is one, which will be evenly divisible by all of the numbers from 1 to 20. But the problem is to find the smallest positive number. So, I thought, it would be the number which is the product of all the prime numbers within 20. I calculated 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 = 9699690, which is wrong, since a number divisible by 2 might not be divisible by 4 and the same applies for other prime numbers and their multiples. After little thinking, I couldn’t get a way to find the solution. So I applied brute force and my solution got accepted.

Brute Force Solution

i = 21
while True:
    flag = True
    for X in xrange(1, 21):
        if (i % X != 0):
            flag = False
            break
    if flag:
        print i
        break
    i += 1

But the problem with this solution is, the time taken to produce the result. It took 162 seconds or 2 minutes and 32 seconds. I wondered if I can improve my solution to produce solution in a reasonable amount of time. I couldn’t, so I googled and got this page. Its high school maths.

Optimized Solution

LCM

Least Common Divisor (LCM) of numbers A and B, is a number which is the least number divisible by both A & B. Moreover LCM (A, B, C) = LCM (LCM (A, B), C) and so on.

Relation between LCM and GCD

LCM(A,B) = (A * B)/GCD(A*B)

Recursive Euclid's algorithm for GCD

Base Condition

GCD(A, 0) = A

Recurrence Relation

GCD(A, B) = GCD (B, A mod B)

So I modified my code like this and it produced the solution within 1 second.

def GCD(M, N):
    while (N != 0):
        M, N = N, M % N
    return M

LCM = 1
for X in xrange(1, 21):
    LCM = (X * LCM)/GCD(X, LCM)
print LCM