program to display multiplication table of a number .

//program to display multiplication table of a number .
#include<stdio.h>
#include<conio.h>
void main()
{
int k=1,n;
printf("enter value of 'n'(as a number)\n");
scanf("%d",&n);
while(k<=10)
   {
      printf("%d*%d=%d\n",n,k,n*k);
      k++;
   }
getch();
}
-------------------------------- note:
1)You can directly print instead given format.
put
printf("%d\n",n*k);

2) you can use nested loop for this as well.
 like
for(i=1;i<=1;i++)
  {
    for(k=1;k<=10;k++)
      {
          printf("%d\n",i*k);
       }
   }
you can put i =n if you enter this.
------------------------------------------------------------------------------
let's see the program using nested loop
//program to display multiplication table of a number using nested loop.
#include<stdio.h>
#include<conio.h>
void main()
{
int k,n;
printf("enter value of 'n'(a number for multiplication table)\n");
scanf("%d",&n);
for(i=n;i<=n;i++)
  {
    for(k=1;k<=10;k++)
      {
          printf("%d\n",i*k);
       }
          printf("\n");
   }
getch();
}
-----------------------------------------------------------------------------------------
let's see the program to get multiplication table of numbers using nested loop
//program to display multiplication table of  numbers using nested loop.
#include<stdio.h>
#include<conio.h>
void main()
{
int k,n;
printf("for how many numbers, you want t o go for multiplication table\n");
scanf("%d",&n);                       // from 1 to n
for(i=1;i<=n;i++)
  {
    for(k=1;k<=10;k++)
      {
          printf("%d\n",i*k);
       }
          printf("\n");
   }
getch();

}

program to print/display 2,4,6,8,10,12... to 15th term.

//program to print/display 2,4,6,8,10,12... to 15th term.

#include<stdio.h>
#include<conio.h>
void main()
{
int k=2;

while(k<=15)
   {
      printf("%d,",k);
      k=k+2;
   }
getch();
}

Or
you can use following method.

first term(a)=2
common difference(d)=2
number of terms(n)=15
here  we can see that given series is in arithmetic progression so nth term=a+(n-1)*d.It gives us nth term=30 i.e.
last term will be 30.
We put 30 in while loop

//program to print/display 2,4,6,8,10,12... to 15th term.
#include<stdio.h>
#include<conio.h>
void main()
{
int k=2;

while(k<=30)
   {
      printf("%d,",k);
      k=k+2;
   }
getch();
}

program to print/display 1,8,27... to 10th term.

//program to print/display 1,8,27... to 10th term.
#include<stdio.h>
#include<conio.h>
void main()
{
int k=1,n;

while(k<=10)
   {
      printf("%d,",k*k*k);
      k++;
   }
getch();
}

second method:
//program to print/display 1,8,27... to 10th term.
#include<stdio.h>
#include<conio.h>
#include<math.h>void main()
{
int k=1,n;

while(k<=10)
   {
      printf("%d,",pow(k,3);// it means get power 3 of k. pow() is  a function stored in 'math.h'
      k++;
   }
getch();
}


note:
If you want to display 'n' number of times then simply put
  int n;declare this.
printf("enter value of n\n");
scanf("%d",&n) 
put above two lines before while loop.

program to input your name and address and to display them 40 times.

//program  to input your name and address and to display them 40 times.
#include<stdio.h>
#include<conio.h>
void main()
{
char name[40],address[40];
int k=0;printf("enter your name\n");
gets(name);
printf("enter your address");
gets(address);
while(k<=40)
   {
      puts(name);
      puts(address);
      k++;
   }
getch();
}

note:
1)
     you can also use printf(  ) instead 'puts'.(without single quote)   and scanf("%^[\n]",name) instead 'gets' function.
2)
   If you want to display 'n' number of times then enter value of 'n' with its declaration.

program to print/display 100 to 1 natural numbers.

//program  to print/display 100 to 1 natural numbers.
#include<stdio.h>
#include<conio.h>
void main()
{
int k=100;
while(k>=1)
   {
      
       printf("%d,",k);
      
    k--;
   }
getch();
}

note:
If you want to display numbers from 'n' to 1 then
   input 'n' and keep all others same.

program to print/display all odd numbers lying between 1 and 100.

first with while loop
// program  to print/display all odd numbers lying between 1 and 100.
#include<stdio.h>
#include<conio.h>
void main()
{
int k=1;
while(k<=100)
   {
      if(k%2!=0)
      {      
         printf("%d,",k);
      }
    k++;
   }
getch();
}

now with for loop

//program  to print/display all odd numbers lying between 1 and 100.
#include<stdio.h>
#include<conio.h>
void main()
{
int k,n;
for(k=1;k<=100;k++)
   {
     if(k%2!=0)
      {      
       printf("%d,",k);
      }
    
   }
getch();
}

now using do..while loop

//program  to print/display all odd numbers lying between 1 and 100.
#include<stdio.h>
#include<conio.h>
void main()
{
int k=1,n;
do
  {
       if(k%2!=0)
       {
          printf("%d,",k);
      } 
         k++;

   }
   while(k<=100);
getch();
}


We can also use some other methods as shown below using different loop.

//program  to print/display all odd numbers lying between 1 and 100.
#include<stdio.h>
#include<conio.h>
void main()
{
int k=1;
while(k<=100)
   {
      printf("%d,",k);
      
    k=k+2;
   }
getch();
}
You now try same using other loops.

program to print/display all even numbers lying between 1 to n natural numbers.

first with while loop
//program  to print/display all even numbers lying between 1 to n natural numbers.
#include<stdio.h>
#include<conio.h>
void main()
{
int k=1,n;
printf("enter value of 'n'\n");
scanf("%d",&n);
while(k<=n)
   {
      if(k%2==0)
      {      printf("%d,",k);
      }
    k++;
   }
getch();
}

now with for loop

//program  to print/display all even numbers lying between 1 to n natural numbers.
#include<stdio.h>
#include<conio.h>
void main()
{
int k,n;
printf("enter value of 'n'\n");
scanf("%d",&n);
for(k=1;k<=n;k++)
   {
     if(k%2==0)
      {      printf("%d,",k);
      }
    
   }
getch();
}

now using do..while loop

//A program  to print/display 1 to n natural numbers.
#include<stdio.h>
#include<conio.h>
void main()
{
int k=1,n;
printf("enter value of 'n'\n");
scanf("%d",&n);
do
  {
       if(k%2==0)
       {
          printf("%d,",k);
      } 
         k++;

   }
   while(k<=n);
getch();
}


We can also use some other methods as shown below using different loop.

//program  to print/display all even numbers lying between 1 to n natural numbers.
#include<stdio.h>
#include<conio.h>
void main()
{
int k=0,n;
printf("enter value of 'n'\n");
scanf("%d",&n);
while(k<=n)
   {
      printf("%d,",k);
      
    k=k+2;
   }
getch();
}
You now try same using other loops.