Program : 7

PROGRAM - 7

a)  Write a program to calculate the following: 

Sum = 1-x² /2!+x⁴ /4!-…..+x¹⁰ /10!

Program:

#include<stdio.h>

#include<conio.h>

#include<math.h>

main()

{

       double sum=1,x,f=1,j,i,p,k=1;

       clrscr();

       printf("enter the value of x");

       scanf("%lf",&x);

       for(i=2;i<=10;i+=2)

       {

        f=1;

        for(j=2;j<=i;j++)

        {

           p=pow(x,i);

           f=f*j;

        }

        k=-k;

        sum=sum+(k*(p/f));

       }

       printf("the value of expression is %lf",sum);

       getch();

}

/*

INPUT:1.enter the value of x 2

      2.enter the value of x 7

OUTPUT:1.the value of expression is -0.416155

       2.the value of expression is -21.726113

*/

b)  i) A perfect number is a number that is the sum of all its divisors except itself. Six is the perfect number. The only numbers that divide 6 evenly are 1,2,3 and 6 (i.e 1+2+3=6).

ii) An abundant number is one that is less than the sum of its divisors (Ex: 12<1+2+3+4+6).

iii) A Deficient number is one that is greater than the sum of its divisors (Ex: 9>1+3)

      Write a program to classify N integers (Read N from keyboard) each as perfect, abundant or deficient.

Program:

#include<stdio.h>

#include<conio.h>

#include<math.h>

main()

{

   int n,sum=0,i;

   clrscr();

   printf("enter the number");

   scanf("%d",&n);

   for(i=1;i<=n/2;i++)

   {

     if(n%i==0)

     sum=sum+i;

   }

   if(sum==n)

   printf("PERFECT NUMBER");

   if(sum<n)

   printf("DEFICIENT NUMBER");

   if(sum>n)

   printf("ABUNDANT NUMBER");

   getch();

}

/*

INPUT:enter the number6

OUTPUT:PERFECT NUMBER

*/

For the range from 0 to N

Program:

#include<stdio.h>

#include<conio.h>

main()

{

     int n,sum=0,i,j;

     clrscr();

     printf("enter the range ");

     scanf("%d",&n);

     for(i=1;i<=n;i++)

     {

       sum=0;

       for(j=1;j<i;j++)

       {

          if(i%j==0)

          sum=sum+j;

       }

       if(sum==i)

       printf("\nthe perfect number is %d",i);

       if(sum<i)

       printf("\nthe deficient number is %d",i);

       if(sum>i)

       printf("\nthe abundant number is %d",i);

     }

     getch();

}

/*

INPUT:enter the range 10

OUTPUT:

the deficient number is 1

the deficient number is 2

the deficient number is 3

the deficient number is 4

the deficient number is 5

the perfect number is 6

the deficient number is 7

the deficient number is 8

the deficient number is 9

the deficient number is 10

*/