Given a number n, write an efficient function to print all prime factors of n. For example, if the input number is 12, then output should be “2 2 3”. And if the input number is 315, then output should be “3 3 5 7”.

Following are the steps to find all prime factors.

**1)** While n is divisible by 2, print 2 and divide n by 2.

**2)** After step 1, n must be odd. Now start a loop from i = 3 to square root of n. While i divides n, print i and divide n by i, increment i by 2 and continue.

**3)** If n is a prime number and is greater than 2, then n will not become 1 by above two steps. So print n if it is greater than 2.

## C/C++

// Program to print all prime factors # include <stdio.h> # include <math.h> // A function to print all prime factors of a given number n void primeFactors(int n) { // Print the number of 2s that divide n while (n%2 == 0) { printf("%d ", 2); n = n/2; } // n must be odd at this point. So we can skip // one element (Note i = i +2) for (int i = 3; i <= sqrt(n); i = i+2) { // While i divides n, print i and divide n while (n%i == 0) { printf("%d ", i); n = n/i; } } // This condition is to handle the case when n // is a prime number greater than 2 if (n > 2) printf ("%d ", n); } /* Driver program to test above function */ int main() { int n = 315; primeFactors(n); return 0; }

## Java

// Program to print all prime factors import java.io.*; import java.lang.Math; class GFG { // A function to print all prime factors // of a given number n public static void primeFactors(int n) { // Print the number of 2s that divide n while (n%2==0) { System.out.print(2 + " "); n /= 2; } // n must be odd at this point. So we can // skip one element (Note i = i +2) for (int i = 3; i <= Math.sqrt(n); i+= 2) { // While i divides n, print i and divide n while (n%i == 0) { System.out.print(i + " "); n /= i; } } // This condition is to handle the case whien // n is a prime number greater than 2 if (n > 2) System.out.print(n); } public static void main (String[] args) { int n = 315; primeFactors(n); } }

## Python

# Python program to print prime factors import math # A function to print all prime factors of # a given number n def primeFactors(n): # Print the number of two's that divide n while n % 2 == 0: print 2, n = n / 2 # n must be odd at this point # so a skip of 2 ( i = i + 2) can be used for i in range(3,int(math.sqrt(n))+1,2): # while i divides n , print i ad divide n while n % i== 0: print i, n = n / i # Condition if n is a prime # number greater than 2 if n > 2: print n # Driver Program to test above function n = 315 primeFactors(n) # This code is contributed by Harshit Agrawal

## C#

// C# Program to print all prime factors using System; namespace prime { public class GFG { // A function to print all prime // factors of a given number n public static void primeFactors(int n) { // Print the number of 2s that divide n while (n % 2 == 0) { Console.Write(2 + " "); n /= 2; } // n must be odd at this point. So we can // skip one element (Note i = i +2) for (int i = 3; i <= Math.Sqrt(n); i+= 2) { // While i divides n, print i and divide n while (n % i == 0) { Console.Write(i + " "); n /= i; } } // This condition is to handle the case whien // n is a prime number greater than 2 if (n > 2) Console.Write(n); } // Driver Code public static void Main() { int n = 315; primeFactors(n); } } } // This code is contributed by Sam007

Output:

3 3 5 7

**How does this work?**

The steps 1 and 2 take care of composite numbers and step 3 takes care of prime numbers. To prove that the complete algorithm works, we need to prove that steps 1 and 2 actually take care of composite numbers. This is clear that step 1 takes care of even numbers. And after step 1, all remaining prime factor must be odd (difference of two prime factors must be at least 2), this explains why i is incremented by 2.

Now the main part is, the loop runs till square root of n not till. To prove that this optimization works, let us consider the following property of composite numbers.

*Every composite number has at least one prime factor less than or equal to square root of itself.*

This property can be proved using counter statement. Let a and b be two factors of n such that a*b = n. If both are greater than √n, then a.b > √n, * √n, which contradicts the expression “a * b = n”.

In step 2 of the above algorithm, we run a loop and do following in loop

a) Find the least prime factor i (must be less than √n,)

b) Remove all occurrences i from n by repeatedly dividing n by i.

c) Repeat steps a and b for divided n and i = i + 2. The steps a and b are repeated till n becomes either 1 or a prime number.

**Related Article : **

Prime Factorization using Sieve O(log n) for multiple queries

Thanks to **Vishwas Garg** for suggesting the above algorithm. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above