C Program to Find Product of Digits of a Number

Ex: Write a C program to print product of digits of a number. How to write a C program to print product of digits of a number. C program to find product of digits of number.


Input from user:

Enter the number: 12345

Expected output:

Product of given number is: 120





   Step by step logic of the given program:

1. Accept input (number) from user.

2. Use one while loop to check given number is greater than 0 or not.

3. Loop will iterate if number is greater than 0.

4. Inside the while loop product of digit calculate using...

z=no%10;
product*=z;
no=no/10;

5. Print product on the output screen.

   C Program to find product of digits of a number:

#include<stdio.h>
int main()
{
int no,z;
long long int product=1ll;

printf("Enter the number\n");
scanf("%d",&no);

while(no!=0)
{
z=no%10;
product*=z;
no=no/10;
}
printf("product of given number is:%lld",product);

return 0;

}


Above prgram shows the following output:

C Program to Check Given Number is Palindrome or Not

Ex: write a C program to check given number is palindrome or not. How to write a C program to check given number is palindrome or not. C Program to check given number is palindrome or not.


Input from user:


Enter the number: 12321


Expected output:

Given number is palindrome.

   Step by step logic of the given program:

1. Accept input (number) from user.

2. After that use one while loop to check given number is greater than zero or not.

3. If number is greater than 0 then while loop will iterate.

4. Use reverse number logic inside the loop to calculate reverse number.

reverse=reverse*10; reverse=reverse+(no % 10);
no=no/10;

5. After that using if statement check condition
if(reverse==num)

6. Then print given number is palindrome. Otherwise print given numbers is not palindrome.





   C program to check given number is palindrome or not:

#include<stdio.h>
int main()
{
int no,num,reverse=0;

    printf("Enter the number:");
    scanf("%d",&no);
    num=no;
    
   while(no!=0)
    {
reverse=reverse*10; reverse=reverse+(no % 10);
no=no/10;   
    }    
    if(reverse==num)
    {
    printf("Given number is palindrome");
    }
    else
    {
    printf("Given number is not palindrome");
    }
    
    return 0;

}



Above prgram shows the following output:

C program to reverse a given number using while loop

Ex: Write a C program for reverse a given number. How to write a C program to reverse a given number. C program for reverse a given number.

Input from user:

Enter the number: 12345


Expected output:

Given number in reverse order: 54321

   Step by step logic of the given program:

1. Accept input (number) from user.

2. Use one while loop to check given number is greater than 0 or not.

3. After that inside the while loop add reverse number logic.

4. Inside the loop the reverse number computed using..

reverse=reverse*10;
reverse=reverse+no%10;
no=no/10;

5. Print reverse number on the output screen.

   C program to reverse a given number using while loop:

#include<stdio.h>
int main()
{
 int no,reverse=0;

printf("Enter the number\n");
scanf("%d",&no);

    while(no>0)
    {
reverse=reverse*10;
reverse=reverse+no%10;
no=no/10;
    }

printf("Given number in reverse order:%d",reverse);

return 0;
}




Above program shows the following output:

C Program to Print Digit Sum Using While Loop

Ex: write a C program to print digit sum. How to write a C program to print digit sum. C program to print digit sum.


Input from user:

Enter the number: 145

Expected output:

Sum of digit=10

  Step by step logic of the given program:

1. Accept number from user.

2. Use while loop to check given number is greater than zero or not.

3.  Enter given digit sum logic inside the while loop.

z=no%10;
sum=sum+z;
no=no/10;

4. Print sum outside the while loop.

  C program to print digit sum using while loop:

#include<stdio.h>
void main()
{
int no,z,sum=0;

printf("Enter the number:");
scanf("%d",&no);

while(no>0)
{
z=no%10;
sum=sum+z;
no=no/10;
}
printf("sum of digit=%d",sum);

}



Above prgram shows the following output:

C program to print number while is less than 10

Ex: write a C program to print number while is less than 10. How to write a C program to print number while is less than 10. C program to print number while is less than 10.





   Step by step logic of the given program:

1. First we declare intiger variable no=5.

2. After that use one while loop to check number is less than 10 or not.

3. If number is less than 10 we print no inside the while loop.

4. After that we increment number by 1 (no++).

5. It will print (5 6 7 8 9 10) .




   C program to print number while  is less than 10:


#include<stdio.h>
int main()
{
int no=5;

while(no<=10)
{
printf("%d ",no);
no++;
}

return 0;
}



Above prgram shows the following output:

Ex: Write a C program to print multiplication table of given number. How to write a C program to print multiplication table of given number. C program to print multiplication table of given number.


Input from user:

Enter the number: 12

Expected output:

Multiplication table of 12 is:

12x1=12
12x2=24
12x3=36
12x4=48
12x5=60
12x6=72
12x7=84
12x8=96
12x9=108
12x10=120

   Step by step logic of the given program:

1. Accept input number from user.

2. Use for loop to print multiplication table. for loop will iterate 1 to 10.   for(i=1;i<=10;i++)

3. inside the for loop print multiplication table using no*i.

   C program to print multiplication table of given number :

#include<stdio.h>
int main()
{
int no,i;

printf("Enter the number\n");
scanf("%d",&no);

printf("Multiplication table of %d is:\n",no);

for(i=1;i<=10;i++)
{
printf("%dx%d=%d\n",no,i,(no*i));
}

return 0;
}



Above prgram shows the following output:

C program to print odd numbers from 1 to n without using if statement

Ex: Write a C program to print odd numbers between 1 to n. How to write a C program to print odd numbers between 1 to n. C program to print odd number's between 1 to n.


Input from user: 

Enter the limit: 20

Expected output:

Odd numbers between 1 to 20:

1
3
5
7
9
11
13
15
17
19

   Step by step logic to print Odd numbers between 1 to n(without using if-statement):

1. Accept limit from user store it in variable no.

2. Run for loop which iterate upto n number's. for(i=1;i<=no;i+=2)

3. We used i=1 because odd numbers starts from 1.

4.  We increment loop by two to print odd numbers.

5. Print value of i inside the for loop.




   Program to print odd number's without using if-statement:


#include<stdio.h>
int main()
{

 int i,no;

 printf("Enter the limit: ");
 scanf("%d",&no);

  printf("Odd numbers between 1 to %d\n",no);

 for(i=1;i<=no;i+=2)
 {
   printf("%d\n",i);
 }

 return 0;
}

   Program to print odd number's using if-statement:

In below Program we print odd numbers from 1 to n using if-statement.


#include<stdio.h>
int main()
{

 int i,no;

 printf("Enter the limit: ");
 scanf("%d",&no);

  printf("Odd numbers between 1 to %d\n",no);

 for(i=1;i<=no;i++)
 {
   if(i%2==1)
   printf("%d\n",i);

 }

 return 0;

}



Above program's shows the following output:

C program to print Even number's from 1 to n without using if statement

Ex: Write a C program to print even numbers between 1 to n. How to write a C program to print even numbers between 1 to n. C program to print even number's between 1 to n.


Input from user: 

Enter the limit: 20

Expected output:

Even numbers between 1 to 20:

2
4
6
8
10
12
14
16
18
20

  Step by step logic to print even numbers between 1 to n(without using if-statement):

1. Accept limit from user store it in variable no.

2. Run for loop which iterate upto n number's. for(i=2;i<=no;i+=2)

3. We used i=2 because even number's starts from 2.

4.  We increment loop by two to print even numbers.

5. Print value of i inside the for loop.

  Program to print even numbers without using if-statement:

#include<stdio.h>
int main()
{

int i,no;

printf("Enter the limit: ");
scanf("%d",&no);

printf("Even numbers between 1 to %d\n",no);

for(i=2;i<=no;i+=2)
{
printf("%d\n",i);

}

return 0;
}

  Program to print even numbers using if-statement:

In below program we used if-statement to print even numbers from 1 to n.


#include<stdio.h>
int main()
{

 int i,no;

 printf("Enter the limit: ");
 scanf("%d",&no);

  printf("Even numbers between 1 to %d\n",no);

 for(i=1;i<=no;i++)
 {
   if(i%2==0)
   printf("%d\n",i);

 }

 return 0;
}



Above two programs will show the following output:

C program to print sum of 1 to n natural numbers

Ex: Write a C program to print sum of 1 to n natural numbers. C program to print sum of 1 to n natural numbers. How to write a C program to print sum of 1 to n natural numbers.


Input from user:

Enter the limit: 10


Expected output:

Sum is 55

  Step by step logic of the given program:

1.  First declare three intiger variable's no(accept limit), i(use in for loop) and sum(store addition of n natural numbers).
2. Accept limit from user.

3. Use for loop to get 1 to n natural numbers.
 for(i=1;i<=no;++i)
4. Add natural numbers one by one and store this in variable sum (sum=sum+i)
5. Print sum out of the for loop to print sum of all natural numbers.

    C program to print sum of 1 to n natural numbers:

#include<stdio.h>
int main()
{

int no,i,sum=0;

printf("Enter the limit:");
scanf("%d",&no);

for(i=1;i<=no;++i)
{
  sum=sum+i;
}
printf("\nsum is %d",sum);

return 0;
}



Above prgram shows the following output:

C Program to Print 1 to n Natural Numbers

Ex: Write a C program to print 1 to n natural numbers. How to write a C program to print 1 to n natural numbers. C program to print 1 to n natural numbers.

Input from user:

Enter the limit: 20

Expected output:

Natural numbers from 1 to 20:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

     Step by step logic to print 1 to n natural numbers:

1. First declare two intiger variable's no for accept limit from user and i variable use in for loop.

2. Accept limit from user.

3. After that use one for loop to print 1 to n natural numbers.
loop like for(i=1; i<=no;i++)

     How It Works?

i=1 because we  need to print natural numbers from 1.

i<=no because we need to print natural numbers upto n. (n=no=20). 

i++ =i+1 because we need to increment it by one. 

 i=1=i+1=2=i+1=3=i+1=4=i+1=5...upto n.


4. Print value of i inside the for loop. Because we need to print natural numbers from 1 to n. (i will iterate from 1 to n)

    C program to print 1 to n natural numbers :

#include <stdio.h>
int main()
{
  int no,i;
  
  printf("Enter the limit:");
  scanf("%d",&no);
  
  printf("\nNatural numbers from 1 to %d : ",no);
  
  for(i=1;i<=no;i++)
  {
  printf("%d ",i);
  }
  return 0;

}



Above prgram shows the following output: