program to print/display all factors of number 'n'.

 //program to print/display all factors of  number 'n'.
#include<stdio.h>
include<conio.h>
void main()
{
   int number,factors,i;
   printf("enter any positive number\n");
  scanf("%d",&number);
  for(i=1;i<=number;i++)
    {
      if(number%i==0)
        {                printf("the factors are=%d ",i);
        }getch();
}
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logics in mind:
->enter a number whose factors value you want to find
->we have to go for repetitive process from number 1 to entered number 'number' i.e we divide given number 'number by 1,2,3... and itself. This we can apply using loop as shown above.If we get zero as remainder while dividing then we display that.

program to get factorial value of a positive number

//program to get factorial value of a positive number
#include<stdio.h>
include<conio.h>
void main()
{
   int number,fact=1,i;
   printf("enter any positive number\n");
  scanf("%d",&number);
if(number>0)
{ 
 for(i=1;i<=number;i++)
    {
     fact=fact*i;
   }
}
else
{
printf("the number is not +ve\n");
}
 printf("the factorial value for entered number=%d is =%d\n",number,fact);
getch();
}
------------------------------
logics in mind:
->enter a number whose factorial value you want to find
->we have to go for repetitive multiplication from number 1 to entered number 'number'. This we can apply using loop as shown above.
->display the final value of variable 'fact'.

   

program to convert hexadecimal number into binary number system.

//program to convert hexadecimal number into binary number system.

#include<stdio.h>

#define MAX 100

void main()

{

    char binaryNumber[MAX],hexaDecimal[MAX];

    long int i=0;

    printf("Enter any hexadecimal number: ");

    scanf("%s",hexaDecimal);

    printf("\nEquivalent binary value: ");

    while(hexaDecimal[i])

{

         switch(hexaDecimal[i])

      {

             case '0': printf("0000"); 

                              break;

             case '1': printf("0001"); 

                             break;

             case '2': printf("0010"); 

                             break;

             case '3': printf("0011"); 

                            break;

             case '4': printf("0100"); 

                             break;

             case '5': printf("0101"); 

                            break;

             case '6': printf("0110"); 

                           break;

             case '7': printf("0111"); 

                          break;

             case '8': printf("1000"); 

                           break;

             case '9': printf("1001"); 

                          break;

             case 'A': case 'a': printf("1010"); 

                         break;

             case 'B': case 'b': printf("1011"); 

                         break;

             case 'C': case 'c': printf("1100"); 

                         break;

             case 'D': case 'd': printf("1101"); 

                          break;

             case 'E': case 'e': printf("1110"); 

                         break;

             case 'F': case 'f': printf("1111"); 

                        break;

        

             default:  

                    printf("\nInvalid hexadecimal digit %c ",hexaDecimal[i]);

         }                           

         i++;

    }

    getch();

}

logics in mind/applied:
->we enter here hexadecimal number
->we make group of 4 bits equivalent of each digit
->for this, we check each digit.
->To check digit, we use loop and switch with each digit equivalent.

program to convert binary no. to hexadecimal number system.

//program to convert binary no. to hexadecimal number system.

#include<stdio.h>

#include<conio.h>

void main()

{

   

    long int binary_Number,hexadecimalNumber=0,j=1,remainder;

   

    printf("Enter any number any binary number: ");

    scanf("%ld",&binary_Number);

   

    while(binary_Number!=0)

{

    remainder=binary_Number%10;

    hexadecimalNumber=hexadecimalNumber+remainder*j;

        j=j*2;

        binary_Number=binary_Number/10;

      }

    printf("Equivalent hexadecimal value: %lX",hexadecimalNumber);

    getch();

}

-----------------------------------------------------------------------

logics in mind:-

->we have to input a binary number, say 111
->we make group of 3 bits from right side, and for this, we use 

hexdecimalNumber=hexadecimalNumber+remainder*j;

program to convert binary no. into octal number system

we have two methods for this conversion.
either we can convert binary to decimal and then decimal to octal
or we can directly convert into octal using 3 bits system.


//program to convert binary no. into octal number system

#include<stdio.h>
#include<conio.h>

void main()

{

   

    long int binary_Number,octalNumber=0,j=1,remainder;

    printf("Enter any number any binary number: ");

    scanf("%ld",&binary_Number);

    while(binary_Number!=0)
{

         remainder=binary_Number%10;

        octalNumber=octalNumber+remainder*j;

        j=j*2;

        binary_Number=binary_Number/10;

   }

    printf("Equivalent octal value: %lo",octalNumber);

    getch();

}

logics in mind:-

->we have to input a binary number, say 111
->we make group of 3 bits from right side, and for this, we use 

octalNumber=octalNumber+remainder*j;

        j=j*2;

program to convert octal no. into decimal number system.

//program to convert octal no. into decimal number system.
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
int k=1,m=0,n,rem,decimal=0;
printf("enter octal value/number for 'n'\n");
scanf("%d",&n);
while(n!=0)
   {
      rem=n%10;
      decimal=decimal+rem*pow(8,m);
      n=n/10;
      m++;
   }
printf("decimal is=%d\n",decimal);
getch();
}

logics in mind or applied:-
->if octal number is 234
-> we have to get first 4 and we multiply this with power( you can say increasing or decreasing)of 8
-> we have to add all 
->we display the last value

program to convert decimal into octal

//program to convert decimal number into octal number system.
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
clrscr();
int k=1,m=0,n,rem,octal=0;
printf("enter decimal value/number of 'n'\n");
scanf("%d",&n);
while(n!=0)
   {
      rem=n%8;
      octal=octal+rem*pow(10,m);
      n=n/2;
      m++;
   }
printf("octal is=%d\n",octal);
getch();
}