program to display all prime numbers between 0 and 100.

//prime numbers in given range 0 and 100
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i,j,count;
for(i=1;i<=100;i++)
{
      if(i<=3)
{
 printf("%d\n",i);
}
      else
      {
     count=0;
       for(j=2;j<=i-1;j++)
 {
   if(i%j==0)
     {
      count=1;
     }
 }
      }
 if(count==0)
  {
   printf("%d\n",i);
  }
}
getch();
}
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logics in mind:-
---------------
->we have to display prime numbers in given range 0 and 100
->number s are 1,2,3,,5,7,....
  we can see that first three numbers are 1,2,3 prime so for them we have used loop with 'if'
-> as the number(i) exceeds 3 it transfers the control to next loop where that number is divided from 2 to  i-1. We are not taking 1 because we want to know , is there any number which can divide with remainder '0' or not.
->We want to know here , how many numbers are there which can divide completely with remainder '0'.
->To know that, we have used one variable 'count'. If it is divisible by by some other numbers then 'count' becomes '1'. otherwise it remains '0'.
->If the variable 'count' has same value '0' then we display that value using variable 'i'.

program to count number of digits present in a number.

//program to count number of digits present in a number.

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

void main()
{
   int n,rem,count=0;
   printf("enter a number for 'n'\n");
  scanf("%d",&n);

  while(n!=0)
    {

        rem=n%10;

       count++;

        n=n/10;    

   }

printf("total digits =%d",count);

getch();

}

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

logics in mind:-

>first we enter a number(for 'n').

->If we have number 123 then it has 3 digits and they are 3, 2 and 1 or 1,2 and 3. It means, first we have to get 3 then 2 and then 1 and we have to count them.        

                  ->For this, 

                 ->we have to divide by 10 to get last digit as remainder 

                 ->we also use 'count' to start counting  digits.

                -> to get second digit, we get first 12 and for this , we use 123/10.It is done to get integer                           part  only.

->We repeat this until the value reaches 0. We use loop for this as shown above in the code.     

->at last we display total number of digits stored in variable 'count'.   

program to get sum of given series with................nth terms.

//program to get sum of x+ x²/2+x³/3+x⁴/4+.............................nth terms.
#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()
{
  float i,n,x,m=1,sum=0;

   printf("enter  positive number for 'n'\n");
  scanf("%f",&n);

printf("enter value for 'x'\n");

scanf("%f",&x);

  for(i=1;i<=n;i++)
    {               

          sum=sum+(pow(x,m)/i);

           m=m+1;

     }

  printf("the sum=%f",sum);

getch();
}

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logics in mind:
->enter a number for range for which you want to get sum, say 'n'.

->let a variable 'sum' with initial value '0'

->here we also need value for 'x' so let's input that

->we get sum using formula sum=sum+(pow(x,m)/i); because in terms x+ x²/2+x³/3+x⁴/4+..............  ,

    ->we have two components namely numerator and denominator. Let's look at numerator;they are

           x¹,x²,x³,x⁴.....it can be formulated as pow(x,m) .  .here the power (m) goes on increasing.
   ->similarly for denominator, it is simply 1,2,3,4.... so it can be used from loop (variable i).

->we find sum inside loop

->at last we display the final sum.

   here 'pow' means finding power. it exists inside math.h

program to get sum of 1+3/4+7/16+15/64..............................nth terms.

//program to get sum of 1+3/4+7/16+15/64..............................nth terms.
#include<stdio.h>

#include<conio.h>

void main()
{
   float i,n,k=2,sum=0,m=0;

   printf("enter  positive number for 'n'\n");
  scanf("%f",&n);

  for(i=1;i<=n;i++)
    {               

          sum=sum+(pow(k,i)-1)/(pow(k,m));

          m=m+2;    

     }

  printf("the sum=%f",sum);

getch();
}

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

logics in mind:
->enter a number for range for which you want to get sum, say 'n'.

->let a variable 'sum' with initial value '0'

->we get sum using formula sum=sum+(pow(k,i)-1)/(pow(k,m)) because in terms 1+3/4+7/16+15/64...    ,

    ->we have two components namely numerator and denominator. Let's look at numerator;they are

           1,3,7,16.....it can be formulated as pow(k,i)-1 that is    2¹-1  .here the power goes on increasing
   ->similarly for denominator, it has power of 2 i.e.
                 2⁰,2²,2^{4. we can see here that power is increasing by 2.}

                for this we have,pow(k,m) then m=m+2
->we find sum inside loop

->at last we display the final sum.

program to get sum of given series with ..................nth terms.

//program to get sum of 1²x1+2³x3+3⁴x5+..............................nth terms.
#include<stdio.h>

#include<conio.h>

void main()
{
   float i,n,k=2,m=1sum=0;

   printf("enter  positive number for 'n'\n");
  scanf("%f",&n);

  for(i=1;i<=n;i++)
    {               

          sum=sum+(pow(i,k)*m);

           k=k+1;

           m=m+2;

     }

  printf("the sum=%f",sum);

getch();
}

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

logics in mind:
->enter a number for range for which you want to get sum, say 'n'.

->let a variable 'sum' with initial value '0'

->we get sum using formula sum=sum+(pow(i,k)*m) because in terms 1²x1+2³x3+3⁴x5+..     ,

        the first value is i(1),second is k(2) and third is m(1).the values in second and all other terms are changing 

   i----by 1
   k by 1
  m by 2     
  so

        we put these values  inside loop.

->at last we display final sum.

program to get sum of (2x3)/5+(4x5)/7+(6x7)/9+,.... to nth term.

//program to get sum of (2x3)/5+(4x5)/7+(6x7)/9+,....           to nth term.
#include<stdio.h>

#include<conio.h>

void main()
{
   float i,n,k=2,m=3,p=5,sum=0;

      printf("enter  positive number for 'n'\n");
  scanf("%f",&n);

  for(i=1;i<=n;i++)
    {               

          sum=sum+(k*m)/p;

           k=k+2;

           m=m+2;

          p=p+2;        }

  printf("the sum=%f",sum);

getch();
}

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logics in mind:
->enter a number for range for which you want to get sum, say 'n'.

->let a variable 'sum' with initial value '0'

->we get sum using formula sum=sum+(k*m)/p because in terms (2x3)/5+(4x5)/7+(6x7)/9+,...     ,

        the first value is k(2),second is m(3) and third is p(5).the values in second and all other terms are changing 

       by +2 so

        we put

          k=k+2;

           m=m+2;

          p=p+2; inside loop.

->at last we display final sum.

program to get 0.9,0.99,0.999,0.9999.....nth terms.

//program to get 0.9,0.99,0.999,0.9999.....nth terms.
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
float n,j, k=0.9;
clrscr();
printf("enter number of terms\n");
scanf("%f",&n);
for(j=1;j<=n;j++)
{
printf("%f,",k);
k=k+0.9/(pow(10,j));
}
getch();
}
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logics in mind for this:-
->0.9,0.99,0.999............. can be written as
             0.9,0.9+0.09,0.99+0.009.......
->         for second term 0.99, it can be written as 0.9+0.09
 ->further it can be as(taking only two terms)
            0.9+( 0.9+0.9/10)...
->it can be as
          0.9+(0.9+0.9/10^{1)...
->now we have to convert it in a formula as}
        k=k+0.9/(pow(10,j)).here k=0.9 is taken
->and to display all the terms we have used loop

or

we can apply following technique:
//getting 0.9,0.99,0.999.....nth term
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
float k=9,n,p=10,j,dis,m=1;
clrscr();
printf("enter number of terms\n");
scanf("%f",&n);
for(j=1;j<=n;j++)
{
dis=k/(pow(10,j));
printf("%f,",dis);
m++;
k=pow(p,m)-1;
}
getch();
}

->take 9 as a fixed value
->we enter no. of terms for 'n'.
->to display 0.9, we divide it (9) by 10<sup>1
->to get 0.99, we take here 99, and for this we use 100-1 concept. we do this using k=pow(p,m)-1
        here pow finds power of 10(p).
->in next execution we get 0.99 and so on..

-> this is how loop continues and we get all terms

we can use different technique for same program.</sup>

         similarly we can also get 
1)9,99,999,9999......
2)2,22,222,2222....
3)5,55,555,5555......
4)0.99999,0.9999,0.999,0.99,0.9
    for this,
      there are 5 nines so it can be written as
         (100000-1)/10⁵,(10000-1)/10⁴........
         (10⁵-1)/10⁵,(10⁴-1)/10⁴........  
       now make a formula to get this one. And it is easy task...