$2520$ 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 (divisible with no remainder) by all of the numbers from $1$ to $20$?

This problem can be solved by taking the prime factors of all the numbers from $1$ to $20$ and using them to find the least common multiple.

$1 : 1$

$2 : 2$

$3 : 3$

$4 : 2^2$

$5 : 5$

$6 : 2 * 3$

$7 : 7$

$8 : 2^3$

$9 : 3^2$

$10 : 2 * 5$

$11 : 11$

$12 : 2^2 * 3$

$13 : 13$

$14 : 2 * 7$

$15 : 3 * 5$

$16 : 2^4$

$17 : 17$

$18 : 2 * 3^2$

$19 : 19$

$20 : 2^2 * 5$

$-> 2^4 * 3^2 * 5 * 7 * 11 * 13 * 17 * 19 $

pow(2, 4) * pow(3, 2) * 5 * 7 * 11 * 13 * 17 * 19

232792560

Another clever way to solve this for the range $1$ to $x$ would be to find all of the primes less than or equal to than $x$, find out each prime's maximum exponent less or equal to $x$, then multiplying all of those numbers together to find their least common multiple.

import math
def isPrime(n):
    if n == 1 or n == 2:
        return True
    elif n > 1:
        # check for factors
        for i in range(2,n):
            if (n % i) == 0:
                return False
        return True
    else:
        return False


def primesLessThan(x):
    i = 2
    primeList = []
    while(i <= x):
        if (isPrime(i)):
            primeList.append(i)
        i += 1
    return primeList

num = 20
primeList = primesLessThan(num)
powList = [0] * len(primeList)


print(primesLessThan(num))

for i in range(0, len(primeList)):
    for j in range(1, math.floor(math.sqrt(num)) + 1):
        if (pow(primeList[i], j) <= num):
            powList[i] = j

print(powList)

lcm = 1
for i in range(0, len(primeList)):
    lcm = lcm * pow(primeList[i], powList[i])

print("The least common multiple of all numbers less than ", num, " is ", lcm)
        

[2, 3, 5, 7, 11, 13, 17, 19]
[4, 2, 1, 1, 1, 1, 1, 1]
The least common multiple of all numbers less than  20  is  232792560