program to get/display digits which are even present in a number.

//program to get/display digits which are even position 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++;

       if(count%2==0)

        {

          printf("%d",rem);

          }

       n=n/10;    

   }

getch();

}

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logics in mind:-

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

->If we have number 123 then its even positioned digits are 2(from both side). It means, first we have to get 3 then 2 and then 1.        

                  ->For this, 

                 ->we have to divide by 10 to get last digit. 

                 ->we also use 'count' to to know position  of digits.

                 ->we divide that count by 2 to get remainder.If count is 2 then we understand that the digit is in even position and if not then that is not.

                 ->Then we display that.

                -> 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.        

program to get sum of all digits present in a number.

//program to get sum of all digits present in a number.

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

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

  while(n!=0)
    {

      rem=n%10;

       sum=sum+rem;

        n=n/10;    

   }

   printf("the sum=%d",sum);

getch();

}

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logics in mind:-

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

->to get sum of digits,first we have to get all digits one by one.If we have number 123 then its sum is 3+2+1=6. It means, first we have to get 3 then 2 and then 1.        

     ->For this, 

                 ->we have to divide by 10 to get last digit. 

                ->Then we go for sum using sum=sum+digit.

                -> 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 and last sum is received. We use loop for this as shown above in the code.             

->then final sum is displayed.    

program to reverse a number.

//program to reverse a number.

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

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

  while(n!=0)
    {

      rem=n%10;

       printf("%d",rem);

       n=n/10;    

   }

getch();

}

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logics in mind:-

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

->If we have number 123 then its reverse is 321. It means, first we have to get 3 then 2 and then 1.        ->For this, 

                 ->we have to divide by 10 to get last digit. 

                 ->Then we display that.

                -> 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.                 

program to get series (Fibonacci)0,1,1,2,3,5,8....15th term.

//program to get series 0,1,1,2,3,5,8....15th term.

#include<stdio.h>

#include<conio.h>

void main()
{
    int i=0,j=1,n,k;

     printf("%d\n",i);

     printf("%d\n",j);

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

  for(i=1;i<=n-2;i++)
    {          

         k=i+j;     

          printf( "%d\n",k);

        i=j;

       j=k;

    }

getch();
}
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logics in mind:

->take two initial numbers as 'i' and 'j' for first two values '0' and '1'. If you want then you can input these numbers and display them.

->since we need to display 15th terms so we input 15 or 16 for variable 'n'. We can also fix it inside loop.

->Fibonacci series is that where we get terms by adding two previous terms.

->We have displayed first two terms (0 and 1). now we need to display only 13th terms.

->We display that using loop.

->inside loop, we get next term by adding fist two terms (look above).

     ->we display that.

     ->now we assign value of 'j' to 'i' and 'k' to 'j'. it works and goes on working for next cycle/iteration

program to test whether a number is palindrome or no

//program to test whether a number is palindrome or not

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

#include<math.h>
void main()
{
   int n,n1,n2,rem,sum=0,count=0;
   printf("enter a number for 'n'\n");
  scanf("%d",&n);

n1=n;

n2=n;
  while(n!=0)
    {

       count=count+1;

       n=n/10;    

   }

count--;

while(n1!=0)
    {

       rem=n1%10;
       sum=sum+rem*(pow(10,count));

       n1=n1/10;

       count--;     

   }

if(sum==n2)
{
printf("it's palindrome\n");
}
else
{
printf("it's not palindrome");
}
getch();
}
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logics in mind:
a palindrome number is that which is same from both the sides i.e. from left to right and from right . e.g. 101 or 121 or 1221 etc.

->first we enter a number.

->before we go for further process,we first need to get total digits present in that number. For this, we have used loop (first,look above in code) with a variable 'count'. We have used logic 

                                                                                              ->count=count+1

                                                                                             ->n=n/10

->We decrease the value of 'count' by one. Let's take an example, 121.

                                                   ->we have to reverse it. for this, we first get last digit '1'. To get '1', we divide it                                                                     by 10 and get remainder. 

                                                  ->121 can be written as 100+20+1(in reverse order)

                                                                                        1x10²+2x10¹+1x10⁰

                                                  ->If we look at this expression then we can see the power is decreasing from 2                                                             to 0. So we have to start power from 2 and not from 3. So we have written                                                        count --
->We use again loop to get sum after reversing the entered number. We have also used power(count) in   decreasing order.

->as we get sum then we check that with original value.

program to get 1+1,1+2,1+3,1+4,1+5,.... to nth term.

//program to get of 1+1,1+2,1+3,1+4,1+5,....           to nth term.
#include<stdio.h>

#include<conio.h>

void main()
{
   int i,n,k=1;

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

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

         printf("%d+%d,",k,i);
        }

  

getch();
}
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logics in mind:
->enter a number for range for which you want to get/print , say 'n'.

->let a variable k=1 with initial value '1'. We take this because first term has value '1'.

->Now we display each of the term using special format as shown above using printf.

->For first term we take variable k then we put characters '+' and ten 'i' because its value is changing 

  as the loop changes. We get that term.

program to get sum of 1+1,1+2,1+3,1+4,1+5,.... to nth term.

//program to get sum of 1+1,1+2,1+3,1+4,1+5,....           to nth term.
#include<stdio.h>

#include<conio.h>

void main()
{
   float i,n,k=1,sum=0;

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

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

          sum=sum+(k+i);
        }

  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=0 with initial value '0'

->we get sum using formula sum=sum+k+i because in terms 1+1,1+2,1+3,1+4,1+5,....        ,the first value is same 

 and second value is changing so for first, k is working and for second, 'i' is working

->at last we display final sum.