C Program to Check Whether a Given Number is Even or Odd

C and C++ programs to check whether a given number is Odd or even

C Program

#include<stdio.h>
void main()
{
int n;
printf("Enter an integer\n");
scanf("%d",&n);
if ( n%2 == 0 )
printf("Even");
else
printf("Odd");
}

Program to Check Armstrong Number

An Armstrong number is a positive integer for which the sum of 'n'th power of its digits is equal to the number itself where n is the number of digits in the number. This is a c program to check whether a given number is an Armstrong number or not. For example, 153 is a 3 digit Armstrong number because 1³+5³+3³=153.

# include <stdio.h>
# include <conio.h>
# include <math.h>
void main ()
{
int a,b=0,sum=0;
long int n;
printf("Enter the number to check\n ");
scanf("%i",&n);
if(n<1)
{
printf ("\nThe number should be greater than 0");
}
else
{
a=n;
//counting the digits
while (a>0)
  {
  a=a/10;
  b++;
  }
a=n;
//adding up bth power of digits
while(a>0)
  {
  sum=sum+pow(( a%10) ,b);
  a=a/10;
  }
if(sum==n)
  printf ("\nThe number is an ARMSTRONG number");
else
  printf ("\nThe number is NOT an ARMSTRONG number");
}
getch();
}

Program to Display All Armstrong Numbers upto 1000

This post is for C program to display all Armstrong numbers less than 1000. But before all, I should tell what an Armstrong number is. An Armstrong number is a number that is the sum of its own digits each raised to the power of the number of digits. Consider an example. 153 is an Armstrong number. Since it is a 3 digit number, to check whether it is an Armstrong number or not, we should add the 3rd power digits together. So, in the program, we should have a code to determine number of digits in number, then add the n^th powers (where n is the number of digits) of each digit in the number and check whether both the sum and original number are the same.

#include<stdio.h>
#include<math.h>
void main()
{
int n,sum,r,digits;
for(n=1;n<=1000;n++)
  {
  digits=0;
  r=n;
  sum=0;
  while(r!=0)  //Counting Digits
    {
    digits++;
    r/=10;
    }
  r=n;
  while(r!=0)  /*Adding Digits Raised To a Power That Is Equal To Number of Digits *
    {
    sum+=pow((r%10),digits);
    r=r/10;
    }
  if(sum==n)
    {
    printf("%5d",sum);
    }
      }
}