## Problem:

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

What is the largest prime factor of the number 600851475143?

*Source: http://projecteuler.net/index.php?section=problems&id=3*

## Solution:

Before we can being to solved the problem above we first must answer the questions raised by it. In other words:

- What is a prime number?
- What is a factor?
- What is a prime factor?
- How do we find the prime factors for a number?

### What is a prime number?

A prime number is any natural number that can only be divided by 1 or itself. E.G. 7 is a prime number because it can only be divided by 1 or 7 while 10 is not a prime number because it’s divisors are 1,2,5,10.

### What is a factor?

A factor (also known as a divisor) is any number that that can divide in to another number and not leave a remainder. In the previous section about prime numbers you can see the for 10 it’s factors are:

### What is a prime factor?

A prime factor is any factor that is also a prime number.

### How do we find the prime factors for a number?

The method of finding prime factors is called Prime Factorisation. It involves taking a number and dividing it by the smallest prime possible, then taking what’s remaining and dividing that by the smallest prime possible; keep doing this until the remainder is a prime. This Khans Academy video offers a great introduction to this concept.

## Code:

*Source: https://bitbucket.org/TWith2Sugars/project-euler/src/8892d0a39946/python/3.py*

def primeNumbers(n):
# Found this example on Wikipedia: http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
# Create a candidate list within which non-primes will be
# marked as None; only candidates below sqrt(n) need be checked.
candidates = list(range(n+1))
# Square root of n
fin = int(n**0.5)
# Loop over the candidates, marking out each multiple.
for i in xrange(2, fin+1):
if candidates[i]:
candidates[2*i::i] = [None] * (n//i - 1)
# Filter out non-primes and return the list.
return [i for i in candidates[2:] if i]
def primeFactors(n):
# Generators all of the prime factors for n
# The largest number that a prime factor could be
maxPossiblePrimeNumber = int(n**0.5)
# List of all prime numbers up to the square root of n
primes = primeNumbers(maxPossiblePrimeNumber)
while n > 1:
for p in primes:
# This is a prime factor
if n % p == 0:
# Set n to be the result of the division of n by the current prime number (p)
n = n / p
yield p
maxPrime = max(primeFactors(600851475143))
print(maxPrime)

The code is broken down in to two parts:

- Generating Prime Numbers (
*primeNumbers*)
- Performing Prime Factorisation (
*primeFactors*)

### Generating Prime Numbers

There are many methods of generating prime numbers and in my search I’ve decided to use a method called Sieve of Eratosthenes which is pretty efficient are generating prime numbers below 100. It also comes with a python example that we can use. You’ll see the code below.

### Performing Prime Factorisation

This function just implemented the method of finding prime factors as mentioned earlier, there is probably a more efficient/mathematical way of doing this but for now it’s a good start.