Project Euler Problem 3 Solution: Largest prime factor

Project Euler Problem 3 Statement

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

What is the largest prime factor of the number 600851475143?

Solution

A reasonable way to solve this problem is to use trial division to factor an integer, $n$. In this instance, we create a set of possible integer factors, $d$, from the set $\{2\} \cup \{3, 5, 7, 9, 11, \dots, \sqrt{n}\}$ and try to divide $n$.

For all $d$ that divide $n$, we remove $d$ and all its factors. Once our set of factors are exhausted, the remainder becomes the largest prime factor.

Below is a walk–through of this algorithm to show the process of removing factors and leaving the remainder as the largest prime factor of $n$.

Finding the largest prime factor

n = 6435
d = 2
while (d*d <= n):
    if (n % d == 0):
        n //= d
    else:
       d += 2 if d>2 else 1  # after 2, consider only odd d

Example: n = 6435
Step  d        d²    n   n%d==0(d divides n)?
  1   2  and   4 ≤ 6435   N: 2 is not a factor of 6435, increment d to 3
  2   3  and   9 ≤ 6435   Y: 3 is a factor of 6435: n=n//3 = 2145
  3   3  and   9 ≤ 2145   Y: 3 is a factor of 2145: n=n//3 = 715
  4   3  and   9 ≤  715   N: 3 is not a factor of 715, increment d to 5
  5   5  and  25 ≤  715   Y: 5 is a factor of 715: n=n//5 = 143
  6   5  and  25 ≤  143   N: 5 is not a factor of 143, increment d to 7
  7   7  and  49 ≤  143   N: 7 is not a factor of 143, increment d to 9
  8   9  and  81 ≤  143   N: 9 is not a factor of 143, increment d to 11
  9  11  and 121 ≤  143   Y 11 is a factor of 143: n=n//11 = 13
 10  11  and 121 >   13: the while loop terminates and 13 is the largest prime factor¹
¹The while’s conditional, d² ≤ n, is checked after executing the loop’s last statement.

If the given number, n, is already prime then n is returned as the largest prime factor, which is nice, because it is, in fact, itself the largest prime factor.

HackerRank version

Solves all test cases for Hackerrank Project Euler Problem 3

HackerRank Project Euler 3 runs up to 10 test cases with 10 ≤ N ≤ 10¹².

Python Source Code

for _ in range(int(input())):
    N = int(input())
    while N%2==0: N//= 2 
    p = 3
    while (p*p <= N):      
        if (N % p == 0):
            N //= p
        else: 
            p+= 2
    print(2 if N==1 else N)

Last Word

Trial division, used here, works well for smaller numbers less than 10²¹, for larger numbers there are several efficient algorithms available.

A more favorable method for factoring numbers under 100 digits long is the Quadratic sieve. It is a much simpler algorithm to implement compared to others and lends itself to parallelization. Because it is a sieve, large amounts of memory may be required.

Another is Pollard’s Rho algorithm. It’s easy to program and fast for numbers with small factors.

For a practical application of this algorithm see the JavaScript prime factor calculator. It finds all the prime factors of a number.