P.E.5: Smallest Multiple

2520 is the least number which is divisible by any no's i 1to 10.

now to find the least number which is divisible by any no's 1 to 20.

2520 is not div. by 11,13,16,17,19 without any remainder.and all other no's will divide 2520.

so we multiply in 2520 by 11,13,17,19 and 2 bcoz 8 divides , now only 2 remains...... 16=8*2

..........so now the answer is

                          =2520*2*11*13*17*19
                          =232792560.

It is my first solution.I hope ,you guy's will like this......

My python code is as follows:

    """
    Euler Problem #5:
    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?
    """
    numcheck = [11, 13, 14, 16, 17, 18, 19, 20]
    def answer(fin_step):
        for num in xrange(fin_step, 999999999, fin_step):
            if all(num % n == 0 for n in numcheck):
                return num
        return None
    solution = answer(20)
    if solution is not None:
       print "the smallest positive number that is evenly divisible by all of the numbers from 1 to 20 is"
       print solution
    raw_input()

Hope this helps!

just find the lcm of 1 to 20

Find all prime factor with max. power in that given range like for 1 to 10, power of 2 can be max. 3. hence (2^3)x(3^2)x(5^1)x(7^1)=2520 Similarly for range 1 to 20 (2^4)x(3^2)x(5^1)x(7^1)x(11^1)x(13^1)x(17^1)x(19^1)=232792560