Stuck on Project Euler #3 in python

Question

The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143 ?

Ok, so i am working on projec

Answer 1

Here is my python code

a=2
list1=[]
while a<=13195:                     #replace 13195 with your input number
    if 13195 %a ==0:                #replace 13195 with your input number
        x , count = 2, 0
        while x <=a:
            if a%x ==0:
                count+=1
            x+=1
        if count <2:
            list1.append(a)
    a+=1

print (max(list1))

Answer 2

Here is my Python code:

num=600851475143
i=2 # Smallest prime factor
for k in range(0,num):
    if i >= num: # Prime factor of the number should not be greater than the number
        break
    elif num % i == 0: # Check if the number is evenly divisible by i
        num = num / i
    else:
        i= i + 1
print ("biggest prime number is: "+str(num))   

Answer 3

Here is my python code:

import math
ma = 600851475143
mn = 2
s = []
while mn < math.sqrt(ma):
    rez = ma / mn
    mn += 1
    if ma % mn == 0:
        s.append(mn)
print(max(s))

Answer 4

def prime_max(x):
a = x
i = 2
while i in range(2,int(a+1)):
    if a%i == 0:
        a = a/i
        
        if a == 1:
             print(i)
        i = i-1
    i = i+1       

Answer 5

'''
The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?
'''
import math

def isPrime(x):
    if x<2:
        return False
    for i in range(2,int(math.sqrt(x))):
        if not x%i:
           return False
    return True

def largest_factor(number):
    for i in range (2, int(math.sqrt(number))):
        if number%i == 0:
            number = number/i
            if isPrime(number) is True:
                max_val = number
                break
                print max_val
        else:
            i += 1
    return max_val

largest_factor(600851475143)

This actually compiles very fast. It checks for the number formed for being the prime number or not. Thanks

Answer 6

Another solution to this problem using Python.

def lpf(x):
lpf=2
while (x>lpf):
    if (x%lpf==0):
        x=x/lpf

    else:
        lpf+=1
return x
print(lpf(600851475143))