# Find all factors of a natural number | Set 1

• Difficulty Level : Easy
• Last Updated : 02 Mar, 2022

Given a natural number n, print all distinct divisors of it.

Examples:

Input : n = 10
Output: 1 2 5 10

Input:  n = 100
Output: 1 2 4 5 10 20 25 50 100

Input:  n = 125
Output: 1 5 25 125

Note that this problem is different from finding all prime factors.

A Naive Solution would be to iterate all the numbers from 1 to n, checking if that number divides n and printing it. Below is a program for the same:

## C++

 // C++ implementation of Naive method to print all // divisors #include using namespace std;   // function to print the divisors void printDivisors(int n) {     for (int i = 1; i <= n; i++)         if (n % i == 0)             cout <<" " << i; }   /* Driver program to test above function */ int main() {     cout <<"The divisors of 100 are: \n";     printDivisors(100);     return 0; }   // this code is contributed by shivanisinghss2110

## C

 // C implementation of Naive method to print all // divisors #include   // function to print the divisors void printDivisors(int n) {     for (int i=1;i<=n;i++)         if (n%i==0)             printf("%d ",i); }   /* Driver program to test above function */ int main() {     printf("The divisors of 100 are: \n");     printDivisors(100);     return 0; }

## Java

 // Java implementation of Naive method to print all // divisors   class Test {     // method to print the divisors     static void printDivisors(int n)     {         for (int i=1;i<=n;i++)             if (n%i==0)                 System.out.print(i+" ");     }       // Driver method     public static void main(String args[])     {         System.out.println("The divisors of 100 are: ");         printDivisors(100);;     } }

## Python3

 # Python implementation of Naive method # to print all divisors   # method to print the divisors def printDivisors(n) :     i = 1     while i <= n :         if (n % i==0) :             print (i,end=" ")         i = i + 1           # Driver method print ("The divisors of 100 are: ") printDivisors(100)   #This code is contributed by Nikita Tiwari.

## C#

 // C# implementation of Naive method // to print all divisors using System;   class GFG {           // method to print the divisors     static void printDivisors(int n)     {         for (int i = 1; i <= n; i++)             if (n % i == 0)                 Console.Write( i + " ");     }       // Driver method     public static void Main()     {         Console.Write("The divisors of",                           " 100 are: ");         printDivisors(100);;     } }   // This code is contributed by nitin mittal.



## Javascript



Output:

The divisors of 100 are:
1 2 4 5 10 20 25 50 100

Time Complexity : O(n)
Auxiliary Space : O(1)

Can we improve the above solution?
If we look carefully, all the divisors are present in pairs. For example if n = 100, then the various pairs of divisors are: (1,100), (2,50), (4,25), (5,20), (10,10)
Using this fact we could speed up our program significantly.
We, however, have to be careful if there are two equal divisors as in the case of (10, 10). In such case, we’d print only one of them.

Below is an implementation for the same:

## C++

 // A Better (than Naive) Solution to find all divisors #include #include using namespace std;   // Function to print the divisors void printDivisors(int n) {     // Note that this loop runs till square root     for (int i=1; i<=sqrt(n); i++)     {         if (n%i == 0)         {             // If divisors are equal, print only one             if (n/i == i)                 cout <<" "<< i;               else // Otherwise print both                 cout << " "<< i << " " << n/i;         }     } }   /* Driver program to test above function */ int main() {     cout <<"The divisors of 100 are: \n";     printDivisors(100);     return 0; }   // this code is contributed by shivanisinghss2110

## C

 // A Better (than Naive) Solution to find all divisors #include #include   // Function to print the divisors void printDivisors(int n) {     // Note that this loop runs till square root     for (int i=1; i<=sqrt(n); i++)     {         if (n%i == 0)         {             // If divisors are equal, print only one             if (n/i == i)                 printf("%d ", i);               else // Otherwise print both                 printf("%d %d ", i, n/i);         }     } }   /* Driver program to test above function */ int main() {     printf("The divisors of 100 are: \n");     printDivisors(100);     return 0; }

## Java

 // A Better (than Naive) Solution to find all divisors   class Test {     // method to print the divisors     static void printDivisors(int n)     {         // Note that this loop runs till square root         for (int i=1; i<=Math.sqrt(n); i++)         {             if (n%i==0)             {                 // If divisors are equal, print only one                 if (n/i == i)                     System.out.print(" "+ i);                        else // Otherwise print both                     System.out.print(i+" " + n/i + " " );             }         }     }       // Driver method     public static void main(String args[])     {         System.out.println("The divisors of 100 are: ");         printDivisors(100);;     } }

## Python3

 # A Better (than Naive) Solution to find all divisors import math   # method to print the divisors def printDivisors(n) :           # Note that this loop runs till square root     i = 1     while i <= math.sqrt(n):                   if (n % i == 0) :                           # If divisors are equal, print only one             if (n / i == i) :                 print (i,end=" ")             else :                 # Otherwise print both                 print (i , int(n/i), end=" ")         i = i + 1   # Driver method print ("The divisors of 100 are: ") printDivisors(100)   #This code is contributed by Nikita Tiwari.

## C#

 // A Better (than Naive) Solution to // find all divisors using System;   class GFG {           // method to print the divisors     static void printDivisors(int n)     {                   // Note that this loop runs         // till square root         for (int i = 1; i <= Math.Sqrt(n);                                       i++)         {             if (n % i == 0)             {                                   // If divisors are equal,                 // print only one                 if (n / i == i)                     Console.Write(i + " ");                                   // Otherwise print both                 else                     Console.Write(i + " "                             + n / i + " ");             }         }     }       // Driver method     public static void Main()     {         Console.Write("The divisors of "                           + "100 are: \n");         printDivisors(100);     } }   // This code is contributed by Smitha



## Javascript



Output:

The divisors of 100 are:
1 100 2 50 4 25 5 20 10

Time Complexity: O(sqrt(n))
Auxiliary Space : O(1)

However there is still a minor problem in the solution, can you guess?
Yes! the output is not in a sorted fashion which we had got using the brute-force technique. Please refer below for an O(sqrt(n)) time solution that prints divisors in sorted order.
Find all divisors of a natural number | Set 2