# C Program to Find and Print all Factors of given Number

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In this tutorial, we will learn about how to create a program in C that will ask from the user to enter any number (at run-time) as input and find and print all the factors of that given number as output. Here is the program:

```#include<stdio.h>
#include<conio.h>
int main()
{
int num, rem, i;
printf("Enter any number: ");
scanf("%d", &num);
printf("\nFactors of %d:\n", num);
for(i=1; i<num; i++)
{
if(num%i==0)
printf("%d\n", i);
}
getch();
return 0;
}```

The program was written under Code::Blocks IDE, here is the sample run after successful build and run. This is the first snapshot of the sample run:

Supply any number say 24 to see all the factors of 24 after pressing the `ENTER` key:

Here are some of the main steps used in above program:

• Receive any number as input say 24
• Create a for loop that runs from 1 to one less than the number say 23 here (if input is 24)
• Inside the for loop, use if block to apply divisibility test of 24 with each and every number from 1 to 23
• That is, if any of the number totally divides the given number say 24, then print it out as a factor of the given number and continue to check for the next number
• At first run of the for loop, i holds 1. 1 totally divides 24 without leaving any remainder, therefore 1 will be one of the factor of 24
• At second run of the for loop, i holds 2. 2 totally divides 24 without leaving any remainder, therefore 2 will also be one of the factor of 24
• Again at third run, i holds 3. It also divides 24 without leaving any remainder, therefore 3 will also be one of the factor of 24
• In same way, check for all the number upto 23 and print it out if the number totally divides the given number say 24 without leaving any remainder

Now let's modify the above program in such way, the program will print the total number of factors with number-wise list. Here is the program. The question is, write a program in C to print all the factors of any given number by user at run-time. For example, if user has provided 24 as input. Then the output must looks like:

```There are 7 factors of 24:
 -> 1
 -> 2
 -> 3
 -> 4
 -> 6
 -> 8
 -> 12```

The program given below is the answer to above question:

```#include<stdio.h>
#include<conio.h>
int main()
{
int num, rem, i, count=0;
printf("Enter any number: ");
scanf("%d", &num);
for(i=1; i<num; i++)
{
if(num%i==0)
count++;
}
printf("\nThere are %d factors of %d:\n", count, num);
count=1;
for(i=1; i<num; i++)
{
if(num%i==0)
{
printf("[%d] -> %d\n", count, i);
count++;
}
}
getch();
return 0;
}```

Here final snapshot of the sample run of above program:

C Online Test

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