program to test whether a number is prime or not.

//program to test whether a number is prime or not.
#include<stdio.h>
include<conio.h>
void main()
{
   int n,count=0,i;
   printf("enter a number for 'n'\n");
  scanf("%d",&n);
  for(i=2;i<n;i++)
    {
      if(n%i==0)
        {                
            count=1;
        }
   }
if(count==0)
{
printf("it's prime\n");
}
else
{
printf("it's not");
}
getch();
}
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logics in mind:
a prime number is that which is divisible by '1' and itself. It means there are only two numbers which can divide that given number. And any 3rd number is not there.
so here we check whether 3rd number is there or not.
->enter a number to be checked
->declare a variable with an initial value 0(u can use any other value)
-> use repetitive process from number 2 to one less than entered number 'n' i.e we divide given number 'n by 2,3... and n-1. This we can apply using loop as shown above.If we get zero as remainder while dividing then we assign value '1' to count.
->later, as the loop terminates then we check whether 'count' has got value 0 or 1.If it has '0' it means it is prime and if it has 1 then it is not prime.

we can also use following technique.
//program to test whether a number is prime or not.
#include<stdio.h>
include<conio.h>
void main()
{
   int n,count=0,i;
   printf("enter a number for 'n'\n");
  scanf("%d",&n);
  for(i=1;i<=n;i++)
    {
      if(n%i==0)
        {                
            count++;
        }
   }
if(count==2)
{
printf("it's prime\n");
}
else
{
printf("it's not");
}
getch();

}
----------------------------------------
logics in mind:
->here, we have used same concept but in different way.
->we have used value from 1 to entered number 'n' in loop to count total numbers which can divide entered number
->obviously if count is 2 then it is prime and if not then it is not

program to print/display 1+1/1,1+1/2,1+1/3,1+1/4,1+1/5,.... to nth term.

//program to print/display 1+1/1,1+1/2,1+1/3,1+1/4,1+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);

  printf("%d+%d/%d,",k,k,k);
  for(i=2;i<=n;i++)
    {               

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

  

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

->let a number k with initial value '1'

->display this using printf with characters '/' and others. so or first term, we displayed it.

->for rest of terms,     ->we have to go for repetitive process from number 2 to entered number 'n'.

     ->we display each of term using printf and all other characters as shown above.

     ->If we get same value then then we display that.

     

program to print/display all perfect square numbers lying between 1 and 'n'.

//program to print/display all perfect square numbers lying between 1 and 'n'.
#include<stdio.h>

#include<conio.h>

#include<math.h>void main()
{
   int factors,i,n;
   printf("enter  positive number for 'n'\n");
  scanf("%d",&n);
  for(i=1;i<=n;i++)
    {
      if(sqrt(i)==int(sqrt(i))
        {                

          printf("the perfect squares are=%d ",i);
        }

  }

getch();
}
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logics in mind:
->enter a number for range to get perfect square
->we have to go for repetitive process from number 1 to entered number 'n'.

->we get square root of number and then its integer value . This we can apply using loop as shown above.

->If we get same value then then we display that.

     for this, we have used 

                if(sqrt(i)==int(sqrt(i))

              it checks square root and its integer value