program to display following numeric/star pattern.

//program to display following numeric pattern.
//      1
//      12
//      123
//      1234
//      12345 
#include<stdio.h>
#include<conio.h>
void main()
{
int i,j;
for(i=1;i<=5;i++)
  {
     for(j=1;j<=i;j++)
          {
                printf("%d",j);
          }
       printf("\n");
 }
getch();
}
--------------------------------------------------------
logics in mind:
----------------------
1)we have to display 1,then 12, then 1233,so on.  we use loop fo this.
 2)since the displays are in repetitive form ,,we set up the loop in nested form such that
        internal loop executes first one  time with display of numbers(above),then two times  and it goes on increasing.
3)First loop(outer) takes value 1 and assigns to inner.The inner executes one times with display.
4)next time, the outer loop takes the value 2 and transfers to inner loop, it displays again and again.
and it goes on
............................................................................................
if you want to display "*" then use "*" in printf function.

program to display star/number pattern

//program to display given star/number pattern
//      12345
//        1234
//          123
//            12
//              1
#include<stdio.h>
#include<stdio.h>
void main()
{
clrscr();
int i,j,k,sp=5;
 for(i=5;i>=1;i--)
   {
               for(k=sp;k>=i;k--)
                 {
                  printf(" ");
                  }
                        for(j=1;j<=i;j++)
                           {
                            printf("%d",j);
                          }
           printf("\n");
  }
getch();
}
---------------------------------
logics in mind
-------------------------------
1)we have to display 1,2,3,4,5 so we use loop.
 2)since the displays are in repetitive form ,,we set up the loop in nested form such that
        internal loop executes 5 times with display of numbers(above).
3)if we look carefully to given pattern we can see there space increased after each display. for this, we have used loop. in each display, it increases space and then it goes for next display.
3)First loop(outer) takes value 5 and assigns to inner.The inner executes 5 times with display.
4)next time, the outer loop takes the value 4 and transfers to inner loop, it displays again and again.
and it goes on

...................................................
if you wan to display star then put "*" in printf

program to display following numeric pattern.

//program to display following numeric pattern.
//      12345
//      1234
//      123
//      12
//      1
#include<stdio.h>
#include<conio.h>
void main()
{
int i,j;
for(i=5;i>=1;i--)
  {
     for(j=1;j<=i;j++)
          {
                printf("%d",j);
          }
       printf("\n");
 }
getch();
}
--------------------------------------------------------
logics in mind:
----------------------
1)we have to display 1,2,3,4,5 so we use loop.
 2)since the displays are in repetitive form ,,we set up the loop in nested form such that
        internal loop executes 5 times with display of numbers(above).
3)First loop(outer) takes value 5 and assigns to inner.The inner executes 5 times with display.
4)next time, the outer loop takes the value 4 and transfers to inner loop, it displays again and again.
and it goes on



        

program to display first 5 Armstrong numbers between 0 and 1000.

//program to display first 5 Armstrong numbers between 0 and 1000

//we are taking numbers with three digits
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
clrscr();
int i=1,rem,k,sum=0,count=0;
while(i<=1000)
{
     k=i;
     sum=0;
while(k!=0)
  {
    rem=k%10;
    sum=sum+pow(rem,3);
    k=k/10;
  }

 if(sum==i)
 {

          count++;

           if(count<=5)

               {   printf("%d\n",i);

                }  }
 sum=0;
 i++;

}
getch();
}
-----------------------------------------------------------------------------------------------------------------------------------------------------------
logics in mind:-
-------------------
->we have to display Armstrong numbers in given range 0 and 1000.Armstrong can be for 3/4 digits numbers.
   Here we taking 3 digits only.

->First  we take first number (k) then we apply following method        

                  ->For this, 

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

                 ->we find its cube and then go for sum

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

                  ->then we compare the number 'i' with that sum. If they are same then we display that.

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

->as the first number is checked, we , once again, assign value '0' to sum to check next value.And this goes on. 

->while displaying Armstrong number, we also count it.If it is less than/equal to 5 then we display that.

program to display all Armstrong numbers between 0 and 100 solution here we are taking Armstrong number with 3 digit. but more digit also can be taken

//armstrong numbers lying in given range 0 and 1000
//we are taking numbers with three digits
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
clrscr();
int i=1,rem,k,sum=0;
while(i<=1000)
{
     k=i;
     sum=0;
while(k!=0)
 {
   rem=k%10;
   sum=sum+pow(rem,3);
   k=k/10;
 }

if(sum==i)
{
      printf("%d\n",i);

}
sum=0;
 i++;

}
getch();
}
-----------------------------------------------------------------------------------------------------------------------------------------------------------
logics in mind:-
-------------------
->we have to display Armstrong numbers in given range 0 and 1000.Armstrong can be for 3/4 digits numbers.
   Here we taking 3 digits only.

->First  we take first number (k) then we apply following method        

                  ->For this, 

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

                 ->we find its cube and then go for sum

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

                  ->then we compare the number 'i' with that sum. If they are same then we display that.

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

->as the first number is checked, we , once again, assign value '0' to sum to check next value.And this goes on.  

program to display first 20 prime numbers between 0 and 500

-----------------------------------------------------------------------------------------------------
// first twenty prime numbers lying in given range 0 and 500
#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int i,j,count,first=0,total=0;
for(i=1;i<=500;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 )

  {
    total++;

if(total<=17 )

{
 printf("%d\n",i);
}
  }
}
getch();
}
--------------------------------------------------------
logics in mind:-
-------------------
->we have to display first twenty prime numbers in given range 0 and 100
->numbers 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'
  If we count then we have three numbers  as a total between 1 and 3.
-> 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'.And to display first 17 numbers(3 already gone) we have used another variable as as a counter named 'total'.
->If the variable 'count' has same value '0' and total is less than 17   then we display that value using variable 'i'.

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();
}
----------------------------------------------------------------------------------------------------------------------------------
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'.