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  Armstrong Numbers Between Two Integers # include <math.h> # include <stdio.h> int main () { int low, high, number, originalNumber, rem, count = 0 ; double result = 0.0 ; printf ( "Enter two numbers(intervals): " ); scanf ( "%d %d" , &low, &high); printf ( "Armstrong numbers between %d and %d are: " , low, high); // iterate number from (low + 1) to (high - 1) // In each iteration, check if number is Armstrong for (number = low + 1 ; number < high; ++number) { originalNumber = number; // number of digits calculation while (originalNumber != 0 ) { originalNumber /= 10 ; ++count; } originalNumber = number; // result contains sum of nth power of individual digits while (originalNumber != 0 ) { rem = originalNumber % 10 ; result += pow (rem, count); originalNumber /= 10 ; } // check if number is e...

  Check Armstrong Number of three digits # include <stdio.h> int main () { int num, originalNum, remainder, result = 0 ; printf ( "Enter a three-digit integer: " ); scanf ( "%d" , &num); originalNum = num; while (originalNum != 0 ) { // remainder contains the last digit remainder = originalNum % 10 ; result += remainder * remainder * remainder; // removing last digit from the orignal number originalNum /= 10 ; } if (result == num) printf ( "%d is an Armstrong number." , num); else printf ( "%d is not an Armstrong number." , num); return 0 ; } Output Enter a three-digit integer: 371 371 is an Armstrong number.

  Display Prime Numbers Between Two Intervals # include <stdio.h> int main () { int low, high, i, flag; printf ( "Enter two numbers(intervals): " ); scanf ( "%d %d" , &low, &high); printf ( "Prime numbers between %d and %d are: " , low, high); // iteration until low is not equal to high while (low < high) { flag = 0 ; // ignore numbers less than 2 if (low <= 1 ) { ++low; continue ; } // if low is a non-prime number, flag will be 1 for (i = 2 ; i <= low / 2 ; ++i) { if (low % i == 0 ) { flag = 1 ; break ; } } if (flag == 0 ) printf ( "%d " , low); // to check prime for the next number // increase low by 1 ++low; } return 0 ; } Output Enter two numbers(intervals): 20 50 Prime numbers between 20 and 50 are: 23 29 31 37 41 43 47

  Program to Check Prime Number # include <stdio.h> int main () { int n, i, flag = 0 ; printf ( "Enter a positive integer: " ); scanf ( "%d" , &n); for (i = 2 ; i <= n / 2 ; ++i) { // condition for non-prime if (n % i == 0 ) { flag = 1 ; break ; } } if (n == 1 ) { printf ( "1 is neither prime nor composite." ); } else { if (flag == 0 ) printf ( "%d is a prime number." , n); else printf ( "%d is not a prime number." , n); } return 0 ; } Output Enter a positive integer: 29 29 is a prime number.

  Program to Check Palindrome # include <stdio.h> int main () { int n, reversedN = 0 , remainder, originalN; printf ( "Enter an integer: " ); scanf ( "%d" , &n); originalN = n; // reversed integer is stored in reversedN while (n != 0 ) { remainder = n % 10 ; reversedN = reversedN * 10 + remainder; n /= 10 ; } // palindrome if orignalN and reversedN are equal if (originalN == reversedN) printf ( "%d is a palindrome." , originalN); else printf ( "%d is not a palindrome." , originalN); return 0 ; } Output Enter an integer: 1001 1001 is a palindrome.

  Reverse an Integer # include <stdio.h> int main () { int n, rev = 0 , remainder; printf ( "Enter an integer: " ); scanf ( "%d" , &n); while (n != 0 ) { remainder = n % 10 ; rev = rev * 10 + remainder; n /= 10 ; } printf ( "Reversed number = %d" , rev); return 0 ; } Output Enter an integer: 2345 Reversed number = 5432

  LCM using while and if # include <stdio.h> int main () { int n1, n2, max; printf ( "Enter two positive integers: " ); scanf ( "%d %d" , &n1, &n2); // maximum number between n1 and n2 is stored in min max = (n1 > n2) ? n1 : n2; while ( 1 ) { if (max % n1 == 0 && max % n2 == 0 ) { printf ( "The LCM of %d and %d is %d." , n1, n2, max); break ; } ++max; } return 0 ; } Output Enter two positive integers: 72 120 The LCM of 72 and 120 is 360.

There are many ways to find the greatest common divisor in C programming. Example #1: GCD Using for loop and if Statement # include <stdio.h> int main () { int n1, n2, i, gcd; printf ( "Enter two integers: " ); scanf ( "%d %d" , &n1, &n2); for (i= 1 ; i <= n1 && i <= n2; ++i) { // Checks if i is factor of both integers if (n1%i== 0 && n2%i== 0 ) gcd = i; } printf ( "G.C.D of %d and %d is %d" , n1, n2, gcd); return 0 ; } Example #2: GCD Using while loop and if...else Statement # include <stdio.h> int main () { int n1, n2; printf ( "Enter two positive integers: " ); scanf ( "%d %d" ,&n1,&n2); while (n1!=n2) { if (n1 > n2) n1 -= n2; else n2 -= n1; } printf ( "GCD = %d" ,n1); return 0 ; } Output Enter two positive integers:...

  The Fibonacci sequence is a sequence where the next term is the sum of the previous two terms. The first two terms of the Fibonacci sequence are 0 followed by 1. The Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21 Visit this page to learn about  the Fibonacci sequence . Fibonacci Series up to n terms # include <stdio.h> int main () { int i, n, t1 = 0 , t2 = 1 , nextTerm; printf ( "Enter the number of terms: " ); scanf ( "%d" , &n); printf ( "Fibonacci Series: " ); for (i = 1 ; i <= n; ++i) { printf ( "%d, " , t1); nextTerm = t1 + t2; t1 = t2; t2 = nextTerm; } return 0 ; } Output Enter the number of terms: 10 Fibonacci Series: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, Fibonacci Sequence Up to a Certain Number # include <stdio.h> int main () { int t1 = 0 , t2 = 1 , nextTerm = 0 , n; printf ( "Enter a positive number: " ); scanf ( "%d...

  The program below takes an integer input from the user and generates the multiplication tables up to 10. Multiplication Table Up to 10 # include <stdio.h> int main () { int n, i; printf ( "Enter an integer: " ); scanf ( "%d" , &n); for (i = 1 ; i <= 10 ; ++i) { printf ( "%d * %d = %d \n" , n, i, n * i); } return 0 ; } Output Enter an integer: 9 9 * 1 = 9 9 * 2 = 18 9 * 3 = 27 9 * 4 = 36 9 * 5 = 45 9 * 6 = 54 9 * 7 = 63 9 * 8 = 72 9 * 9 = 81 9 * 10 = 90

  The factorial of a positive number   n   is given by: factorial of n (n!) = 1 * 2 * 3 * 4. ...n The factorial of a negative number doesn't exist. And, the factorial of 0 is 1. Factorial of a Number # include <stdio.h> int main () { int n, i; unsigned long long fact = 1 ; printf ( "Enter an integer: " ); scanf ( "%d" , &n); // shows error if the user enters a negative integer if (n < 0 ) printf ( "Error! Factorial of a negative number doesn't exist." ); else { for (i = 1 ; i <= n; ++i) { fact *= i; } printf ( "Factorial of %d = %llu" , n, fact); } return 0 ; } Output Enter an integer: 10 Factorial of 10 = 3628800

  The positive numbers   1, 2, 3...   are known as natural numbers. The sum of natural numbers up to 10 is: sum = 1 + 2 + 3 + ... + 10 Sum of Natural Numbers Using for Loop # include <stdio.h> int main () { int n, i, sum = 0 ; printf ( "Enter a positive integer: " ); scanf ( "%d" , &n); for (i = 1 ; i <= n; ++i) { sum += i; } printf ( "Sum = %d" , sum); return 0 ; } The above program takes input from the user and stores it in the variable  n . Then,  for  loop is used to calculate the sum up to  n . Sum of Natural Numbers Using while Loop # include <stdio.h> int main () { int n, i, sum = 0 ; printf ( "Enter a positive integer: " ); scanf ( "%d" , &n); i = 1 ; while (i <= n) { sum += i; ++i; } printf ( "Sum = %d" , sum); return 0 ; } Output Enter a positive integer: 100 Sum = 5050