Project Euler: Problem 3


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

What is the largest prime factor of the number 600851475143?

Source: //


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:

  • 1
  • 2
  • 5
  • 10


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.



Source: //

def primeNumbers(n):
    # Found this example on Wikipedia: //
    # 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))

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.



One comment

  • sujith
    September 21, 2011 - 11:42 pm | Permalink
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