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; ` `} ` |

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## 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); ` ` ` `} ` `} ` |

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## 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 ` |

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## 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 ` |

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## PHP

`<?php ` `// PHP Efficient program to print all ` `// prime factors of a given number ` ` ` `// function to print all prime ` `// factors of a given number n ` `function` `primeFactors(` `$n` `) ` `{ ` ` ` ` ` `// Print the number of ` ` ` `// 2s that divide n ` ` ` `while` `(` `$n` `% 2 == 0) ` ` ` `{ ` ` ` `echo` `2,` `" "` `; ` ` ` `$n` `= ` `$n` `/ 2; ` ` ` `} ` ` ` ` ` `// n must be odd at this ` ` ` `// point. So we can skip ` ` ` `// one element (Note i = i +2) ` ` ` `for` `(` `$i` `= 3; ` `$i` `<= sqrt(` `$n` `); ` ` ` `$i` `= ` `$i` `+ 2) ` ` ` `{ ` ` ` ` ` `// While i divides n, ` ` ` `// print i and divide n ` ` ` `while` `(` `$n` `% ` `$i` `== 0) ` ` ` `{ ` ` ` `echo` `$i` `,` `" "` `; ` ` ` `$n` `= ` `$n` `/ ` `$i` `; ` ` ` `} ` ` ` `} ` ` ` ` ` `// This condition is to ` ` ` `// handle the case when n ` ` ` `// is a prime number greater ` ` ` `// than 2 ` ` ` `if` `(` `$n` `> 2) ` ` ` `echo` `$n` `,` `" "` `; ` `} ` ` ` ` ` `// Driver Code ` ` ` `$n` `= 315; ` ` ` `primeFactors(` `$n` `); ` ` ` `// This code is contributed by aj_36 ` `?> ` |

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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

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