A solution to the 5th Euler Challenge

Euler5

"""I wish to create a function lcm(x) which gives the lowest common multiple of an array of numbers x"""

# I will need the factor function

def factor(x):
    array=[]
    i=1
    while (i<=x):
        if x%i==0:
            array.append(i)
        i+=1    
    return array

# And the prime function

def prime(x):
    if factor(x)==[1,x]:
        return True
    else:
        return False

# now combine these to give a prime factor function

def primeFactor(x):
    ans=[]
    y=factor(x)
    for z in y:
        if prime(z):
            ans.append(z)
    return ans        
            

#A function output how many times x divides y:

def powerDivide(x,y):
    ans=0
    n=0
    while y%(pow(x,n))==0:
        n+=1
    return n-1

#Now define a function that outputs the highest power of 2 that divides any element of an array x:

def powerTwo(x):
    ans=0
    for y in x:
        if powerDivide(2,y)>ans:
            ans=powerDivide(2,y)
    return ans

#Now generalise the above to any prime p:

def power(p,x):
    ans=0
    for y in x:
        if powerDivide(p,y)>ans:
            ans=powerDivide(p,y)
    return ans

#Define a function that gives the primes dividing the elements of x:

def primes(x):
    ans=[]
    for y in x:
        ans+=primeFactor(y)
    ans=list(set(ans))    
    return ans

#Define a function copy(x,n) giving an array containing n copies of x:

def copy(x,n):
    ans=[]
    m=0
    while m<n:
        ans.append(x)
        m+=1
    return ans    

#Define a function that computes the prime decomposition as an array of lcm(x):

def primeDecompLcm(x):
    ans=[]
    for y in primes(x):
        ans+=copy(y,power(y,x))
    return ans

#Define a function that multiplies the elements of an array:

def mult(x):
    ans=1
    for y in x:
        ans*=y
    return ans

#define the lcm function

def lcm(x):
    return mult(primeDecompLcm(x))

print(lcm(range(1,20)))