HarishGadade

Programs-02

1. Write a C program to check whether a given number (N) is a perfect number or not?

[Perfect Number - A perfect number is a positive integer number which is equals to the sum of its proper positive divisors. For example 6 is a perfect number because its proper divisors are 1, 2, 3 and it’s sum is equals to 6.]

int main()
{
  int n,i=1,sum=0; 
  printf("Enter a number: ");
  scanf("%d",&n); 
  while(i<n){
      if(n%i==0)
           sum=sum+i;
          i++;
  }
  if(sum==n)
      printf("%d is a perfect number",i);
  else
      printf("%d is not a perfect number",i);
    return 0;
}

Here is Output

Enter a number: 6
6 is a perfect number

Enter a number: 87
87 is not a perfect number

2. Write a C program to count total number of digits of an Integer number (N).

int main()
{
        int num,tNum,cnt;
	printf("Enter a number: ");
	scanf("%d",&num);
	cnt=0;
	tNum=num;
	while(tNum>0){
		cnt++;
		tNum/=10;
	}
	printf("Total numbers of digits are: %d in number: %d.",cnt,num);
        return 0;
}

Here is the output

Enter a number: 30001                                                            
Total numbers of digits are: 5 in number: 30001.

3. Write a C program to check whether the given number(N) can be expressed as Power of Two (2) or not. For example 8 can be expressed as 2^3.

int main()
{
    int num;
    int tempNum,flag;     
    printf("Enter an integer number: ");
    scanf("%d",&num);     
    tempNum=num;
    flag=0;
    /*check power of two*/
    while(tempNum!=1)
    {
        if(tempNum%2!=0){
            flag=1;
            break;
        }
        tempNum=tempNum/2;
    }  
    if(flag==0)
        printf("%d is a number that is the power of 2.",num);
    else
        printf("%d is not the power of 2.",num);	
    return 0;
}

Here is the output

Enter an integer number: 8                                                       
8 is a number that is the power of 2.

4. Write a C program to find sum of following series where the value of N is taken as input

1+ 1/2 + 1/3 + 1/4 + 1/5 + .. 1/N

int main()
{
     double number, sum = 0, i;
    printf("\n Enter the number : ");
    scanf("%lf", &number);
    for (i = 1; i <= number; i++)
    {
        sum = sum + (1 / i);
        if (i == 1)
            printf("\n 1 +");
        else if (i == number)
            printf(" (1 / %lf)", i);
        else
            printf(" (1 / %lf) + ", i);
    }
    printf("\n The sum of the given series is %.2lf", sum);	
    return 0;
}

Here is the output

Enter the number:  6                                                              
                                                                                 
 1 + (1 / 2.000000) +  (1 / 3.000000) +  (1 / 4.000000) +  (1 / 5.000000) +  (1 /
 6.000000)                                                                       
 The sum of the given series is 2.45

C Programming

Programs-02